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Ru2$IIcontent.xmlElementary Statistical MethodsElementary Statistical MethodsTim Contreras, Odessa CollegeElementary Statistical Methods by Lumen Learning is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.ContentsSampling and DataIntroduction1.1 Definitions of Statistics and Key Terms1.2 Data: Quantitative Data & Qualitative Data1.3 Sampling1.4 Levels of Measurement1.5 Frequency & Frequency Tables1.6 Experimental Design & EthicsDescriptive StatisticsIntroduction: Descriptive Statistics2.1 Stem-and-Leaf Graphs, Line Graphs, Bar Graphs, and Pie Charts2.2 Histograms, Frequency Polygons, and Time Series Graphs2.3 Measures of the Location of the Data2.4 Box Plots2.5 Measures of the Center of the Data2.6 Skewness and the Mean, Median, and Mode2.7 Measures of the Spread of Data2.8 When to use each measure of Central TendencyProbabilityIntroduction: Probability Topics3.1 The Terminology of Probability3.2 Independent and Mutually Exclusive Events3.3 Two Basic Rules of Probability3.4 Contingency Tables3.5 Tree and Venn DiagramsDiscrete Random VariablesIntroduction: Discrete Random Variables4.1 Probability Distribution Function (PDF) for a Discrete Random Variable4.2 Mean / Expected Value and Standard Deviation of Discrete Random Variable4.3 Binomial DistributionGeometric DistributionPoisson DistributionNormal DistributionIntroduction to the Normal Distribution (Need Pic)6.1 The Standard Normal Distribution6.2 Using the Normal DistributionThe Central Limit TheoremIntroduction: The Central Limit Theorem7.1 The Central Limit Theorem for Sample Means (Averages)7.2 The Central Limit Theorem for Sums7.3 Using the Central Limit TheoremConfidence IntervalsIntroduction: Confidence Intervals8.1 A Single Population Mean using the Normal Distribution8.2 A Single Population Mean using the Student t Distribution8.3 Confidence Interval for Population ProportionHypothesis Testing With One SampleIntroduction: Hypothesis Testing with One Sample9.1 Null and Alternative Hypotheses9.2 Outcomes, Type I and Type II Errors9.3 Distribution Needed for Hypothesis Testing9.4 Rare Events, the Sample, Decision and Conclusion9.5 Additional Information and Full Hypothesis Test ExamplesLinear Regression and CorrelationIntroduction: Linear Regression and Correlation12.1 Linear Equations12.2 Scatter Plots12.3 The Regression Equation12.4 Prediction12.5 Testing the Significance of the Correlation CoefficientPART ISampling and DataChapter 1IntroductionThis photo shows a large open news room with enough space to seat about 200 employees.We encounter statistics in our daily lives more often than we probably realize and from many different sources, like the news. (credit: David Sim)Learning ObjectivesBy the end of this chapter, the student should be able to:Recognize and differentiate between key terms.Apply various types of sampling methods to data collection.Create and interpret frequency tables. You are probably asking yourself the question, “When and where will I use statistics?” If you read any newspaper, watch television, or use the Internet, you will see statistical information. There are statistics about crime, sports, education, politics, and real estate. Typically, when you read a newspaper article or watch a television news program, you are given sample information. With this information, you may make a decision about the correctness of a statement, claim, or “fact.” Statistical methods can help you make the “best educated guess.”Since you will undoubtedly be given statistical information at some point in your life, you need to know some techniques for analyzing the information thoughtfully. Think about buying a house or managing a budget. Think about your chosen profession. The fields of economics, business, psychology, education, biology, law, computer science, police science, and early childhood development require at least one course in statistics.Included in this chapter are the basic ideas and words of probability and statistics. You will soon understand that statistics and probability work together. You will also learn how data are gathered and what “good” data can be distinguished from “bad.”Chapter 21.1 Definitions of Statistics and Key TermsThe science of statistics deals with the collection, analysis, interpretation, and presentation of data. We see and use data in our everyday lives.Important Terms in StatisticsIn statistics, we generally want to study a population. You can think of a population as a collection of persons, things, or objects under study. To study the population, we select a sample. The idea of sampling is to select a portion (or subset) of the larger population and study that portion (the sample) to gain information about the population. Data are the result of sampling from a population.Example:If you wished to compute the overall grade point average at your school, it would make sense to select a sample of students who attend the school. The data collected from the sample would be the students’ grade point averages. Show the population and sample. Population: All the students at your schoolSample: a sample of 50 students In presidential elections, opinion poll samples of 1,000–2,000 people are taken. The opinion poll is supposed to represent the views of the people in the entire country. Show the population and sample. Population: All the people in the entire countrySample: Sample of 1000-2000 people City of Houston wants to know if the annual household income in the city is higher than national average. The statisticians collect data from 1500 families. Show the population and sample. Population: All the households in HoustonSample: 1500 families who are being sampled. An automobile manufacturer wanted to know if more than 50% of the US drivers own at least a domestic car. This company surveyed 10,000 drivers over US. Show the population and sample.Population: All the drivers in the US Sample: 10,000 drivers who were being surveyed From the sample data, we can calculate a statistic. A statistic is a number that represents a property of the sample. For example, if we consider one math class to be a sample of the population of all math classes, then the average number of points earned by students in that one math class at the end of the term is an example of a statistic. The statistic is an estimate of a population parameter. A parameter is a number that is a property of the population. Since we considered all math classes to be the population, then the average number of points earned per student over all the math classes is an example of a parameter.Population: all math classesSample: One of the math classesParameter: Average number of points earned per student over all math classesOne of the main concerns in the field of statistics is how accurately a statistic estimates a parameter. The accuracy really depends on how well the sample represents the population. The sample must contain the characteristics of the population in order to be a representative sample. We are interested in both the sample statistic and the population parameter in inferential statistics. In a later chapter, we will use the sample statistic to test the validity of the established population parameter.A variable, notated by capital letters such as X and Y, is a characteristic of interest for each person or thing in a population. Variables may be numerical or categorical. Numerical variables take on values with equal units such as weight in pounds and time in hours. Categorical variables place the person or thing into a category.Example:If we assume that X is equal to the number of points earned by one math student at the end of a term, then X is a numerical variable.If we let Y be a person’s party affiliation, then some examples of Y include Republican, Democrat, and Independent. Y is a categorical variable.We could do some math with values of X (calculate the average number of points earned, for example), but it makes no sense to do math with values of Y (calculating an average party affiliation makes no sense).Data are the actual values of the variable. They may be numbers or they may be words. Datum is a single value.Two words that come up often in statistics are mean and proportion.If you were to take three exams in your math classes and obtain scores of 86, 75, and 92, you would calculate your mean score by adding the three exam scores and dividing by three (your mean score would be 84.3 to one decimal place). If, in your math class, there are 40 students and 22 are men and 18 are women, then the proportion of men students is 2240 and the proportion of women students is 1840. Mean and proportion are discussed in more detail in later chapters.NOTE (We will learn the meaning of mean in next chapter! )The words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean,” and “average” is technically a center location. However, in practice among non-statisticians, “average” is commonly accepted for “arithmetic mean.”Determine what the key terms refer to in the following study.A study was conducted at a local college to analyze the average cumulative GPA’s of students who graduated last year. Fill in the letter of the phrase that best describes each of the items below.Populationall students who attended the college last yearSamplea group of students who graduated from the college last year, randomly selectedData3.65, 2.80, 1.50, 3.90Statisticsthe cumulative GPA of one student who graduated from the college last yearVariablethe average cumulative GPA of students who graduated from the college last yearParameter: the average cumulative GPA of students in the study who graduated from the college last yearTry ItDetermine what the key terms refer to in the following study. We want to know the average (mean) amount of money first year college students spend at ABC College on school supplies that do not include books. We randomly survey 100 first year students at the college. Three of those students spent $150, $200, and $225, respectively.Show Answer: Population: All the first year college students at ABC College Sample: 100 first year students who are surveyed at the college. Parameter: Average amount of money a first year college student spent on school supplies that do not include books. Statistics: Average amount of money these 100 first year college student spent on school supplies that do not include books. Variable: The amount of money a first year ABC College student spend on school supplies that do not include books. Data: The amount that we collected from these 100 students, like $150, $200, and $225. Example 2Determine what the key terms refer to in the following study. We want to know the average (mean) amount of money spent on school uniforms each year by families with children at Knoll Academy. We randomly survey 100 families with children in the school. Three of the families spent $65, $75, and $95, respectively.Solution:Population: All families with children at Knoll Academy Sample: 100 families with children who are surveyed in the school. Parameter: average (mean) amount of money spent on school uniforms by families with children at Knoll Academy. Statistics: average (mean) amount of money spent on school uniforms by families in the sample. Variable: the amount of money spent by one family Data: The amount that we collected from these 100 families, like $65, $75, and $95.Example 3 As part of a study designed to test the safety of automobiles, the National Transportation Safety Board collected and reviewed data about the effects of an automobile crash on test dummies. Here is the criterion they used:Speed at which Cars CrashedLocation of “drive” (i.e. dummies)35 miles/hourFront SeatCars with dummies in the front seats were crashed into a wall at a speed of 35 miles per hour. We want to know the proportion of dummies in the driver’s seat that would have had head injuries, if they had been actual drivers. We start with a simple random sample of 75 cars.Solution:Population: all cars containing dummies in the front seat.Sample: the 75 cars, selected by a simple random sample.Parameter: the proportion of driver dummies (if they had been real people) who would have suffered head injuries in the population.Statistics: proportion of driver dummies (if they had been real people) who would have suffered head injuries in the sample.Variable: the number of driver dummies (if they had been real people) who would have suffered head injuries.Data: yes, had head injury, or no, did not.Try ItAn insurance company would like to determine the proportion of all medical doctors who have been involved in one or more malpractice lawsuits. The company selects 500 doctors at random from a professional directory and determines the number in the sample who have been involved in a malpractice lawsuit.Population: all medical doctors listed in the professional directory.Sample: the 500 doctors selected at random from the professional directory.Parameterthe proportion of medical doctors who have been involved in one or more malpractice suits in the population.Statisticsthe proportion of medical doctors who have been involved in one or more malpractice suits in the sample.Variable: the number of medical doctors who have been involved in one or more malpractice suits.DataYes, was involved in one or more malpractice lawsuits; or no, was not. Watch the following video for a brief introduction to statistics.Thumbnail for the embedded element "Introduction to Statistics"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=19 ReferencesThe Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/CrashTestDummies.html (accessed May 1, 2013).Concept ReviewThe mathematical theory of statistics is easier to learn when you know the language. This module presents important terms that will be used throughout the text.GlossaryAveragealso called mean; a number that describes the central tendency of the dataCategorical Variablevariables that take on values that are names or labelsDataa set of observations (a set of possible outcomes); most data can be put into two groups: qualitative (an attribute whose value is indicated by a label) or quantitative (an attribute whose value is indicated by a number). Quantitative data can be separated into two subgroups: discrete and continuous. Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage)Numerical Variablevariables that take on values that are indicated by numbersParametera number that is used to represent a population characteristic and that generally cannot be determined easilyPopulationall individuals, objects, or measurements whose properties are being studiedProbabilitya number between zero and one, inclusive, that gives the likelihood that a specific event will occurProportionthe number of successes divided by the total number in the sampleRepresentative Samplea subset of the population that has the same characteristics as the populationSamplea subset of the population studiedStatistica numerical characteristic of the sample; a statistic estimates the corresponding population parameter.Variablea characteristic of interest for each person or object in a populationChapter 31.2 Data: Quantitative Data & Qualitative DataData may come from a population or from a sample. Small letters like x or y generally are used to represent data values. Most data can be put into the following categories:QualitativeQuantitativeQuantitative DataQualitative DataDefinition Quantitative data are the result of counting or measuring attributes of a population.Qualitative data are the result of categorizing or describing attributes of a population.Data that you will seeQuantitative data are always numbers.Qualitative data are generally described by words or letters.Examples Amount of money you have Height Weight Number of people living in your town Number of students who take statisticsHair color Blood type Ethnic group The car a person drives The street a person lives onResearchers often prefer to use quantitative data over qualitative data because it lends itself more easily to mathematical analysis. For example, it does not make sense to find an average hair color or blood type.Quantitative data may be either discrete or continuous.All data that are the result of counting are called quantitative discrete data. These data take on only certain numerical values. If you count the number of phone calls you receive for each day of the week, you might get values such as zero, one, two, or three.All data that are the result of measuring are quantitative continuous data assuming that we can measure accurately. Measuring angles in radians might result in such numbers as , and so on. If you and your friends carry backpacks with books in them to school, the numbers of books in the backpacks are discrete data and the weights of the backpacks are continuous data. Example of Quantitative Discrete DataThe data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data.Try ItThe data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this?Show AnswerIt is quantitative discrete data Example of Quantitative Continuous DataThe data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data because weights are measured.Try ItThe data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this?Show AnswerIt is quantitative continuous data.Examples of Qualitative DataThe data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative data.Try ItA statistics professor collects information about the classification of her students as freshmen, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart. What type of data does this graph show?Show AnswerThis pie chart shows the students in each year, which is qualitative data. Example 1Determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words “the number of.”The number of pairs of shoes you ownThe type of car you driveThe place where you go on vacationThe distance it is from your home to the nearest grocery storeThe number of classes you take per school year.The tuition for your classesThe type of calculator you useMovie ratingsPolitical party preferencesWeights of sumo wrestlersAmount of money (in dollars) won playing pokerNumber of correct answers on a quizPeoples’ attitudes toward the governmentIQ scoresShow AnswerItems a, e, f, k, and l are quantitative discrete; items d, j, and n are quantitative continuous; items b, c, g, h, i, and m are qualitative.Omitting Categories and Missing DataThe table displays Ethnicity of Students but is missing the “Other/Unknown” category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart.FrequencyPercentAsian8,79436.1%Black1,4125.8%Filipino1,2985.3%Hispanic4,18017.1%Native American1460.6%Pacific Islander2361.0%White5,97824.5%TOTAL22,044 out of 24,38290.4% out of 100%Figure 1. Ethnicity of StudentsThe following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us. This particular bar graph in Figure 2 can be difficult to understand visually.Figure 2. Bar Graph with Other/Unknown CategoryThe graph in Figure 3 is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret.Figure 3. Pareto Chart with Bars Sorted by SizeChapter 41.3 SamplingSamplingThe following video introduces the different methods that statisticians use collect samples of data.Thumbnail for the embedded element "Sampling Methods"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=21 Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling. In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons.The 5 different types of random sampling methods are the simple random sample, the stratified sample, the cluster sample, and the systematic sample.TypeRandom Sampling1. Simple Random Sample2. Stratified Sample3. Cluster Sample4. Systematic Sample5. Convenient SampleSimple Random SampleThe easiest method to describe is called a simple random sample. Any group of n individuals is equally likely to be chosen by any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus class, which has 31 members not including Lisa. To choose a simple random sample of size three from the other members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out three names. A more technological way is for Lisa to first list the last names of the members of her class together with a two-digit number, as in the following table.Class RosterIDNameIDNameIDName00Anselmo11King21Roquero01Bautista12Legeny22Roth02Bayani13Lundquist23Rowell03Cheng14Macierz24Salangsang04Cuarismo15Motogawa25Slade05Cuningham16Okimoto26Stratcher06Fontecha17Patel27Tallai07Hong18Price28Tran08Hoobler19Quizon29Wai09Jiao20Reyes30Wood10KhanLisa can use a table of random numbers (found in many statistics books and mathematical handbooks), a calculator, or a computer to generate random numbers. For this example, suppose Lisa chooses to generate random numbers from a calculator. The numbers generated are as follows:0.94360; 0.99832; 0.14669; 0.51470; 0.40581; 0.73381; 0.04399Lisa reads two-digit groups until she has chosen three class members (that is, she reads 0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could have generated more random numbers.The random numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However the third random number, 0.14669, contains 14 (the fourth random number also contains 14), the fifth random number contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist of Marcierz, Cuningham, and Cuarismo.Generating Random NumbersPress MATH.Arrow over to PRB.Press 5:randInt(. Enter 0, 30).Press ENTER for the first random number.Press ENTER two more times for the other 2 random numbers. If there is a repeat press ENTER again.Note: randInt(0, 30, 3) will generate 3 random numbers.Stratified SampleTo choose a stratified sample, divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample.Cluster Sample To choose a cluster sample, divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample.Systematic SampleTo choose a systematic sample, randomly select a starting point and take every nth piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method.Convenience SampleA type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others. Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents. True random sampling is done with replacement. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low.In a college population of 10,000 people, suppose you want to pick a sample of 1,000 randomly for a survey. For any particular sample of 1,000, if you are sampling with replacement,the chance of picking the first person is 1,000 out of 10,000 (0.1000);the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999);the chance of picking the same person again is 1 out of 10,000 (very low).If you are sampling without replacement,the chance of picking the first person for any particular sample is 1000 out of 10,000 (0.1000);the chance of picking a different second person is 999 out of 9,999 (0.0999);you do not replace the first person before picking the next person.Compare the fractions and . For accuracy, carry the decimal answers to four decimal places. To four decimal places, these numbers are equivalent (0.0999).Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. For example, if the population is 25 people, the sample is ten, and you are sampling with replacement for any particular sample, then the chance of picking the first person is ten out of 25, and the chance of picking a different second person is nine out of 25 (you replace the first person).If you sample without replacement, then the chance of picking the first person is ten out of 25, and then the chance of picking the second person (who is different) is nine out of 24 (you do not replace the first person).Compare the fractions and . To four decimal places, = 0.3600 and = 0.3750. To four decimal places, these numbers are not equivalent.When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors. A defective counting device can cause a nonsampling error.In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error.In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied.Watch the following video to learn more about sources of sampling bias.Thumbnail for the embedded element "Statistics: Sources of Bias"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=21 Chapter 51.4 Levels of MeasurementLevels of MeasurementThe way a set of data is measured is called its level of measurement. Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level):Nominal scale levelOrdinal scale levelInterval scale levelRatio scale level Nominal Scale LevelData that is measured using a nominal scale is qualitative. Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. Nominal scale data cannot be used in calculations.Example: To classify people according to their favorite food, like pizza, spaghetti, and sushi. Putting pizza first and sushi second is not meaningful.Smartphone companies are another example of nominal scale data. Some examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list and there is no agreed upon order. Some people may favor Apple but that is a matter of opinion. Ordinal Scale LevelData that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. Like the nominal scale data, ordinal scale data cannot be used in calculations.Example: A list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data.A cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Interval Scale LevelData that is measured using the interval scale is similar to ordinal level data because it has a definite ordering but there is a difference between data. The differences between interval scale data can be measured though the data does not have a starting point.Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not because, in both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0.Interval level data can be used in calculations, but comparison cannot be done. 80° C is not four times as hot as 20° C (nor is 80° F four times as hot as 20° F). There is no meaning to the ratio of 80 to 20 (or four to one).Example:Monthly income of 2000 part-time students in TexasHighest daily temperature in Odessa Ratio Scale LevelData that is measured using the ratio scale takes care of the ratio problem and gives you the most information. Ratio scale data is like interval scale data, but it has a 0 point and ratios can be calculated. You will not have a negative value in ratio scale data.For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points) (given that the exams are machine-graded.) The data can be put in order from lowest to highest: 20, 68, 80, 92. There is no negative point in the final exam scores as the lowest score is 0 point.The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The smallest score is 0. So 80 is four times 20. If one student scores 80 points and another student scores 20 points, the student who scores higher is 4 times better than the student who scores lower.Example:Weight of 200 cancer patients in the past 5 monthsHeight of 549 newborn babiesDiameter of 150 donuts Thumbnail for the embedded element "Nominal, ordinal, interval and ratio data: How to Remember the differences"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=22 References“State & County QuickFacts,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/download_data.html (accessed May 1, 2013).“State & County QuickFacts: Quick, easy access to facts about people, business, and geography,” U.S. Census Bureau. http://quickfacts.census.gov/qfd/index.html (accessed May 1, 2013).“Table 5: Direct hits by mainland United States Hurricanes (1851-2004),” National Hurricane Center, http://www.nhc.noaa.gov/gifs/table5.gif (accessed May 1, 2013).“Levels of Measurement,” http://infinity.cos.edu/faculty/woodbury/stats/tutorial/Data_Levels.htm (accessed May 1, 2013).Courtney Taylor, “Levels of Measurement,” about.com, http://statistics.about.com/od/HelpandTutorials/a/Levels-Of-Measurement.htm (accessed May 1, 2013).David Lane. “Levels of Measurement,” Connexions, http://cnx.org/content/m10809/latest/ (accessed May 1, 2013).Concept ReviewSome calculations generate numbers that are artificially precise. It is not necessary to report a value to eight decimal places when the measures that generated that value were only accurate to the nearest tenth. Round off your final answer to one more decimal place than was present in the original data. This means that if you have data measured to the nearest tenth of a unit, report the final statistic to the nearest hundredth.In addition to rounding your answers, you can measure your data using the following four levels of measurement.Nominal scale level: data that cannot be ordered nor can it be used in calculationsOrdinal scale level: data that can be ordered; the differences cannot be measuredInterval scale level: data with a definite ordering but no starting point; the differences can be measured, but there is no such thing as a ratio.Ratio scale level: data with a starting point that can be ordered; the differences have meaning and ratios can be calculated.When organizing data, it is important to know how many times a value appears. How many statistics students study five hours or more for an exam? What percent of families on our block own two pets? Frequency, relative frequency, and cumulative relative frequency are measures that answer questions like these.Chapter 61.5 Frequency & Frequency TablesTwenty students were asked how many hours they worked per day. Their responses, in hours, are as follows: 5, 6, 3, 3, 2, 4, 7, 5, 2, 3, 5, 6, 5, 4, 4, 3, 5, 2, 5, 3.The following table lists the different data values in ascending order and their frequencies.Frequency Table of Student Work HoursDATA VALUEFREQUENCY233543566271In this research, 3 students studied for 2 hours. 5 students studies for 3 hours.A frequency is the number of times a value of the data occurs. According to the table, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.Relative frequency = \frac{\text{frequency of the class}}{\text{total}}Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in the table below.Cumulative relative frequency = sum of previous relative frequencies + current class frequency Example 1Frequency Table of Student Work Hours with Relative and Cumulative Relative FrequenciesDATA VALUEFREQUENCYRELATIVE FREQUENCYCUMULATIVE RELATIVE FREQUENCY23\frac{3}{20} or 0.150.1535\frac{5}{20} or 0.250.15 + 0.25 = 0.4043\frac{3}{20} or 0.150.40 + 0.15 = 0.5556\frac{6}{20} or 0.300.55 + 0.30 = 0.8562\frac{2}{20} or 0.100.85 + 0.10 = 0.9571\frac{1}{20} or 0.050.95 + 0.05 = 1.00The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.Example 2We sample the height of 100 soccer players. The result is shown below.Height (inches)Frequency59.95 – 61.95561.95 – 63.95363.95 – 65.951565.95 – 67.954067.95 – 69.951769.95 – 71.951271.95 – 73.95773.95 – 75.951Total = 100Find:a. the relative frequency for each class.Show AnswerHeight (Inches)FrequencyRelative FrequencyCumulative Relative Frequency59.95 – 61.955\frac{5}{100} or 0.050.0561.95 – 63.953\frac{3}{100} or 0.030.05 + 0.03 = 0.0863.95 – 65.9515\frac{15}{100} or 0.150.08 + 0.15 = 0.2365.95 – 67.9540\frac{4}{100} or 0.040.23 + 0.40 = 0.6367.95 – 69.9517\frac{17}{100} or 0.170.63 + 0.17 = 0.8069.95 – 71.9512\frac{12}{100} or 0.120.80 + 0.12 = 0.9271.95 – 73.957\frac{7}{100} or 0.070.92 + 0.07 = 0.9973.95 – 75.951\frac{1}{100} or 0.010.99 + 0.01 = 1.00Total = 100Total = 1b. the percentage for height that is less than 63.95 inches.Show Answer = 0.08 = 8%c. the percentage for height that is between 69.95 inches and 73.95 inches.Show Answer + = 0.12 + 0.07 = 0.19In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints.Example 3The table shows the amount, in inches, of annual rainfall in a sample of towns.Rainfall (inches)Frequency2.95 – 4.9764.97 – 6.9976.99 – 9.01159.01 – 11.03811.03 – 13.05913.05 – 15.075Findthe relative frequency and cumulative relative frequency for each class. Show AnswerTotal = sum of all frequencies = 6 + 7 + 15 + 8 + 9 + 5 = 50Rainfall (inches)FrequencyRelative frequencyCumulative relative frequency2.95 – 4.976\frac{6}{50} = 0.120.124.97 – 6.997\frac{7}{50} = 0.140.12 + 0.14 = 0.266.99 – 9.0115\frac{15}{50} = 0.300.26 + 0.30 = 0.569.01 – 11.038\frac{8}{50} = 0.160.56 + 0.16 = 0.7211.03 – 13.059\frac{9}{50} = 0.180.72 + 0.18 = 0.9013.05 – 15.075\frac{5}{50} = 0.100.90 + 0.10 = 1.00 the percentage of rainfall that is less than 9.01 inches. Show AnswerThe percentage of rainfall that is less than 9.01 inches = 0.12 + 0.14 + 0.30 = 0.56 the percentage of heights that fall between 61.95 and 65.95 inches. Show AnswerThe percentage of heights that fall between 6.99 inches and 11.03 inches = + = 0.26 Try ItThe table contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.YearTotal Number of Deaths2000231200121,357200211,685200333,8192004228,802200588,00320066,6052007712200888,01120091,7902010320,120201121,9532012768Total823,356 What is the frequency of deaths measured from 2006 through 2009?Show Answer97,118 What percentage of deaths occurred after 2009? Show Answer41.6% What is the relative frequency of deaths that occurred in 2003 or earlier? Show Answer = 0.081 What is the percentage of deaths that occurred in 2004? Show Answer27.8% What kind of data are the numbers of deaths? Show AnswerQuantitative discrete The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers? Show AnswerQuantitative continuous Example 4The table contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.YearTotal Number of CrashesYearTotal Number of Crashes199436,254200438,444199537,241200539,252199637,494200638,648199737,324200737,435199837,107200834,172199937,140200930,862200037,526201030,296200137,862201129,757200238,491Total653,782200338,477What is the frequency of deaths measured from 2000 through 2004? Show Answer37,526 + 37,862 + 38,491 + 38,477 + 38,444 = 190,800 What percentage of deaths occurred after 2006? Show Answer or 24.9% What is the relative frequency of deaths that occurred in 2000 or before? Show Answer or 39.8% What is the percentage of deaths that occurred in 2011? Show Answer or 4.6% What is the cumulative relative frequency for 2006? Explain what this number tells you about the data. Show Answer75.1% of all fatal traffic crashes for the period from 1994 to 2011 happened from 1994 to 2006. Chapter 71.6 Experimental Design & EthicsExperimental Design Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data.The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the explanatory variable. The affected variable is called the response variable. In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments. An experimental unit is a single object or individual to be measured.The following video explains the difference between collecting data from observations and collecting data from experiments.Thumbnail for the embedded element "Observational Studies and Experiments"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=24 You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention.Additional variables that can cloud a study are called lurking variables. In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables.The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted:Results showed that believing one had taken the substance resulted in [performance] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment.1When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group. This group is given a placebo treatment–a treatment that cannot influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, he does not know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded.SamplingResearchers have a responsibility to verify that proper methods are being followed. The report describing the investigation of Stapel’s fraud states that, “statistical flaws frequently revealed a lack of familiarity with elementary statistics.”3 Many of Stapel’s co-authors should have spotted irregularities in his data. Unfortunately, they did not know very much about statistical analysis, and they simply trusted that he was collecting and reporting data properly.Many types of statistical fraud are difficult to spot. Some researchers simply stop collecting data once they have just enough to prove what they had hoped to prove. They don’t want to take the chance that a more extensive study would complicate their lives by producing data contradicting their hypothesis.Professional organizations, like the American Statistical Association, clearly define expectations for researchers. There are even laws in the federal code about the use of research data.When a statistical study uses human participants, as in medical studies, both ethics and the law dictate that researchers should be mindful of the safety of their research subjects. The U.S. Department of Health and Human Services oversees federal regulations of research studies with the aim of protecting participants. When a university or other research institution engages in research, it must ensure the safety of all human subjects. For this reason, research institutions establish oversight committees known as Institutional Review Boards (IRB). All planned studies must be approved in advance by the IRB. Key protections that are mandated by law include the following:Risks to participants must be minimized and reasonable with respect to projected benefits.Participants must give informed consent. This means that the risks of participation must be clearly explained to the subjects of the study. Subjects must consent in writing, and researchers are required to keep documentation of their consent.Data collected from individuals must be guarded carefully to protect their privacy.These ideas may seem fundamental, but they can be very difficult to verify in practice. Is removing a participant’s name from the data record sufficient to protect privacy? Perhaps the person’s identity could be discovered from the data that remains. What happens if the study does not proceed as planned and risks arise that were not anticipated? When is informed consent really necessary? Suppose your doctor wants a blood sample to check your cholesterol level. Once the sample has been tested, you expect the lab to dispose of the remaining blood. At that point the blood becomes biological waste. Does a researcher have the right to take it for use in a study?It is important that students of statistics take time to consider the ethical questions that arise in statistical studies. How prevalent is fraud in statistical studies? You might be surprised—and disappointed. There is a website (www.retractionwatch.com) dedicated to cataloging retractions of study articles that have been proven fraudulent. A quick glance will show that the misuse of statistics is a bigger problem than most people realize.Vigilance against fraud requires knowledge. Learning the basic theory of statistics will empower you to analyze statistical studies critically.References“Vitamin E and Health,” Nutrition Source, Harvard School of Public Health, http://www.hsph.harvard.edu/nutritionsource/vitamin-e/ (accessed May 1, 2013).Stan Reents. “Don’t Underestimate the Power of Suggestion,” athleteinme.com, http://www.athleteinme.com/ArticleView.aspx?id=1053 (accessed May 1, 2013).Ankita Mehta. “Daily Dose of Aspiring Helps Reduce Heart Attacks: Study,” International Business Times, July 21, 2011. Also available online at http://www.ibtimes.com/daily-dose-aspirin-helps-reduce-heart-attacks-study-300443 (accessed May 1, 2013).The Data and Story Library, http://lib.stat.cmu.edu/DASL/Stories/ScentsandLearning.html (accessed May 1, 2013).M.L. Jacskon et al., “Cognitive Components of Simulated Driving Performance: Sleep Loss effect and Predictors,” Accident Analysis and Prevention Journal, Jan no. 50 (2013), http://www.ncbi.nlm.nih.gov/pubmed/22721550 (accessed May 1, 2013).“Earthquake Information by Year,” U.S. Geological Survey. http://earthquake.usgs.gov/earthquakes/eqarchives/year/ (accessed May 1, 2013).“Fatality Analysis Report Systems (FARS) Encyclopedia,” National Highway Traffic and Safety Administration. http://www-fars.nhtsa.dot.gov/Main/index.aspx (accessed May 1, 2013).Data from www.businessweek.com (accessed May 1, 2013).Data from www.forbes.com (accessed May 1, 2013).“America’s Best Small Companies,” http://www.forbes.com/best-small-companies/list/ (accessed May 1, 2013).U.S. Department of Health and Human Services, Code of Federal Regulations Title 45 Public Welfare Department of Health and Human Services Part 46 Protection of Human Subjects revised January 15, 2009. Section 46.111:Criteria for IRB Approval of Research.“April 2013 Air Travel Consumer Report,” U.S. Department of Transportation, April 11 (2013), http://www.dot.gov/airconsumer/april-2013-air-travel-consumer-report (accessed May 1, 2013).Lori Alden, “Statistics can be Misleading,” econoclass.com, http://www.econoclass.com/misleadingstats.html (accessed May 1, 2013).Maria de los A. Medina, “Ethics in Statistics,” Based on “Building an Ethics Module for Business, Science, and Engineering Students” by Jose A. Cruz-Cruz and William Frey, Connexions, http://cnx.org/content/m15555/latest/ (accessed May 1, 2013).Concept ReviewA poorly designed study will not produce reliable data. There are certain key components that must be included in every experiment. To eliminate lurking variables, subjects must be assigned randomly to different treatment groups. One of the groups must act as a control group, demonstrating what happens when the active treatment is not applied. Participants in the control group receive a placebo treatment that looks exactly like the active treatments but cannot influence the response variable. To preserve the integrity of the placebo, both researchers and subjects may be blinded. When a study is designed properly, the only difference between treatment groups is the one imposed by the researcher. Therefore, when groups respond differently to different treatments, the difference must be due to the influence of the explanatory variable.“An ethics problem arises when you are considering an action that benefits you or some cause you support, hurts or reduces benefits to others, and violates some rule.”4 Ethical violations in statistics are not always easy to spot. Professional associations and federal agencies post guidelines for proper conduct. It is important that you learn basic statistical procedures so that you can recognize proper data analysis.PART IIDescriptive StatisticsChapter 8Introduction: Descriptive Statistics This photo shows about 26 rolls of paper piled together. The rolls are different sizes.When you have large amounts of data, you will need to organize it in a way that makes sense. These ballots from an election are rolled together with similar ballots to keep them organized. (credit: William Greeson)Try ItBy the end of this chapter, the student should be able to:Display data graphically and interpret graphs: stemplots, histograms, and box plots.Recognize, describe, and calculate the measures of location of data: quartiles and percentiles.Recognize, describe, and calculate the measures of the center of data: mean, median, and mode.Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation, and range.Once you have collected data, what will you do with it? Data can be described and presented in many different formats. For example, suppose you are interested in buying a house in a particular area. You may have no clue about the house prices, so you might ask your real estate agent to give you a sample data set of prices. Looking at all the prices in the sample often is overwhelming. A better way might be to look at the median price and the variation of prices. The median and variation are just two ways that you will learn to describe data. Your agent might also provide you with a graph of the data.In this chapter, you will study numerical and graphical ways to describe and display your data. This area of statistics is called “Descriptive Statistics.” You will learn how to calculate, and even more importantly, how to interpret these measurements and graphs.A statistical graph is a tool that helps you learn about the shape or distribution of a sample or a population. A graph can be a more effective way of presenting data than a mass of numbers because we can see where data clusters and where there are only a few data values. Newspapers and the Internet use graphs to show trends and to enable readers to compare facts and figures quickly. Statisticians often graph data first to get a picture of the data. Then, more formal tools may be applied.Some of the types of graphs that are used to summarize and organize data are the dot plot, the bar graph, the histogram, the stem-and-leaf plot, the frequency polygon (a type of broken line graph), the pie chart, and the box plot. In this chapter, we will briefly look at stem-and-leaf plots, line graphs, and bar graphs, as well as frequency polygons, and time series graphs. Our emphasis will be on histograms and box plots.NOTE:This book contains instructions for constructing a histogram and a box plot for the TI-83+ and TI-84 calculators. The Texas Instruments (TI) website provides additional instructions for using these calculators.Chapter 92.1 Stem-and-Leaf Graphs, Line Graphs, Bar Graphs, and Pie ChartsStem-and-leaf GraphsOne simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem. Example 1For Susan Dean’s spring pre-calculus class, scores for the first exam were as follows (smallest to largest): 33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100 Form a stem-and-leaf graph.Show AnswerStemLeaf3342 9 953 5 561 3 7 8 8 9 972 3 4 880 3 8 8 890 2 4 4 4 4 6100The stem-and-leaf plot shows that most scores fell in the 60s, 70s, 80s, and 90s.Eight out of the 31 scores or approximately 26% were in the 90s or 100, a fairly high number of As. Try ItFor the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61Construct a stem plot for the data.[practice-area rows=”3″][/practice-area]Show AnswerStemLeaf32 2 3 4 840 2 2 3 4 6 7 7 8 8 8 950 0 1 2 2 2 3 4 6 7 760 1The stemplot is a quick way to graph data and gives an exact picture of the data.Try ItThe following data show the distances (in miles) from the homes of off-campus statistics students to the college.0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0Create a stem plot using the data and identify any outliers:Show AnswerStemLeaf05 711 2 2 3 3 5 5 7 7 8 920 2 5 6 8 8 835 844 8 952 5 7 86780The value 8.0 may be an outlier. Values appear to concentrate at one and two miles. Watch this video to see an example of how to create a stem plot.Thumbnail for the embedded element "Stem and Leaf Plots aka Stemplots"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=27 Example 2A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. The two following tables show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.Presidential Ages at InaugurationPresidentAgePresidentAgePresidentAgeWashington57Lincoln52Hoover54J. Adams61A. Johnson56F. Roosevelt51Jefferson57Grant46Truman60Madison57Hayes54Eisenhower62Monroe58Garfield49Kennedy43J. Q. Adams57Arthur51L. Johnson55Jackson61Cleveland47Nixon56Van Buren54B. Harrison55Ford61W. H. Harrison68Cleveland55Carter52Tyler51McKinley54Reagan69Polk49T. Roosevelt42G.H.W. Bush64Taylor64Taft51Clinton47Fillmore50Wilson56G. W. Bush54Pierce48Harding55Obama47Buchanan65Coolidge51Presidential Age at DeathPresidentAgePresidentAgePresidentAgeWashington67Lincoln56Hoover90J. Adams90A. Johnson66F. Roosevelt63Jefferson83Grant63Truman88Madison85Hayes70Eisenhower78Monroe73Garfield49Kennedy46J. Q. Adams80Arthur56L. Johnson64Jackson78Cleveland71Nixon81Van Buren79B. Harrison67Ford93W. H. Harrison68Cleveland71Reagan93Tyler71McKinley58Polk53T. Roosevelt60Taylor65Taft72Fillmore74Wilson67Pierce64Harding57Buchanan77Coolidge60Show AnswerAges at InaugurationAges at InaugurationAges at Death9 9 8 7 7 7 6 3 246 98 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 1 1 1 1 1 053 6 6 7 7 89 5 4 4 2 1 1 1 060 0 3 3 4 4 5 6 7 7 7 870 0 1 1 1 4 7 8 8 980 1 3 5 890 0 3 3 Example 3The table shows the number of wins and losses the Atlanta Hawks have had in 42 seasons. Create a side-by-side stem-and-leaf plot of these wins and losses.LossesWinsYearLossesWinsYear34481968–196941411989–199034481969–197039431990–199146361970–197144381991–199246361971–197239431992–199336461972–197325571993–199447351973–197440421994–199551311974–197536461995–199653291975–197626561996–199751311976–197732501997–199841411977–197819311998–199936461978–197954281999–200032501979–198057252000–200151311980–198149332001–200240421981–198247352002–200339431982–198354282003–200442401983–198469132004–200548341984–198556262005–200632501985–198652302006–200725571986–198745372007–200832501987–198835472008–200930521988–198929532009–2010Show AnswerNumber of lossesAtalanta Hawks Wins and Leaves Number of Wins Number of Loses3199 8 8 6 525 5 98 7 6 6 5 5 4 3 1 1 1 1 030 2 2 2 2 4 4 5 6 6 6 9 9 98 8 7 6 6 6 3 3 3 2 2 1 1 040 0 1 1 2 4 5 6 6 7 7 8 97 7 6 3 2 0 0 0 051 1 1 2 3 4 4 6 769 Line GraphAnother type of graph that is useful for specific data values is a line graph.Example 1In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores.Number of times teenager is remindedFrequency0215283144754Show AnswerTry ItIn a survey, 40 people were asked how many times per year they had their car in the shop for repairs.Number of times in shopFrequency0711021439Construct a line graph.Show Answer Bar Graphs & Pie ChartsBar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal.Pie Charts is another method to “visualize” the data. Each proportion on a pie chart represents the proportion of the corresponding class in the data set.Example 1By the end of 2011, Facebook had over 146 million users in the United States. The table shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph and a pie chart using this data.Age groupsNumber of Facebook usersProportion (%) of Facebook users13–2565,082,28045%26–4453,300,20036%45–6427,885,10019%Bar GraphThe bar graph has age groups represented on the x-axis and proportions on the y-axis.Pie ChartTry ItThe population in Park City is made up of children, working-age adults, and retirees. The table shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group.Age groupsNumber of peopleProportion of populationChildren67,05919%Working-age adults152,19843%Retirees131,66238%Construct a bar graph and a pie chart showing the proportions.Bar GraphPie ChartExample 2The columns in the table contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population.Race/EthnicityAP Examinee PopulationOverall Student Population1 = Asian, Asian American or Pacific Islander10.3%5.7%2 = Black or African American9.0%14.7%3 = Hispanic or Latino17.0%17.6%4 = American Indian or Alaska Native0.6%1.1%5 = White57.1%59.2%6 = Not reported/other6.0%1.7%Create a bar graph and a pie chart with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis.Show AnswerPie Chart Try ItPark city is broken down into six voting districts. The table shows the percentage of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district.DistrictRegistered voter populationOverall city population115.5%19.4%212.2%15.6%39.8%9.0%417.4%18.5%522.8%20.7%622.3%16.8%Construct a bar graph and a pie chart that shows the registered voter population by district.Bar GraphPie Chart Chapter 102.2 Histograms, Frequency Polygons, and Time Series GraphsFor most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.HistogramTo construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places.For example:If the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision.If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495).If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995).If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5).Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.Watch the following video for an example of how to draw a histogram.Thumbnail for the embedded element "Histograms | Applying mathematical reasoning | Pre-Algebra | Khan Academy"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=28 Example 1The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured.60; 60.5; 61; 61; 61.563.5; 63.5; 63.564; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.566; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.568; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.570; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 7172; 72; 72; 72.5; 72.5; 73; 73.5; 74Construct a relative frequency table and histogram.Show AnswerThe smallest data value is 60.Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places.Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place.The starting point is, then, 59.95.The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.Next, calculate the width of each bar or class interval.To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire).Suppose you choose eight bars.The boundaries are:59.9559.95 + 2 = 61.9561.95 + 2 = 63.9563.95 + 2 = 65.9565.95 + 2 = 67.9567.95 + 2 = 69.9569.95 + 2 = 71.9571.95 + 2 = 73.9573.95 + 2 = 75.95The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.Class IntervalFrequencyRelative Frequency59.95 – 61.9550.0561.95 – 63.9530.0363.95 – 65.95150.1565.95 – 67.9540.0467.95 – 69.95170.1769.95 – 71.95120.1271.95 – 73.9570.0773.95 – 75.9510.01The following histogram displays the heights on the x-axis and relative frequency on the y-axis.Histogram consists of 8 bars with the y-axis in increments of 0.05 from 0-0.4 and the x-axis in intervals of 2 from 59.95-75.95. Note:We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.Example 2The following data are the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted.1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 12; 2; 2; 2; 2; 2; 2; 2; 2; 23; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 34; 4; 4; 4; 4; 45; 5; 5; 5; 56; 6Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .Solution:3.5 to 4.54.5 to 5.565.5 to 6.5Calculate the number of bars as follows:where 1 is the width of a bar. Therefore, bars = 6.The following histogram displays the number of books on the x-axis and the frequency on the y-axis.Histogram consists of 6 bars with the y-axis in increments of 2 from 0-16 and the x-axis in intervals of 1 from 0.5-6.5. Create the histogram for Example 2 by using TI-Calculator:Press Y=. Press CLEAR to delete any equations.Press STAT 1:EDIT. If L1 has data in it, arrow up into the name L1, press CLEAR and then arrow down. If necessary, do the same for L2.Into L1, enter 1, 2, 3, 4, 5, 6.Into L2, enter 11, 10, 16, 6, 5, 2.Press WINDOW. Set Xmin = .5, Xscl = (6.5 – .5)/6, Ymin = –1, Ymax = 20, Yscl = 1, Xres = 1.Press 2nd Y=. Start by pressing 4:Plotsoff ENTER.Press 2nd Y=. Press 1:Plot1. Press ENTER. Arrow down to TYPE. Arrow to the 3rd picture (histogram). Press ENTER.Arrow down to Xlist: Enter L1 (2nd 1). Arrow down to Freq. Enter L2 (2nd 2).Press GRAPH.Use the TRACE key and the arrow keys to examine the histogram. Example 3Using this data set, construct a histogram.Number of Hours My Classmates Spent Playing Video Games on Weekends9.95102.2516.75019.522.57.51512.755.5111020.7517.52321.92423.7518201522.918.820.5Show Answer This is a histogram that matches the supplied data. The x-axis consists of 5 bars in intervals of 5 from 0 to 25. The y-axis is marked in increments of 1 from 0 to 10. The x-axis shows the number of hours spent playing video games on the weekends, and the y-axis shows the number of students.Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram. Practice Problems 1:The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.511; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.512; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14Option 1: Smallest value: 9Largest value: 14Convenient starting value: 9 – 0.05 = 8.95Convenient ending value: 14 0.05 = 14.05The calculations suggests using 0.85 as the width of each bar or class interval.You can also use an interval with a width equal to one.Class IntervalFrequency8.95 – 9.8049.80 – 10.651410.65 – 11.501311.50 – 12.351412.35 – 13.20413.20 – 14.051 Option 2: Smallest value: 9Largest value: 14Convenient starting value: 9 – 0.05 = 8.95Convenient ending value: 14 0.05 = 14.05You can also use an interval with a width equal to one.Class IntervalFrequency8.95 – 9.9549.95 – 10.951410.95 – 11.952011.95 – 12.951112.95 – 13.95013.95 – 14.951Practice Problem 2:The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted.1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 12; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 23; 3; 3; 3; 3; 3; 3; 3Solution20 student athletes play one sport. 22 student athletes play two sports. Eight student athletes play three sports.Fill in the blanks for the following sentence. Since the data consist of the numbers 1, 2, 3, and the starting point is 0.5, a width of one places the 1 in the middle of the interval 0.5 to _____, the 2 in the middle of the interval from _____ to _____, and the 3 in the middle of the interval from _____ to _____.Show Answer1.51.5 to 2.52.5 to 3.5 Frequency PolygonsFrequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons.To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x-axis and y-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them.Example 4A frequency polygon was constructed from the frequency table below.Frequency Distribution for Calculus Final Test ScoresLower BoundUpper BoundFrequencyCumulative Frequency49.559.55559.569.5101569.579.5304579.589.5408589.599.515100A frequency polygon was constructed from the frequency table below.The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.Example 5We will construct an overlay frequency polygon comparing the scores with the students’ final numeric grade.Frequency Distribution for Calculus Final Test ScoresLower BoundUpper BoundFrequencyCumulative Frequency49.559.55559.569.5101569.579.5304579.589.5408589.599.515100Frequency Distribution for Calculus Final GradesLower BoundUpper BoundFrequencyCumulative Frequency49.559.5101059.569.5102069.579.5305079.589.5459589.599.55100This is an overlay frequency polygon that matches the supplied data. The x-axis shows the grades, and the y-axis shows the frequency.Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with this data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.Practice Problem 3:Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in the table.Age at InaugurationFrequency41.5–46.5446.5–51.51151.5–56.51456.5–61.5961.5–66.5466.5–71.52Show AnswerFrequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets.Constructing a Time Series GraphOne feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur. Example 6 The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only.YearJanFebMarAprMayJunJul2003181.7183.1184.2183.8183.5183.7183.92004185.2186.2187.4188.0189.1189.7189.42005190.7191.8193.3194.6194.4194.5195.42006198.3198.7199.8201.5202.5202.9203.52007202.416203.499205.352206.686207.949208.352208.2992008211.080211.693213.528214.823216.632218.815219.9642009211.143212.193212.709213.240213.856215.693215.3512010216.687216.741217.631218.009218.178217.965218.0112011220.223221.309223.467224.906225.964225.722225.9222012226.665227.663229.392230.085229.815229.478229.104YearAugSepOctNovDecAnnual2003184.6185.2185.0184.5184.3184.02004189.5189.9190.9191.0190.3188.92005196.4198.8199.2197.6196.8195.32006203.9202.9201.8201.5201.8201.62007207.917208.490208.936210.177210.036207.3422008219.086218.783216.573212.425210.228215.3032009215.834215.969216.177216.330215.949214.5372010218.312218.439218.711218.803219.179218.0562011226.545226.889226.421226.230225.672224.9392012230.379231.407231.317230.221229.601229.594Show Answer Try ItThe following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO2 emissions for the United States.CO2 EmissionsUkraineUnited KingdomUnited States2003352,259540,6405,681,6642004343,121540,4095,790,7612005339,029541,9905,826,3942006327,797542,0455,737,6152007328,357528,6315,828,6972008323,657522,2475,656,8392009272,176474,5795,299,563Show Answer Uses of a Time Series GraphTime series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot.ReferencesData on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker“Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013).“Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013).“Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013).“Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013).“CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013).“Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013).“Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013).Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach. CRC Press: 1980.“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013).Concept ReviewA histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time.Chapter 112.3 Measures of the Location of the DataThe common measures of location are quartiles and percentiles. Quartiles are special percentiles.The first quartile, Q1, is the same as the 25th percentile. 25% of data will be less than 25th percentile; 75% of data will be more than 25th percentile.The second quartile, Q2, is the same as the 50th percentile / median. 50% of data will be less than 50th percentile; 50% of data will be more than 50th percentile.The third quartile, Q3, is the same as the 75th percentile. 75% of data will be less than 75th percentile; 25% of data will be more than 75th percentile.The general form is :n % of data will be less than nth percentile and (100% – n%) of data will be more than nth percentile.The following video gives an introduction to Median, Quartiles and Interquartile Range, the topic you will learn in this section.Thumbnail for the embedded element "Median, Quartiles and Interquartile Range : ExamSolutions"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=29 To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75th percentile. That translates into a score of at least 1220. To be admitted as Duke student, your SAT score has to be at least 1220.Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.The median is a number that measures the “center” of the data. You can think of the median as the “middle value,” but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger.For example, consider the following data.1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1Ordered from smallest to largest:1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.The median is seven. 50% of the values are smaller than 7 and 50% of the values are larger than 7.Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data.To find the quartiles, Find the median or second quartile (Q2) first.Find the first quartile (Q1), the median of the lower half of the data.Find the third quartile (Q3), the median, of the upper half of the data. To get the idea, consider the same data set:1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5The median or second quartile fall between the 7th and 8th data. The median is the mean of 6.8 and 7.2.Hence, the median = = 7The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is 2. 1; 1; 2; 2; 4; 6; 6.8 The number 2, which is part of the data, is the first quartile. 25% of the entire sets of values are the same as or less than 2. 75% of the values are more than 2.The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is 9. The third quartile, Q3, is 9. 75% of the ordered data set are less than 9. 25% of the ordered data set are greater than 9.The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile (Q3) and the first quartile (Q1).IQR = Q3 – Q1The IQR can help determine outliers.A data is a potential outlier if and only if the data is \begin{cases}\text{smaller than Q1 - 1.5 * IQR}\\\text{or}\\\text{larger than Q3 + 1.5 * IQR}\end{cases}Example 1For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars.$33,000 $64,500 $28,000 $54,000 $72,000 $68,500 $69,000 $42,000 $54,000 $120,000 $40,500Find the median, Q1, and Q3. Show AnswerOrder the data from smallest to largest.$28,000 $33,000 $40,500 $42,000 $54,000 $54,000 $64,500 $68,500 $69,000 $72,000 $120,000Median = $54,000Q1 = $40,500Q3 = $69,000 Find the interquartile range, IQR. Show AnswerIQR = $69,000 – $40,500 = $28,500 Find the potential outlier. Show Answer(1.5)( IQR) = (1.5)($28,500) = $42,750Q1 – (1.5)(IQR) = $40,500 – $42,750 = –$2,250Q3 + (1.5)(IQR) = $69,000 + $42,750 = $111,750No salary is less than $2,250. However, $120,000 is more than $11,750, so $120,000 is a potential outlier. Try ItTest Scores for Class A69; 96; 81; 79; 34; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94Test Scores for Class B90; 72; 80; 92; 90; 97; 92; 75; 79; 39; 70; 80; 129; 95; 78; 73; 71; 68; 95; 1341. Find the interquartile range (IQR) for the following two data sets and compare them.Show AnswerClass AOrder the data from smallest to largest.34; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99IQR = 90.5 – 72.5 = 18Class BOrder the data from smallest to largest.39; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 129; 134IQR = 93.5 – 72.5 = 21The data for Class B has a larger IQR, so the scores between Q3 and Q1 (middle 50%) for the data for Class B are more spread out and not clustered about the median. 2. Is there any outlier in class A?Show AnswerQ1 – (1.5)(IQR) = 72.5 – (1.5)(18) = 45.5 Q3 + (1.5)(IQR) = 90.5 + (1.5)(18) = 117.5In class A, we have these data:34; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99.34 is less than Q1 – (1.5)(IQR), 45.5. No data is greater than Q3 + (1.5)(IQR) = 117.5. So, 34 is the only potential outlier in class A. 3. Is there any outlier in class B?Show AnswerQ1 – (1.5)(IQR) = 72.5 – (1.5)(21) = 41 Q3 + (1.5)(IQR) = 93.5 + (1.5)(21) = 125In class B, we have these data:39; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 129; 134.39 is less than Q1 – (1.5)(IQR), 41. 129 and 134 are greater than Q3 + (1.5)(IQR), 125. So 39, 129 and 134 are the potential outliers in class B. Example 2The following 13 real estate prices. (Prices are in dollars.)389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000Calculate the IQR and determine if any prices are potential outliers.Show AnswerOrder the data from smallest to largest.114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000Median = 488,800Q1 = 230,500+387,000}{2}" title="\frac{230,500+387,000}{2}" class="latex mathjax"> = 308,750Q3 =}{2}" title="\frac{639,000+659,000}{2}" class="latex mathjax"> = 649,000IQR = 649,000 – 308,750 = 340,250(1.5)(IQR) = (1.5)(340,250) = 510,375Q1 – (1.5)(IQR) = 308,750 – 510,375 = –201,625Q3 + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier. A Formula for Finding the kthPercentileIf you were to do a little research, you would find several formulas for calculating the kth percentile. Here is one of them.k = the kth percentile. It may or may not be part of the data.i = the index (ranking or position of a data value)n = the total number of dataOrder the data from smallest to largest.Calculate \displaystyle{i}=\frac{{k}}{{100}}{({n}+{1})}If i is an integer, then the kth percentile is the data value in the ith position in the ordered set of data.If i is not an integer, then round i up and round i down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example.Example 3Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were:Amount of Sleep per School Night (Hours)FrequencyRelative FrequencyCumulative Relative Frequency420.040.04550.100.14670.140.287120.240.528140.280.80970.140.941030.061.00(a) Find the 28th percentile. Show AnswerNotice the 0.28 in the “cumulative relative frequency” column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28th percentile is between the last six and the first seven. The 28th percentile is 6.5.(b) Find median. Show AnswerLook again at the “cumulative relative frequency” column and find 0.52. The median is the 50th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50th percentile is between the 25th, or seven, and 26th, or seven, values. The median is seven.(c) Find the third quartile. Show AnswerThe third quartile is the same as the 75th percentile. You can “eyeball” this answer. If you look at the “cumulative relative frequency” column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75th percentile, then, must be an eight. Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q3, is the 38th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above. Try ItForty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour).Amount of time spent on route (hours)FrequencyRelative FrequencyCumulative Relative Frequency2120.300.303140.350.654100.250.90540.101.00Find the 65th percentile.[practice-area rows=”2″][/practice-area]Show AnswerThe 65th percentile is 3.5.The 65th percentile is between the last three and the first four.Example 4Amount of Sleep per School Night (Hours)FrequencyRelative FrequencyCumulative Relative Frequency420.040.04550.100.14670.140.287120.240.528140.280.80970.140.941030.061.00Find the 80th percentile. Show AnswerThe 80th percentile is between the last eight and the first nine in the table (between the 40th and 41st values). Therefore, we need to take the mean of the 40th an 41st values. The 80th percentile = Find the 90th percentile. Show AnswerThe 90th percentile will be the 45th data value (location is 0.90(50) = 45) and the 45th data value is nine. Find the first quartile. What is another name for the first quartile? Show AnswerQ1 is also the 25th percentile. The 25th percentile location calculation: 25th percentile = 0.25(50) = 12.5 ≈ 13 the 13th data value. Thus, the 25th percentile is six. Try ItAmount of time spent on route (hours)FrequencyRelative FrequencyCumulative Relative Frequency2120.300.303140.350.654100.250.90540.101.00Find the third quartile. What is another name for the third quartile?Show Answer[practice-area rows=”2″][/practice-area] Show AnswerThe third quartile is the 75th percentile, which is four. The 65th percentile is between three and four, and the 90th percentile is between four and 5.75. The third quartile is between 65 and 90, so it must be four. Example 5Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77Find the 70th percentile.Show Answerk = 70th percentile, i = the index, n = 29 . Twenty-one is an integer, and the data value in the 21st position in the ordered data set is 64. The 70th percentile is 64 years.Find the 83rd percentile.Show Answerk = 83rd percentile, i = the index, n = 29 , which is NOT an integer. Round it down to 24 and up to 25. The age in the 24th position is 71 and the age in the 25th position is 72. We will find the average of 71 and 72. The 83rd percentile is 71.5 years. Try ItListed are 29 ages for Academy Award winning best actors in order from smallest to largest.18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77Calculatethe 20th percentile Show Answerk = 20. Index = The age in the sixth position is 27. The 20th percentile is 27 years. the 55th percentile.Show Answerk = 55. Index, i = . Round down to 16 and up to 17. The age in the 16th position is 52 and the age in the 17th position is 55. The average of 52 and 55 is 53.5. The 55th percentile is 53.5 years. A Formula for Finding the Percentile of a Value in a Data SetOrder the data from smallest to largest.x = the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile.y = the number of data values equal to the data value for which you want to find the percentile.n = the total number of data.Calculate \displaystyle\frac{{{x}+{0.5}{y}}}{{n}}{({100})}. Then round to the nearest integer.Example 6Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77Find the percentile for 58.Find the percentile for 25.Solution:Counting from the bottom of the list, there are 18 data values less than 58. There is one value of 58. Number of data values counting from the bottom of the data list up to but not including the data value 58, x = 18 Number of data values equal to the data value 58, y = 1 \frac{{{x}+{0.5}{y}}}{{n}}{({100})}=\frac{{{18}+{0.5}{({1})}}}{{29}}{({100})}={63.80}}.}58 is the 64th percentile.Counting from the bottom of the list, there are three data values less than 25. There is one value of 25. Number of data values counting from the bottom of the data list up to but not including the data value 25, x = 18 Number of data values equal to the data value 25, y = 1 \frac{{{x}+{0.5}{y}}}{{n}}{({100})}=\frac{{{3}+{0.5}{({1})}}}{{29}}{({100})}={12.07}}.}25 is the 12th percentile.Try ItListed are 30 ages for Academy Award winning best actors in order from smallest to largest.18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77Find the percentiles for 47.Show AnswerPercentile for 47: Counting from the bottom of the list, there are 15 data values less than 47. There is one value of 47.47 is the 53rd percentile.Find the percentiles for 31. Show AnswerPercentile for 31: Counting from the bottom of the list, there are eight data values less than 31. There are two values of 31.31 is the 31st percentile. Interpreting Percentiles, Quartiles, and MedianA percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the pth percentile. For example, 15% of data values are less than or equal to the 15th percentile.Low percentiles always correspond to lower data values.High percentiles always correspond to higher data values.A percentile may or may not correspond to a value judgment about whether it is “good” or “bad.” The interpretation of whether a certain percentile is “good” or “bad” depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered “good;” in other contexts a high percentile might be considered “good”. In many situations, there is no value judgment that applies.Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.GuidelineWhen writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.information about the context of the situation being consideredthe data value (value of the variable) that represents the percentilethe percent of individuals or items with data values below the percentilethe percent of individuals or items with data values above the percentile.Example 7On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.Solution:25% of students finished the exam in 35 minutes or less.75% of students finished the exam in 35 minutes or more.A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)Try ItFor the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.Show Answer25% of runners finished the race in 11.5 seconds or more. 75% of runners finished the race in 11.5 seconds or less. A lower percentile is good because finishing a race more quickly is desirable.Example 8On a 20 question math test, the 70th percentile for number of correct answers was 16. Interpret the 70th percentile in the context of this situation.Solution:70% of students answered 16 or fewer questions correctly.30% of students answered 16 or more questions correctly.A higher percentile could be considered good, as answering more questions correctly is desirable.Try ItOn a 60 point written assignment, the 80th percentile for the number of points earned was 49. Interpret the 80th percentile in the context of this situation.Show Answer80% of students earned 49 points or fewer. 20% of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable. Example 9At a community college, it was found that the 30th percentile of credit units that students are enrolled for is 7 units. Interpret the 30th percentile in the context of this situation.Solution:30% of students are enrolled in 7 or fewer credit units.70% of students are enrolled in 7 or more credit units.In this example, there is no “good” or “bad” value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs.Try ItDuring a season, the 40th percentile for points scored per player in a game is eight. Interpret the 40th percentile in the context of this situation.Show Answer40% of players scored eight points or fewer. 60% of players scored eight points or more. A higher percentile is good because getting more points in a basketball game is desirable. Example 10Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutesDetermine the following five values.Min = 0Q1 = 20Med = 40Q3 = 60Max = 300Solution:If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.However, the principal needs to be careful. The value 300 appears to be a potential outlier.Q3 + 1.5(IQR) = 60 + (1.5)(40) = 120.The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:Min = 0Q1 = 20Q3 = 60Max = 120We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results. Concept ReviewThe values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50th percentile would be greater than 50 percent of the other obeservations in the set. Quartiles divide data into quarters. The first quartile (Q1) is the 25th percentile,the second quartile (Q2 or median) is 50th percentile, and the third quartile (Q3) is the the 75th percentile. The interquartile range, or IQR, is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q1 from Q3, and can help determine outliers by using the following two expressions.Q3 + IQR(1.5)Q1 – IQR(1.5)Formula Reviewwherei = the ranking or position of a data value,k = the kth percentile,n = total number of data.Expression for finding the percentile of a data value:wherex = the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,y = the number of data values equal to the data value for which you want to find the percentile,n = total number of dataReferencesCauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at http://usatoday30.usatoday.com/news/nation/story/2012-05-17/minority-birthscensus/55029100/1 (accessed April 3, 2013).Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013).“1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013).Data from San Jose Mercury News.Data from Time Magazine; survey by Yankelovich Partners, Inc.Chapter 122.4 Box PlotsBox plots (also called box-and-whisker plots or box-whisker plots) give a good graphical image of the concentration of the data. They also show how far the extreme values are from most of the data. A box plot is constructed from five values: the minimum value, the first quartile (Q1), the median (Q2), the third quartile (Q3), and the maximum value. We use these values to compare how close other data values are to them.To construct a box plot, use a horizontal or vertical number line and a rectangular box. The smallest and largest data values label the endpoints of the axis. The first quartile marks one end of the box and the third quartile marks the other end of the box. Approximately the middle 50 percent of the data fall inside the box. The “whiskers” extend from the ends of the box to the smallest and largest data values. The median or second quartile can be between the first and third quartiles, or it can be one, or the other, or both. The box plot gives a good, quick picture of the data.Note:You may encounter box-and-whisker plots that have dots marking outlier values. In those cases, the whiskers are not extending to the minimum and maximum values.Consider, again, this dataset.1 1 2 2 4 6 6.8 7.2 8 8.3 9 10 10 11.5The first quartile, Q1 is 2. The median, Q2 is 7. The third quartile, Q3 is 9. The smallest value is 1. The largest value is 11.5.The following image shows the constructed box plot.The two whiskers extend from the first quartile to the smallest value and from the third quartile to the largest value. The median is shown with a dashed line.Note:It is important to start a box plot with a scaled number line. Otherwise the box plot may not be useful.Example 1The following data are the heights of 40 students in a statistics class.59 60 61 62 62 63 63 64 64 64 65 65 65 65 65 65 65 65 65 66 66 67 67 68 68 69 70 70 70 70 70 71 71 72 72 73 74 74 75 77Construct a box plot with the following properties; the calculator instructions for the minimum and maximum values as well as the quartiles follow the example.Minimum value = 59Maximum value = 77Q1, First quartile = 64.5Q2, Second quartile or median= 66Q3, Third quartile = 70Each quarter has approximately 25% of the data.The spreads of the four quarters are 64.5 – 59 = 5.5 (first quarter), 66 – 64.5 = 1.5 (second quarter), 70 – 66 = 4 (third quarter), and 77 – 70 = 7 (fourth quarter). So, the second quarter has the smallest spread and the fourth quarter has the largest spread.Range = maximum value – the minimum value = 77 – 59 = 18Interquartile Range: IQR = Q3 – Q1 = 70 – 64.5 = 5.5.The interval 59–65 has more than 25% of the data so it has more data in it than the interval 66 through 70 which has 25% of the data.The middle 50% (middle half) of the data has a range of 5.5 inches.TI-Calculator: To find the minimum, maximum, and quartiles:Enter data into the list editor (Pres STAT 1:EDIT). If you need to clear the list, arrow up to the name L1, press CLEAR, and then arrow down.Put the data values into the list L1.Press STAT and arrow to CALC. Press 1:1-VarStats. Enter L1.Press ENTER.Use the down and up arrow keys to scroll. Smallest value = 59. Largest value = 77. Q1: First quartile = 64.5. Q2: Second quartile or median = 66. Q3: Third quartile = 70.To construct the box plot:Press 4:Plotsoff. Press ENTER.Arrow down and then use the right arrow key to go to the fifth picture, which is the box plot. Press ENTER.Arrow down to Xlist: Press 2nd 1 for L1Arrow down to Freq: Press ALPHA. Press 1.Press Zoom. Press 9: ZoomStat.Press TRACE, and use the arrow keys to examine the box plot. Try ItThe following data are the number of pages in 40 books on a shelf.136 140 178 190 205 215 217 218 232 234 240 255 270 275 290 301 303 315 317 318 326 333 343 349 360 369 377 388 391 392 398 400 402 405 408 422 429 450 475 512Construct a box plot using a graphing calculator, and state the interquartile range.Show AnswerIQR = 158 This video explains what descriptive statistics are needed to create a box and whisker plot.Thumbnail for the embedded element "Box and Whisker Plot"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=30 For some sets of data, some of the largest value, smallest value, first quartile, median, and third quartile may be the same. For instance, you might have a data set in which the median and the third quartile are the same. In this case, the diagram would not have a dotted line inside the box displaying the median. The right side of the box would display both the third quartile and the median. For example, if the smallest value and the first quartile were both one, the median and the third quartile were both five, and the largest value was seven, the box plot would look like:In this case, at least 25% of the values are equal to one. 25% of the values are between one and five, inclusive. At least 25% of the values are equal to five. The top 25% of the values fall between five and seven, inclusive.Example 2Test scores for a college statistics class held during the day are:99 56 78 55.5 32 90 80 81 56 59 45 77 84.5 84 70 72 68 32 79 90Test scores for a college statistics class held during the evening are:98 78 68 83 81 89 88 76 65 45 98 90 80 84.5 85 79 78 98 90 79 81 25.5Find the smallest and largest values, the median, and the first and third quartile for the day class. Show AnswerMin = 32 Q1 = 56 Q2 = 74.5 Q3 = 82.5 Max = 99 Find the smallest and largest values, the median, and the first and third quartile for the night class. Show AnswerMin = 25.5 Q1 = 78 Q2 = 81 Q3 = 89 Max = 98 Create a box plot for each set of data. Use one number line for both box plots. Show Answer Which box plot has the widest spread for the middle 50% of the data (the data between the first and third quartiles)? What does this mean for that set of data in comparison to the other set of data? Show AnswerThe first data set has the wider spread for the middle 50% of the data. The IQR for the first data set is greater than the IQR for the second set. This means that there is more variability in the middle 50% of the first data set. Try ItTry ItThe following data set shows the heights in inches for the boys in a class of 40 students.66; 66; 67; 67; 68; 68; 68; 68; 68; 69; 69; 69; 70; 71; 72; 72; 72; 73; 73; 74The following data set shows the heights in inches for the girls in a class of 40 students.61; 61; 62; 62; 63; 63; 63; 65; 65; 65; 66; 66; 66; 67; 68; 68; 68; 69; 69; 69Construct a box plot using a graphing calculator for each data set, and state which box plot has the wider spread for the middle 50% of the data.Show AnswerIQR for the boys = 4IQR for the girls = 5The box plot for the heights of the girls has the wider spread for the middle 50% of the data. Example 3Graph a box-and-whisker plot for the data values shown.10 10 10 15 35 75 90 95 100 175 420 490 515 515 790Show AnswerThe five numbers used to create a box-and-whisker plot are:Min: 10Q1: 15Med: 95Q3: 490Max: 790The following graph shows the box-and-whisker plot. Try ItTry ItGraph a box-and-whisker plot for the data values shown.0 5 5 15 30 30 45 50 50 60 75 110 140 240 330Show AnswerThe data are in order from least to greatest. There are 15 values, so the eighth number in order is the median: 50. There are seven data values written to the left of the median and 7 values to the right. The five values that are used to create the boxplot are:Min: 0Q1: 15Med: 50Q3: 110Max: 330 ReferencesData from West Magazine.Concept ReviewBox plots are a type of graph that can help visually organize data. To graph a box plot the following data points must be calculated: the minimum value, the first quartile, the median, the third quartile, and the maximum value. Once the box plot is graphed, you can display and compare distributions of data.Additional ResourcesUse the online imathAS box plot tool to create box and whisker plots.Chapter 132.5 Measures of the Center of the DataMean & Median The “center” of a data set is also a way of describing location. The two most widely used measures of the “center” of the data are the mean (average) and the median. To calculate the mean weight of 50 people, add the 50 weights together and divide by 50. To find the median weight of the 50 people, order the data and find the number that splits the data into two equal parts. The median is generally a better measure of the center when there are extreme values or outliers because it is not affected by the precise numerical values of the outliers. The mean is the most common measure of the center.NoteThe words “mean” and “average” are often used interchangeably. The substitution of one word for the other is common practice. The technical term is “arithmetic mean” and “average” is technically a center location. However, in practice among non-statisticians, “average” is commonly accepted for “arithmetic mean.”When each value in the data set is not unique, the mean can be calculated by multiplying each distinct value by its frequency and then dividing the sum by the total number of data values. The letter used to represent the sample mean is an x with a bar over it (read “x bar”): .The Greek letter μ (pronounced “mew”) represents the population mean. One of the requirements for the sample mean to be a good estimate of the population mean is for the sample taken to be truly random.To see that both ways of calculating the mean are the same, consider the sample:1; 1; 1; 2; 2; 3; 4; 4; 4; 4; 4In the second example, the frequencies are 3(1) + 2(2) + 1(3) + 5(4).You can quickly find the location of the median by using the expression .The letter n is the total number of data values in the sample. If n is an odd number, the median is the middle value of the ordered data (ordered smallest to largest). If n is an even number, the median is equal to the two middle values added together and divided by two after the data has been ordered.For example, if the total number of data values is 97, then . The median occurs midway between the 50th and 51st values. The location of the median and the value of the median are not the same. The upper case letter M is often used to represent the median. The next example illustrates the location of the median and the value of the median.Example 1AIDS data indicating the number of months a patient with AIDS lives after taking a new antibody drug are as follows (smallest to largest):3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47Calculate the mean and the median.MeanThe calculation for the mean is MedianTo find the median, M, first use the formula for the location. The location is: Starting at the smallest value, the median is located between the 20th and 21stvvalues (the two 24s):3; 4; 8; 8; 10; 11; 12; 13; 14; 15; 15; 16; 16; 17; 17; 18; 21; 22; 22; 24; 24; 25; 26; 26; 27; 27; 29; 29; 31; 32; 33; 33; 34; 34; 35; 37; 40; 44; 44; 47;Finding the Mean and the MedianUsing the TI-83, 83+, 84, 84+ CalculatorClear list L1. Pres STAT 4:ClrList. Enter 2nd 1 for list L1. Press ENTER.Enter data into the list editor. Press STAT 1:EDIT.Put the data values into list L1.Press STAT and arrow to CALC. Press 1:1-VarStats. Press 2nd 1 for L1 and then ENTER.Press the down and up arrow keys to scroll.\displaystyle\overline{{x}}= 23.6, M = 24Try ItThe following data show the number of months patients typically wait on a transplant list before getting surgery. The data are ordered from smallest to largest.3 4 5 7 7 7 7 8 8 9 9 10 1010 10 10 11 12 12 13 14 1415 15 17 17 18 19 19 19 2121 22 22 23 24 24 24 24Calculate the mean and median.Show AnswerMean: 3 + 4 + 5 + 7 + 7 + 7 + 7 + 8 + 8 + 9 + 9 + 10 + 10 + 10 + 10 + 10 + 11 + 12 + 12 + 13 + 14 + 14 + 15 + 15 + 17 + 17 + 18 + 19 + 19 + 19 + 21 + 21 + 22 + 22 + 23 + 24 + 24 + 24 = 544Median: Starting at the smallest value, the median is the 20th term, which is 13. Example 2Suppose that in a small town of 50 people, one person earns $5,000,000 per year and the other 49 each earn $30,000. Which is the better measure of the “center”: the mean or the median?Show Answer, (There are 49 people who earn $30,000 and one person who earns $5,000,000.)The median is a better measure of the “center” than the mean because 49 of the values are 30,000 and one is 5,000,000. The 5,000,000 is an outlier. The 30,000 gives us a better sense of the middle of the data. Try ItTry ItIn a sample of 60 households, one house is worth $2,500,000. Half of the rest are worth $280,000, and all the others are worth $315,000. Which is the better measure of the “center”: the mean or the median?Show AnswerThe median is the better measure of the “center” than the mean because 59 of the values are $280,000 and one is $2,500,000. The $2,500,000 is an outlier. Either $280,000 or $315,000 gives us a better sense of the middle of the data. ModeAnother measure of the center is the mode. The mode is the most frequent value. There can be more than one mode in a data set as long as those values have the same frequency and that frequency is the highest. A data set with two modes is called bimodal.Example 3Statistics exam scores for 20 students are as follows:50, 53, 59, 59, 63, 63, 72, 72, 72, 72, 72, 76, 78, 81, 83, 84, 84, 84, 90, 93Find the mode.Show AnswerThe most frequent score is 72, which occurs five times. Mode = 72.Try ItThe number of books checked out from the library from 25 students are as follows:0, 0, 0, 1, 2, 3, 3, 4, 4, 5, 5, 7, 7, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 12, 12Find the mode.Show AnswerThe most frequent number of books is 7, which occurs four times. Mode = 7. Five real estate exam scores are 430, 430, 480, 480, 495. The data set is bimodal because the scores 430 and 480 each occur twice.6 credit scores are 590, 680, 680, 700, 720, 720. The data set is bimodal because the scores 680 and 720 each occur twice.When is the mode the best measure of the “center”? Consider a weight loss program that advertises a mean weight loss of six pounds the first week of the program. The mode might indicate that most people lose two pounds the first week, making the program less appealing.Note: The mode can be calculated for qualitative data as well as for quantitative data. For example, if the data set is: red, red, red, green, green, yellow, purple, black, blue, the mode is red.Statistical software will easily calculate the mean, the median, and the mode. Some graphing calculators can also make these calculations. In the real world, people make these calculations using software.Example 4Consider the annual earnings of workers at a factory. The mode is $25,000 and occurs 150 times out of 301. The median is $50,000 and the mean is $47,500. What would be the best measure of the “center”?Show AnswerBecause $25,000 occurs nearly half the time, the mode would be the best measure of the center because the median and mean don’t represent what most people make at the factory.Watch the following video from Kahn Academy on finding the mean, median and mode of a set of data.Thumbnail for the embedded element "Finding mean, median, and mode | Descriptive statistics | Probability and Statistics | Khan Academy"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=31 The Law of Large Numbers and the MeanThe Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean of the sample is very likely to get closer and closer to µ. This is discussed in more detail later in the text.Sampling Distributions and Statistic of a Sampling DistributionYou can think of a sampling distribution as a relative frequency distribution with a great many samples. Suppose thirty randomly selected students were asked the number of movies they watched the previous week. The results are in the relative frequency table shown below.# of moviesRelative Frequency0\displaystyle\frac{{5}}{{30}}\\1\displaystyle\frac{{15}}{{30}}\\2\displaystyle\frac{{6}}{{30}}\\3\displaystyle\frac{{3}}{{30}}\\4\displaystyle\frac{{1}}{{30}}\\If you let the number of samples get very large (say, 300 million or more), the relative frequency table becomes a relative frequency distribution.A statistic is a number calculated from a sample. Statistic examples include the mean, the median and the mode as well as others. The sample mean is an example of a statistic which estimates the population mean μ.Calculating the Mean of Grouped Frequency TablesWhen only grouped data is available, you do not know the individual data values (we only know intervals and interval frequencies); therefore, you cannot compute an exact mean for the data set. What we must do is estimate the actual mean by calculating the mean of a frequency table. A frequency table is a data representation in which grouped data is displayed along with the corresponding frequencies. To calculate the mean from a grouped frequency table we can apply the basic definition of mean:.We simply need to modify the definition to fit within the restrictions of a frequency table.Since we do not know the individual data values we can instead find the midpoint of each interval. The midpoint is where f = the frequency of the interval and m = the midpoint of the interval.Example 5A frequency table displaying professor Blount’s last statistic test is shown.Grade IntervalNumber of Students50–56.5156.5–62.5062.5–68.5468.5–74.5474.5–80.5280.5–86.5386.5–92.5492.5–98.51Find the best estimate of the class mean.Show AnswerFind the midpoints for all intervals.Grade Interval Midpoint50.0–56.553.2556.5–62.559.562.5–68.565.568.5–74.571.574.5–80.577.580.5–86.583.586.5–92.589.592.5–98.595.5Calculate the sum of the product of each interval frequency and midpoint.53.25(1) + 59.5(0) + 65.5(4) + 71.5(4) + 77.5(2) + 83.5(3) + 89.5(4) + 95.5(1) = 1460.2 Try ItMaris conducted a study on the effect that playing video games has on memory recall. As part of her study, she compiled the following data:Hours Teenagers Spend on Video GamesNumber of Teenagers0–3.533.5–7.577.5–11.5 1211.5–15.5715.5–19.59What is the best estimate for the mean number of hours spent playing video games?Show AnswerFind the midpoint of each interval, multiply by the corresponding number of teenagers, add the results and then divide by the total number of teenagersThe midpoints are 1.75, 5.5, 9.5, 13.5,17.5.Mean = (1.75)(3) + (5.5)(7) + (9.5)(12) + (13.5)(7) + (17.5)(9) = 409.75 ReviewThe mean and the median can be calculated to help you find the “center” of a data set. The mean is the best estimate for the actual data set, but the median is the best measurement when a data set contains several outliers or extreme values. The mode will tell you the most frequently occurring datum (or data) in your data set. The mean, median, and mode are extremely helpful when you need to analyze your data, but if your data set consists of ranges which lack specific values, the mean may seem impossible to calculate. However, the mean can be approximated if you add the lower boundary with the upper boundary and divide by two to find the midpoint of each interval. Multiply each midpoint by the number of values found in the corresponding range. Divide the sum of these values by the total number of data values in the set.Formula ReviewWhere f = interval frequencies and m = interval midpoints.ReferencesData from The World Bank, available online at http://www.worldbank.org (accessed April 3, 2013).“Demographics: Obesity – adult prevalence rate.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2228&l=en (accessed April 3, 2013).Chapter 142.6 Skewness and the Mean, Median, and ModeConsider the following data set.4; 5; 6; 6; 6; 7; 7; 7; 7; 7; 7; 8; 8; 8; 9; 10This data set can be represented by following histogram. Each interval has width one, and each value is located in the middle of an interval.Figure 1.This histogram matches the supplied data. It consists of 7 adjacent bars with the x-axis split into intervals of 1 from 4 to 10. The heighs of the bars peak in the middle and taper symmetrically to the right and left.The histogram displays a symmetrical distribution of data. A distribution is symmetrical if a vertical line can be drawn at some point in the histogram such that the shape to the left and the right of the vertical line are mirror images of each other. The mean, the median, and the mode are each seven for these data. In a perfectly symmetrical distribution, the mean and the median are the same. This example has one mode (unimodal), and the mode is the same as the mean and median. In a symmetrical distribution that has two modes (bimodal), the two modes would be different from the mean and median. Consider the following data set: 4 5 6 6 6 7 7 7 7 8. The histogram is not symmetrical.The right-hand side seems “chopped off” compared to the left side. A distribution of this type is called skewed to the left because it is pulled out to the left.Figure 2.This histogram matches the supplied data. It consists of 5 adjacent bars with the x-axis split into intervals of 1 from 4 to 8. The peak is to the right, and the heights of the bars taper down to the left.The mean is 6.3, the median is 6.5, and the mode is seven. Notice that the mean is less than the median, and they are both less than the mode. The mean and the median both reflect the skewing, but the mean reflects it more so. Consider the following data set: 6 7 7 7 7 8 8 8 9 10.The histogram is also not symmetrical. It is skewed to the right.Figure 3.This histogram matches the supplied data. It consists of 5 adjacent bars with the x-axis split into intervals of 1 from 6 to 10. The peak is to the left, and the heights of the bars taper down to the right.The mean is 7.7, the median is 7.5, and the mode is 7.Of the three statistics, the mean is the largest, while the mode is the smallest. Again, the mean reflects the skewing the most.To summarize,if the distribution of data is skewed to the left, the mean is less than the median, which is often less than the mode. (median < median < mode)If the distribution of data is skewed to the right, the mode is often less than the median, which is less than the mean. (mean > median > mode)If the distribution of data is symmetric, the mode = the median = the mean.Skewness and symmetry become important when we discuss probability distributions in later chapters.Here is a video that summarizes how the mean, median and mode can help us describe the skewness of a dataset. Don’t worry about the terms leptokurtic and platykurtic for this course.Thumbnail for the embedded element "Elementary Business Statistics | Skewness and the Mean, Median, and Mode"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=32 Statistics are used to compare and sometimes identify authors. The following lists shows a simple random sample that compares the letter counts for three authors.Terry: 7; 9; 3; 3; 3; 4; 1; 3; 2; 2Davi: 3; 3; 3; 4; 1; 4; 3; 2; 3; 1Mari: 2; 3; 4; 4; 4; 6; 6; 6; 8; 3Make a dot plot for the three authors and compare the shapes. Show AnswerTerry’s distribution is right-skewed. Davi’s distribution is slightly left-skewed.Mari’s distribution is symmetrical. Calculate the mean for each. Show AnswerTerry’s mean is 3.7, Davi’s mean is 2.7, Mari’s mean is 4.6. Calculate the median for each. Show AnswerTerry’s median is three, Davi’s median is three. Mari’s median is four. Describe any pattern you notice between the shape and the measures of center. Show AnswerIt appears that the median is always closest to the high point (the mode), while the mean tends to be farther out on the tail. In a symmetrical distribution, the mean and the median are both centrally located close to the high point of the distribution. Try ItDiscuss the mean, median, and mode for each of the following problems. Is there a pattern between the shape and measure of the center?a.This dot plot matches the supplied data. The plot uses a number line from 0 to 14. It shows two x's over 0, four x's over 1, three x's over 2, one x over 3, two x's over the number 4, 5, 6, and 9, and 1 x each over 10 and 14. There are no x's over the numbers 7, 8, 11, 12, and 13.b.The Ages Former U.S Presidents Died46 953 6 7 7 7 860 0 3 3 4 4 5 6 7 7 7 870 1 1 2 3 4 7 8 8 980 1 3 5 890 0 3 3Key: 8|0 means 80.c.This is a histogram titled Hours Spent Playing Video Games on Weekends. The x-axis shows the number of hours spent playing video games with bars showing values at intervals of 5. The y-axis shows the number of students. The first bar for 0 - 4.99 hours has a height of 2. The second bar from 5 - 9.99 has a height of 3. The third bar from 10 - 14.99 has a height of 4. The fourth bar from 15 - 19.99 has a height of 7. The fifth bar from 20 - 24.99 has a height of 9.Show Answermean = 4.25, median = 3.5, mode = 1; The mean > median > mode which indicates skewness to the right. (data are 0, 1, 2, 3, 4, 5, 6, 9, 10, 14 and respective frequencies are 2, 4, 3, 1, 2, 2, 2, 2, 1, 1)mean = 70.1 , median = 68, mode = 57, 67 bimodal; the mean and median are close but there is a little skewness to the right which is influenced by the data being bimodal. (data are 46, 49, 53, 56, 57, 57, 57, 58, 60, 60, 63, 63, 64, 64, 65, 66, 67, 67, 67, 68, 70, 71, 71, 72, 73, 74, 77, 78, 78, 79, 80, 81, 83, 85, 88, 90, 90 93, 93).These are estimates: mean =16.095, median = 17.495, mode = 22.495 (there may be no mode); The mean < median < mode which indicates skewness to the left. (data are the midponts of the intervals: 2.495, 7.495, 12.495, 17.495, 22.495 and respective frequencies are 2, 3, 4, 7, 9).Concept ReviewLooking at the distribution of data can reveal a lot about the relationship between the mean, the median, and the mode. There are three types of distributions. A right (or positive) skewed distribution has a shape like Figure 2. A left (or negative) skewed distribution has a shape like Figure 3 . A symmetrical distribution looks like Figure 1.Chapter 152.7 Measures of the Spread of DataAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.The standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean.The standard deviation provides a measure of the overall variation in a data set.The standard deviation is always positive or zero.The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread.The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is 5 minutes. At supermarket A, the standard deviation for the wait time is 2 minutes. At supermarket B, the standard deviation for the wait time is 4 minutes.Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.The standard deviation can be used to determine whether a data value is close to or far from the mean.Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for 7 minutes and Binh waits for 1 minute. At supermarket A, the mean waiting time is 5 minutes and the standard deviation is 2 minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.Rosa waits for 7 minutes:7 minutes is 2 minutes longer than the average of 5 minutes; 2 minutes is equal to 1 standard deviation.Rosa’s wait time of 7 minutes is two minutes longer than the average of 5 minutes.Rosa’s wait time of 7 minutes is one standard deviation above the average of 5 minutes.Binh waits for 1 minute.1 minute is 4 minutes less than the average of 5 minutes; 4 minutes is equal to 2 standard deviations.Binh’s wait time of one minute is four minutes less than the average of five minutes.Binh’s wait time of one minute is two standard deviations below the average of five minutes.A data value that is 2 standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than 2 standard deviations away is more of an approximate “rule of thumb” than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than 2 standard deviations. (You will learn more about this in later chapters.)The number line may help you understand standard deviation. If we were to put 5 and 7 on a number line, 7 is to the right of 5. We say, then, that7 is one standard deviation to the right of five because 5 + (1)(2) = 7.If one were also part of the data set, then one is two standard deviations to the left of 5 because 5 + (–2)(2) = 1.In general, a value = mean + (# ofSTDEV)(standard deviation)where # of STDEV = the number of standard deviations# of STDEV does not need to be an integerOne is two standard deviations less than the mean of five because: 1 = 5 + (–2)(2).The equation a value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population.Sample: \displaystyle{x}=\overline{{x}}+(# of STDEV){({s})}Population: \displaystyle{x}=\mu+(# of STDEV){(\sigma)}The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation.The symbol is the sample mean and the Greek symbol μ is the population mean.Calculating the Standard DeviationIf x is a number, then the difference “x – mean” is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is x – μ. For sample data, in symbols a deviation is .The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter σ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ.To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – values for a sample, or the x – μ values for a population).The symbol σ2 represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s2 represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by N, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample. Formulas for the Sample Standard DeviationFor the sample standard deviation, the denominator is n – 1, that is the sample size – 1.Formulas for the Population Standard DeviationFor the population standard deviation, the denominator is N, the number of items in the population.In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three.In the following video an example of calculating the variance and standard deviation of a set of data is presented.Thumbnail for the embedded element "How to calculate Standard Deviation and Variance"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=33 Sampling Variability of a StatisticHow much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean when you learn about The Central Limit Theorem (not now). The notation for the standard error of the mean is where σ is the standard deviation of the population and n is the size of the sample.Note:In practice, use a calculator or computer software to calculate the standard deviation. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation σx or sx from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)Example 1The following data are the ages for a sample of n = 20 fifth grade students. The ages are rounded to the nearest half year:9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.51. The teacher was interested in the average age and the sample standard deviation of the ages of her students.Show AnswerThe average age is 10.525 years, rounded to two places.The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s. DataFreq.DeviationsDeviations2(Freq.)( Deviations2)xf( x – \overline{x})( x –\overline{x})2(f)(x –\overline{x})2919 – 10.525 = –1.525(–1.525) 2 = 2.3256251 × 2.325625 = 2.3256259.529.5 – 10.525 = –1.025(–1.025) 2 = 1.0506252 × 1.050625 = 2.10125010410 – 10.525 = –0.525(–0.525) 2 = 0.2756254 × 0.275625 = 1.102510.5410.5 – 10.525 = –0.025(–0.025) 2 = 0.0006254 × 0.000625 = 0.002511611 – 10.525 = 0.475(0.475) 2 = 0.2256256 × 0.225625 = 1.3537511.5311.5 – 10.525 = 0.975(0.975) 2 = 0.9506253 × 0.950625 = 2.851875The total is 9.7375The sample variance, , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s2 = \frac{9.7375}{20-1} = 0.5125The sample standard deviation s is equal to the square root of the sample variance: which is rounded to two decimal places, s = = 0.72.2. Find the value that is one standard deviation above the mean. (Find + 1s. )Show Answer+ 1s = 10.53 + (1)(0.72) = 11.253. Find the value that is two standard deviations below the mean. (Find – 2s. )Show Answer – 2s = 10.53 – (2)(0.72) = 9.094. Find the values that are 1.5 standard deviations from(below and above) the mean.Show Answer( – 1.5s) = 10.53 – (1.5)(0.72) = 9.45( + 1.5s) = 10.53 + (1.5)(0.72) = 11.61Typically, you do the calculation for the standard deviation on your calculator or computer. The intermediate results are not rounded. This is done for accuracy.For the following problems, recall that value = mean + (# of STDEVs)(standard deviation). Verify the mean and standard deviation or a calculator or computer.For a sample: x = + (# of STDEV)(s) For a population: x = μ + (# of STDEV)(σ)For this example, use x = + (# of STDEV)(s) because the data is from a sample.Verify the mean and standard deviation on your calculator or computer:Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.\displaystyle\overline{x} = 10.525Use Sx because this is sample data (not a population): Sx=0.715891. Try ItOn a baseball team, the ages of each of the players are as follows:Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.Show Answerμ = 30.68 s = 6.09 + 2s = 30.68 + (2)(6.09) = 42.86. Explanation of the standard deviation calculation shown in the tableThe deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For Example 1, there are n = 20 deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.Notice that instead of dividing by n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n – 1). Why not divide by n? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by (n – 1) gives a better estimate of the population variance.Note:Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.The standard deviation, s or σ, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make s or σ very large.The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better “feel” for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot.Example 2Use the following data (first exam scores) from Susan Dean’s spring pre-calculus class:33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places. Show AnswerDataFrequencyRelative FrequencyCumulative Relative Frequency3310.0320.0324210.0320.0644920.0650.1295310.0320.1615520.0650.2266110.0320.2586310.0320.296710.0320.3226820.0650.3876920.0650.4527210.0320.4847310.0320.5167410.0320.5487810.0320.5808010.0320.6128310.0320.6448830.0970.7419010.0320.7739210.0320.8059440.1290.9349610.0320.96610010.0320.998 (Why isn’t this value 1?) Calculate the following to one decimal place using a TI-83+ or TI-84 calculator: The sample mean Show AnswerThe sample mean = 73.5 The sample standard deviation Show AnswerThe sample standard deviation = 17.9 The median Show AnswerThe median = 73 The first quartile Show AnswerThe first quartile = 61 The third quartile Show AnswerThe third quartile = 90 IQR Show Answer IQR = 90 – 61 = 29 Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart. Show AnswerThe x-axis goes from 32.5 to 100.5; y-axis goes from –2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5 – 32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5 + 13.6 = 46.1, 46.1 + 13.6 = 59.7, 59.7 + 13.6 = 73.3, 73.3 + 13.6 = 86.9, 86.9 + 13.6 = 100.5 = the ending value; No data values fall on an interval boundary. The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater (73 – 33 = 40) than the spread in the upper 50% (100 – 73 = 27). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs. Try ItThe following data show the different types of pet food stores in the area carry.Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.Show Answerμ = 9.3s = 2.2 Standard Deviation of Grouped Frequency TablesRecall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:Mean of Frequency Table, = Standard Deviation of Frequency Table, where f = interval frequencies and m = interval midpoints.Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.Example 3ClassFrequency0 – 213 – 566 – 8109 – 11712 – 14015 – 172Find the standard deviation for the data in the table.Show AnswerSteps 1: Find the mean.ClassFrequency, fMidpoint, m0–2113–5646–81079–1171012–1401315–17216For this data set, we have the mean, = = = 7.58.Step 2: Use the mean to find the standard deviation.ClassFrequency, fMidpoint, mm-\overline{x}(m-\overline{x})^2(f)(m-\overline{x})^20–2111 – 7.58 = -6.58(-6.58)^2 = 43.2964(1)(43.296) = 43.29643–5644 – 7.58 = -3.58(-3.58)^2 = 12.8164(6)(12.816) = 76.89846–81077 – 7.58 = -0.58(-0.58)^2 = 0.3364(10)(0.3364) = 3.3649–1171010 – 7.58 = 2.42(2.42)^2 = 5.8564(7)(5.8564) = 40.994812–1401313 – 7.58 = 5.42(5.42)^2 = 29.3764(0)(29.376) = 015–1721616 – 7.58 = 8.42(8.42)^2 = 70.8964(2)(70.896) = 141.7928The standard deviation, . Mean of Frequency Table, = Standard Deviation of Frequency Table, where f = interval frequencies and m = interval midpoints.The calculations are tedious. It is usually best to use technology when performing the calculations.Try ItFind the standard deviation for the data from the previous exampleClassFrequency, f0–213–566–8109–11712–14015–172Use TI-Calculators to find the standard deviation: Screen Shot 2015-06-07 at 6.08.28 PMInput the midpoint values into L1 and the frequencies into L2Screen Shot 2015-06-07 at 6.09.59 PMSelect STAT, CALC, and 1: 1-Var StatsScreen Shot 2015-06-07 at 6.10.46 PMSelect 2nd then 1 then , 2nd then 2 EnterScreen Shot 2015-06-07 at 6.11.19 PMYou will see displayed both a population standard deviation, σx, and the sample standard deviation, sx. Comparing Values from Different Data SetsThe standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.For each data value, calculate how many standard deviations away from its mean the value is.Use the formula: value = mean + (#ofSTDEVs)(standard deviation); solve for #ofSTDEVs.#ofSTDEVs=value−meanstandard deviationCompare the results of this calculation.#ofSTDEVs is often called a ” z-score”; we can use the symbol z. In symbols, the formulas become:Data value, xZ-Score of data valueSamplex = \overline{x}+(z)(s)z = \frac{x-\overline{x}}{s}Populationx = \mu+(z)(\sigma)z = \frac{x-\mu}{\sigma}Example 4Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school.StudentGPASchool Mean GPASchool Standard DeviationJohn2.853.00.7Ali778010Which student had the highest GPA when compared to his school?Show AnswerFor each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.z=# of STDEVs = JohnAliz =\frac{{2.85 - 3.00}}{{0.7}} = −0.21z = \frac{{77 - 80}}{{10}}= −0.3 John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school’s mean while Ali’s GPA is 0.3 standard deviations below his school’s mean.John’s z-score of –0.21 is higher than Ali’s z-score of –0.3.For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. Try ItTwo swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?SwimmerTime (seconds)Team Mean TimeTeam Standard DeviationAngie26.227.20.8Beth27.330.11.4Show AnswerFor Angie: z = = = –1.25For Beth: z = = = –2 The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.For ANY data set, no matter what the distribution of the data is:At least 75% of the data is within two standard deviations of the mean.At least 89% of the data is within three standard deviations of the mean.At least 95% of the data is within 4.5 standard deviations of the mean.This is known as Chebyshev’s Rule.For data having a distribution that is BELL-SHAPED and SYMMETRIC:Approximately 68% of the data is within one standard deviation of the mean.Approximately 95% of the data is within two standard deviations of the mean.More than 99% of the data is within three standard deviations of the mean.This is known as the Empirical Rule.It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the “Normal” or “Gaussian” probability distribution in later chapters.ReferencesData from Microsoft Bookshelf.King, Bill.”Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013).Concept ReviewThe standard deviation can help you calculate the spread of data. There are different equations to use if are calculating the standard deviation of a sample or of a population.The Standard Deviation allows us to compare individual data or classes to the data set mean numerically.\displaystyle{s}_{x}=\sqrt{{\frac{{f{(m-\overline{x})}^{2}}}{{n-1}}}} is the formula for calculating the standard deviation of a sample.To calculate the standard deviation of a population, we would use the population mean, μ, and the formula \displaystyle{\sigma}=\sqrt{{\frac{{f{(x-\mu)}^{2}}}{{N}}}}Formula Reviewwhere =sample standard deviation, = sample meanChapter 162.8 When to use each measure of Central TendencyBy now, everyone should know how to calculate mean, median and mode. They each give us a measure of Central Tendency (i.e. where the center of our data falls), but often give different answers. So how do we know when to use each? Here are some general rules: Mean is the most frequently used measure of central tendency and generally considered the best measure of it. However, there are some situations where either median or mode are preferred.Median is the preferred measure of central tendency when: There are a few extreme scores in the distribution of the data. (NOTE: Remember that a single outlier can have a great effect on the mean). b.There are some missing or undetermined values in your data. c.There is an open ended distribution (For example, if you have a data field which measures number of children and your options are 0, 1, 2, 3, 4, 5 or “6 or more,” then the “6 or more field” is open ended and makes calculating the mean impossible, since we do not know exact values for this field) d.You have data measured on an ordinal scale. Mode is the preferred measure when data are measured in a nominal ( and even sometimes ordinal) scale.PART IIIProbabilityChapter 17Introduction: Probability Topics This is a photo taken of the night sky. A meteor and its tail are shown entering the earth's atmosphere.Meteor showers are rare, but the probability of them occurring can be calculated. (credit: Navicore/flickr)Learning ObjectivesBy the end of this chapter, the student should be able to:Understand and use the terminology of probability.Determine whether two events are mutually exclusive and whether two events are independent.Calculate probabilities using the Addition Rules and Multiplication Rules.Construct and interpret Contingency Tables.Construct and interpret Venn Diagrams.Construct and interpret Tree Diagrams. It is often necessary to “guess” about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.COLLABORATIVE EXERCISEYour instructor will survey your class. Count the number of students in the class today.Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.Raise your hand if you rode a bus within the past month. Record the number of raised hands.Raise your hand if you answered “yes” to BOTH of the first two questions. Record the number of raised hands.Use the class data as estimates of the following probabilities. P(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. P(bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.Find P(change).Find P(bus).Find P(change AND bus). Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month.Find P(change|bus). Find the probability that a randomly chosen student has change given that he or she rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus.Chapter 183.1 The Terminology of ProbabilityProbability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Example of an experiment: Flipping one fair coin twice.A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter S is used to denote the sample space. Example: if you flip one fair coin, S = {H, T} where H = heads and T = tails are the outcomes. An event is any combination of outcomes. Upper case letters like A and B represent events. For example, if the experiment is to flip one fair coin, event A might be getting at most one head. Example: The probability of an event A is probability of getting at most one head. It is also written as P(A). The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, 0 probability of an event 1).P(A) = 0 means the event A can never happen.P(A) = 1 means the event A always happens.P(A) = 0.5 means the event A is equally likely to occur or not to occur.Example: If you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads). Equally likely means that each outcome of an experiment occurs with equal probability. Example:If you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face.If you toss a fair coin, a Head (H) and a Tail (T) are equally likely to occur.If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer. To calculate the probability of an event A when all outcomes in the sample space are equally likely, count the number of outcomes for event A and divide by the total number of outcomes in the sample space.For examples:If you toss a fair dime and a fair nickel, you will see four possible outcomes. These 4 outcomes will form a sample space. Therefore, the sample space is {HH, TH, HT, TT} where T = tails and H = heads. If event A = getting one head, then there are two outcomes that meet this condition {HT, TH}.The probability of event A, P(A) = = = 0.5.Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event E = rolling a number that is at least five.There are two outcomes {5, 6}.P(E) = = as the number of repetitions grows larger and larger. This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)This video gives more examples of basic probabilities.Thumbnail for the embedded element "Introduction to Probability"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=37 It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased.Examples:Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias.Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely. “OR” EventAn outcome is in the event A OR B if the outcome is in A or is in B or is in both A and B. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}.A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. (Notice that 4 and 5 are NOT listed twice.)“AND” EventAn outcome is in the event A AND B if the outcome is in both A and B at the same time. For example, let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, respectively. ThenA AND B = {4, 5}.Complimentary EventThe complement of event A is denoted A′ (read “A prime”). A′ consists of all outcomes that are NOT in A.P(A) + P(A′) = 1.For example:We have sample space S = {1, 2, 3, 4, 5, 6}. If event A = {1, 2, 3, 4}. Then, event A’={5, 6}. P(A) = and P(A’) = P(A) + P(A’) = Conditional Probability of an Event The conditional probability of A given B is written P(A|B). P(A|B) is the probability that event A will occur given that the event B has already occurred. A conditional reduces the sample space. We calculate the probability of A from the reduced sample space B. The formula to calculate P(A|B) is where P(B) is greater than zero.For example:Suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}. Let event A = face is 2 or 3 and B = event that face is even. Event A = {2, 3}, Event B ={2, 4, 6}. To calculate P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}. Then we divide that by the number of outcomes B (rather than S).We get the same result by using the formula. Remember that S has six outcomes.A and B = {2} (as 2 appears in both event A and event B.) Understanding Terminology and SymbolsIt is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.Example 1The sample space S is the whole numbers starting at one and less than 20.S = _____________________________Let event A = the even numbers and event B = numbers greater than 13. Show AnswerS = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19} A = _____________________, B = _____________________ Show AnswerA = {2, 4, 6, 8, 10, 12, 14, 16, 18}, B = {14, 15, 16, 17, 18, 19} P(A) = _____________, P(B) = ________________Show AnswerA AND B = ____________________, A OR B = ________________ Show AnswerA AND B = {14,16,18}, A OR B = 2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19} P(A AND B) = _________, P(A OR B) = _____________ Show Answer A′ = _____________, P(A′) = _____________ Show Answer P(A) + P(A′) = ____________Show AnswerP(A)+P(A’)=P(A|B) = ___________, P(B|A) = _____________; Are the probabilities equal?Show AnswerP(A|B) = ,P(B|A)=, No Try ItThe sample space S is the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).S = _____________________________Let event A = the sum is even and event B = the first number is prime.A = _____________________, B = _____________________P(A) = _____________, P(B) = ________________A AND B = ____________________, A OR B = ________________P(A AND B) = _________, P(A OR B) = _____________B′ = _____________, P(B′) = _____________P(A) + P(A′) = ____________P(A|B) = ___________, P(B|A) = _____________; are the probabilities equal?Show AnswerS = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}A = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3)} B = {(2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}\displaystyle{P}{({A})}=\frac{{1}}{{2}},{P}{({B})}=\frac{{2}}{{3}}A AND B = {(2,2), (2,4), (3,1), (3,3)} A OR B = {(1,1), (1,3), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)}P(A and B)=\displaystyle\frac{{1}}{{3}}, {P(A or B)=\frac{{5}}{{6}}B’={(1,1),(1,2),(1,3),(1,4)}, P(B’)=\frac{{1}}{{3}}P(B) + P(B′) = 1P(A|B)=\displaystyle=\frac{{2}}{{3}}, No. Example 2A fair, six-sided die is rolled. Describe the sample space S, identify each of the following events with a subset of S and compute its probability (an outcome is the number of dots that show up).Event T = the outcome is two. Show AnswerT = {2}, P(T) = Event A = the outcome is an even number. Show AnswerA = {2, 4, 6} , P(A)= Event B = the outcome is less than four. Show AnswerB = {1, 2, 3},P(B)= The complement of A. Show AnswerA’ = {1, 3, 5}, P(A’)= A GIVEN B Show Answer(A | B)={2},P(A | B) = B GIVEN A Show Answer(B | A) ={2},P(B|A)= A AND B Show Answer (A and B)={2}, P(A and B)=\frac{{1}}{{6}} A OR B Show Answer (A orB)={1, 2, 3, 4, 6},P(A or B)=\frac{{5}}{{6}} A OR B′ Show Answer (A or B’)={2, 4, 5, 6},P(A or B’)=\frac{{2}}{{3}} Event N = the outcome is a prime number. Show AnswerN = {2, 3, 5}, P(N)= Event I = the outcome is seven. Show AnswerA six-sided die does not have seven dots. P(7) = 0. Example 3Table describes the distribution of a random sample S of 100 individuals, organized by gender and whether they are right- or left-handed.Right-handedLeft-handedTotalMales43943 + 9 = 52Females44444 + 4 = 48Total43 + 44 = 879 + 4 = 13100Let’s denote the events M = the subject is male, F = the subject is female, R = the subject is right-handed, L = the subject is left-handed. Compute the following probabilities:P(M)Show AnswerP(M) = = = 0.52P(F)Show AnswerP(F) = = = 0.48P(R)Show AnswerP(R) = = = 0.87P(L)Show AnswerP(L) = = = 0.13P(M AND R) Show AnswerP(M AND R) = = = 0.43 P(F AND L) Show AnswerP(F AND L) = = =0.04 P(M OR F) Show AnswerP(M OR F) = = = 1 P(M OR R) Show AnswerP(M OR R) = = = 0.96 P(F OR L) Show AnswerP(F OR L) = = = 0.57 P(M’)Show AnswerP(M’) = 1 – P(M) = 1 – 0.52 = 0.48P(R|M)Show AnswerP(R|M) = = = 0.8269 (rounded to four decimal places)P(F|L)Show AnswerP(F|L) = = 0.3077 (rounded to four decimal places)P(L|F)Show AnswerP(L|F) = = = 0.0833References“Countries List by Continent.” Worldatlas, 2013. Available online at http://www.worldatlas.com/cntycont.htm (accessed May 2, 2013).Concept ReviewIn this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.Formula ReviewA and B are eventsP(S) = 1 where S is the sample space0 ≤ P(A) ≤ 1P(A|B)=Chapter 193.2 Independent and Mutually Exclusive EventsIndependent and mutually exclusive do not mean the same thing.Independent EventsTwo events are independent if the following are true:P(A|B) = P(A)P(B|A) = P(B)P(A AND B) = P(A)P(B)Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions.If two events are NOT independent, then we say that they are dependent.Sampling may be done with replacement or without replacement.With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise.You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit.Sampling with replacement:Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is theQ of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the Q of spades again. Your picks are {Q of spades, ten of clubs, Q of spades}. You have picked the Q of spades twice. You pick each card from the 52-card deck.Samplingwithout replacement:Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is theK of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the J of spades. Your picks are {K of hearts, three of diamonds, J of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice. Example 1You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king) of that suit. Three cards are picked at random.Suppose you know that the picked cards are Q of spades, K of hearts and Q of spades. Can you decide if the sampling was with or without replacement? Show AnswerSampling with replacement Suppose you know that the picked cards are Q of spades, K of hearts, and J of spades. Can you decide if the sampling was with or without replacement? Show AnswerNo, we cannot tell if the sampling was with or without replacement. Example 2You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs.Suppose you pick four cards, but do not put any cards back into the deck. Your cards are QS, 1D, 1C, QD.Suppose you pick four cards and put each card back before you pick the next card. Your cards are KH, 7D, 6D, KH.Which of 1 or 2 did you sample with replacement and which did you sample without replacement?Show AnswerWithout replacementWith replacementThis video provides a brief lesson on finding the probability of independent events.Thumbnail for the embedded element "Probability of Independent Events"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=38 Try ItYou have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), and K (king) of that suit. S = spades, H = Hearts, D = Diamonds, C = Clubs. Suppose that you sample four cards.QS, 7D, 6D, KSKH, 7D, 6D, KHQS, 1D, 1C, QDWhich of the following outcomes are possible for sampling without replacement?Show AnswerWithout replacement:QS, 7D, 6D, KS: PossibleKH, 7D, 6D, KH: ImpossibleQS, 1D, 1C, QD: PossibleWhich of the following outcomes are possible for sampling with replacement?Show AnswerWith replacement:QS, 7D, 6D, KS: PossibleKH, 7D, 6D, KH: PossibleQS, 1D, 1C, QD: Possible Mutually Exclusive EventsA and B are mutually exclusive events if they cannot occur at the same time. This means that A and B do not share any outcomes and P(A AND B) = 0.For example, suppose the sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, and C = {7, 9}.A AND B = {4, 5}. \displaystyle{P}{({A} \text{ AND } {B})}=\frac{{2}}{{10}}and is not equal to zero. Therefore, A and B are not mutually exclusive.A AND C do not have any numbers in common so P(A AND C) = 0. Therefore, A and C are mutually exclusive.If it is not known whether A and B are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.Example 3Flip two fair coins. (This is an experiment.)The sample space is {HH, HT, TH, TT} where T = tails and H = heads. The possible outcomes are HH, HT, TH, and TT. The outcomes HT and TH are different. The HT means that the first coin showed heads and the second coin showed tails. The TH means that the first coin showed tails and the second coin showed heads.Let A = the event of getting at most one tail. (At most one tail means zero or one tail.) Then A can be written as {HH, HT,TH}. The outcome HH shows zero tails. HT and TH each show one tail.Let B = the event of getting all tails. B can be written as {TT}. B is the complement of A, so B = A′. P(A) + P(B) = P(A) + P(A′) = 1.The probabilities for A = P(event A) = \frac{{3}}{{4}}.Let C = the event of getting all heads. C = {HH}. Are eevnt C and event B mutually exclusive? Show AnswerSince B = {TT}, P(B AND C) = 0. B and C are mutually exclusive. (B and C have no members in common because you cannot have all tails and all heads at the same time.) Let D = event of getting more than one tail. D = {TT}. Find P(D). Show AnswerP(D) = Let E = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) Find P(E). Show AnswerE = {HT,HH}. P(E) = Find the probability of getting at least one (one or two) tail in two flips. Show AnswerLet F = event of getting at least one tail in two flips. F = {HT, TH, TT}. P(F) = Try ItDraw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.Show AnswerThe sample space of drawing two cards with replacement from a standard 52-card deck with respect to color is {BB, BR, RB, RR}.Event A = Getting at least one black card = {BB, BR, RB} Example 4Flip two fair coins. Find the probabilities of the events.Let F = the event of getting at most one tail (zero or one tail).Let G = the event of getting two faces that are the same.Let H = the event of getting a head on the first flip followed by a head or tail on the second flip.Are F and G mutually exclusive?Let J = the event of getting all tails. Are J and H mutually exclusive?Show AnswerThe sample space is {HH, HT, TH, TT} where T = tails and H = heads.Zero (0) or one (1) tails occur when the outcomes HH, TH, HT show up. P(F) = \frac{{3}}{{4}}Two faces are the same if HH or TT show up. P(G) = \frac{{2}}{{4}}A head on the first flip followed by a head or tail on the second flip occurs when HH or HT show up. P(H) = \frac{{2}}{{4}}F and G share HH so P(F AND G) is not equal to zero (0). F and G are not mutually exclusive.Getting all tails occurs when tails shows up on both coins (TT). H‘s outcomes are HH and HT. J and H have nothing in common so P(J AND H) = 0. J and H are mutually exclusive. This video provides two more examples of finding the probability of events that are mutually exclusive.Thumbnail for the embedded element "Ex: Probability of Events that are Mutually Exclusive Events"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=38 Try ItA box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement).If event F = the event of getting the white ball twice, find P(F). Show AnswerP(F) = If event G = the event of getting two balls of different colors, find P(G). Show AnswerP(G) = If event H = the event of getting white on the first pick, find P(G). Show AnswerP(H) = Are F and G mutually exclusive? Show AnswerYes Are G and H mutually exclusive? Show AnswerNo Example 5Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event A = a face is odd. Then A = {1, 3, 5}. Let event B = a face is even. Then B = {2, 4, 6}.Find the complement of A, A′. Show AnswerThe complement of A, A′, is B because A and B together make up the sample space. P(A) +P(B) = P(A) + P(A′) = 1. Also, P(A) = . Let event C = odd faces larger than two. Let event D = all even faces smaller than five. Are C and D mutually exclusive events? Why? Show AnswerC = {3, 5}.D = {2, 4}. P(C AND D) = 0 because you cannot have an odd and even face at the same time. Therefore, C and D are mutually exclusive events. Let event E = all faces less than five. Are C and E mutually exclusive events? Why? Show AnswerC = {3, 5} and E = {1, 2, 3, 4}. C and E = {3} P(C AND E) = . No, event C and event E are not mutually exclusive events. Find P(C|A). Show AnswerThis is a conditional probability. Recall that the event C is {3, 5} and event A is {1, 3, 5}. To find P(C|A), find the probability of C using the sample space A. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, P(C|A) = . Try ItLet event A = learning Spanish. Let event B = learning German. Then A AND B = learning Spanish and German. Suppose P(A) = 0.4 and P(B) = 0.2. P(A AND B) = 0.08. Are events A and B independent?Hint:You must show ONE of the following:P(A|B) = P(A)P(B|A)P(A AND B) = P(A)P(B)Show AnswerP(A|B)= The events are independent because P(A|B) = P(A). Example 6Let event G = taking a math class. Let event H = taking a science class. Then, G AND H = taking a math class and a science class. Suppose P(G) = 0.6, P(H) = 0.5, and P(G AND H) = 0.3. Are G and H independent?Hint:If G and H are independent, then you must show ONE of the following: P(G|H) = P(G)P(H|G) = P(H)P(G AND H) = P(G)P(H) Method 1: Show that P(G|H) = P(G). P(G|H) = = = 0.6 P(G) = 0.6 Since P(G|H) = P(G), G and H are independent. Method 2: Show P(G AND H) = P(G)P(H). P(G and H) = 0.3 P(G) * P(H) = 0.6 * 0.5 = 0.3 Since P(G AND H) = P(G)P(H), G and H are independent. Method 3: Show that P(H|G) = P(H). P(H|G) = = = 0.5 P(H) = 0.5 Since P(H|G) = P(H), G and H are independent. Since G and H are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. Try ItIn a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.R = a red marbleG = a green marbleO = an odd-numbered marbleThe sample space is S = {R1, R2, R3, R4, R5, R6, G1, G2, G3, G4}.S has ten outcomes. What is P(G AND O)?Show AnswerEvent G and O = {G1, G3}P(G and O) = = 0.2 Example 7Let event C = taking an English class. Let event D = taking a speech class.Suppose P(C) = 0.75, P(D) = 0.3, P(C|D) = 0.75 and P(C AND D) = 0.225.Justify your answers to the following questions numerically.Are C and D independent? Show AnswerYes, because P(C|D) = P(C). Are C and D mutually exclusive? Show AnswerNo, because P(C AND D) is not equal to zero. What is P(D|C)? Show Answer Try ItA student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(B AND D) = 0.20.Find P(B|D).Find P(D|B).Are B and D independent?Are B and D mutually exclusive?Show AnswerP(B|D) = 0.6667P(D|B) = 0.5NoNo Example 8In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.Let R = red card is drawn, B = blue card is drawn, E = even-numbered card is drawn.The sample space S = R1, R2, R3, B1, B2, B3, B4, B5. S has eight outcomes.Find P(R). Show AnswerP(R) = . Find P(R and B). Show AnswerP(R AND B) = 0. (You cannot draw one card that is both red and blue.) Find P(E). Show AnswerP(E) = . (There are three even-numbered cards, R2, B2, and B4.) Find P(E|B) . Show AnswerP(E|B) = = (There are five blue cards: B1, B2, B3, B4, and B5. Out of the blue cards, there are two even cards; B2 and B4.) Find P(B|E). Show AnswerP(B|E) = = . There are three even-numbered cards: R2, B2, and B4. Out of the even-numbered cards, to are blue; B2 and B4. Are events R and mutually exclusive? Show AnswerThe events R and B are mutually exclusive because P(R AND B) = 0. Let G = card with a number greater than 3. Find P(G). Show AnswerG = {B4, B5}. P(G) = , P(G) = P(G|H), which means that G and H are independent. Try ItIn a basketball arena,70% of the fans are rooting for the home team.25% of the fans are wearing blue.20% of the fans are wearing blue and are rooting for the away team.Of the fans rooting for the away team, 67% are wearing blue.Let A be the event that a fan is rooting for the away team, B be the event that a fan is wearing blue.Are the events of rooting for the away team and wearing blue independent?Are they mutually exclusive?Show AnswerP(B|A) = 0.67, P(B) = 0.25 So P(B) does not equal P(B|A) which means that B and A are not independent (wearing blue and rooting for the away team are not independent).They are also not mutually exclusive, because P(B AND A) = 0.20 \ne 0.Example 9In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let F be the event that a student is female. Let L be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?The following probabilities are given in this example:P(F) = 0.60; P(L) = 0.50P(F AND L) = 0.45P(L|F) = 0.75Solution 1: Check if P(F AND L) = P(F) * P(L). We are given that P(F AND L) = 0.45, but P(F) * P(L) = (0.60)(0.50) = 0.30. P(F AND L) P(F)P(L), the events of being female and having long hair are not independent.Solution 2Check whether P(L|F) equals P(L). We are given that P(L|F) = 0.75, but P(L) = 0.50. Since P(L|F) P(L), the events of being female and having long hair are not independent.Interpretation of ResultsThe events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.Try ItMark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street.P(I) = 0.44 and P(F) = 0.55P(I AND F) = 0 because Mark will take only one route to work.What is the probability of P(I OR F)?Show AnswerP(I AND F) = 0, P(I OR F) = P(I) + P(F) – P(I AND F) = 0.44 + 0.56 – 0 = 1 Example 10Toss one fair coin (the coin has two sides, H and T). The outcomes are ________. Count the outcomes. There are ____ outcomes. Show AnswerH and T; 2 Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are ________________. Count the outcomes. There are ___ outcomes. Show Answer1, 2, 3, 4, 5, 6 ; 6 Multiply the two numbers of outcomes. The answer is _______. Show Answer2(6) = 12 If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in three is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are H1 and T6.) Show AnswerT1, T2, T3, T4, T5, T6, H1, H2, H3, H4, H5, H6 Event A = heads (H) on the coin followed by an even number (2, 4, 6) on the die. A = {_________________}. Find P(A). Show AnswerA = {H2, H4, H6}; P(A) = Event B = heads on the coin followed by a three on the die. B = {________}. Find P(B). Show AnswerB = {H3}; P(B) = Are A and B mutually exclusive? (Hint: What is P(A AND B)? If P(A AND B) = 0, then A and B are mutually exclusive.) Show AnswerYes, because P(A AND B) = 0 Are A and B independent? (Hint: Is P(A AND B) = P(A)P(B)? If P(A AND B) = P(A)P(B), then A and B are independent. If not, then they are dependent). Show AnswerP(A AND B) = 0. P(A) * P(B) = * = . P(A AND B) P(A)P(B), so A and B are dependent. Try ItA box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let T be the event of getting the white ball twice, F the event of picking the white ball first, S the event of picking the white ball in the second drawing.Compute P(T).Compute P(T|F).Are T and F independent?.Are F and S mutually exclusive?Are F and S independent?Show AnswerP(T) = \frac{{1}}{{4}}P(T|F) = \frac{{1}}{{2}}NoNoYes ReferencesLopez, Shane, Preety Sidhu. “U.S. Teachers Love Their Lives, but Struggle in the Workplace.” Gallup Wellbeing, 2013. http://www.gallup.com/poll/161516/teachers-love-lives-struggle-workplace.aspx (accessed May 2, 2013).Data from Gallup. Available online at www.gallup.com/ (accessed May 2, 2013).Concept ReviewTwo events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.Formula ReviewIf A and B are independent, P(A AND B) = P(A)P(B), P(A|B) = P(A) and P(B|A) = P(B).If A and B are mutually exclusive, P(A OR B) = P(A) + P(B) and P(A AND B) = 0.Chapter 203.3 Two Basic Rules of ProbabilityWhen calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.The Multiplication RuleIf A and B are two events defined on a sample space, then: P(A AND B) = P(B)*P(A|B).This rule may also be written as (The probability of A given B equals the probability of A and B divided by the probability of B.)If A and B are independent, then P(A|B) = P(A). Then P(A AND B) = P(A|B)*P(B) becomes P(A AND B) = P(A)*P(B).The Addition RuleIf A and B are defined on a sample space, then: P(A OR B) = P(A) + P(B) – P(A AND B).If A and B are mutually exclusive, then P(A AND B) = 0. Then P(A OR B) = P(A) + P(B) – P(A AND B) becomes P(A OR B) = P(A) + P(B).Example 1Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska. Klaus can only afford one vacation. The probability that he chooses New Zealand is 0.6 and the probability that he chooses Alaska is 0.35. Klaus can only afford to take one vacation. What is the probability that he chooses either New Zealand or Alaska?What is the probability that he does not choose to go anywhere on vacation?Show AnswerLet A be New Zealand, B be Alaska. P(Klaus chooses New Zealand) = P(A) = 0.6 P(Klaus chooses Alaska ) = P(B) = 0.35 P(Klaus chooses both New Zealand and Alaska) = P(A and B) = 0 as he can only afford one vacation. P(A OR B) = P(A) + P(B) – P(A and B) = 0.6 + 0.35 – 0 = 0.95. Therefore, the probability that he chooses either New Zealand or Alaska is 0.95.The probability that he does not choose to go anywhere on vacation = 1 – P( A or B) = 1 – 0.95 = 0.05. Example 2Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. B = the event Carlos is successful on his second attempt. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.P(A) = 0.65, P(B) = 0.65, P(B|A) = 0.90.What is the probability that he makes both goals? Show AnswerThe problem is asking you to find P(A AND B) = P(B AND A). Since P(B|A) = 0.90, P(B AND A) = P(B|A) * P(A) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585. What is the probability that Carlos makes either the first goal or the second goal? Show AnswerThe problem is asking you to find P(A OR B). P(A OR B) = P(A) + P(B) – P(A AND B) = 0.65 + 0.65 – 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715. Are A and B independent? Show AnswerP(B AND A) = 0.585. P(B)P(A) = (0.65)(0.65) = 0.423 Since P(B AND A) P(B)*P(A), A and B are not independent. Are A and B mutually exclusive? Show AnswerP(A and B) = 0.585. To be mutually exclusive, P(A AND B) must equal zero. Therefore, A and are not mutually exclusive events. Watch this video for another example about first determining whether a series of events are mutually exclusive, then finding the probability of a specific outcome.Thumbnail for the embedded element "Ex: Probability of Events that are NOT Mutually Exclusive Events"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=39 Try ItHelen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. The probability that Helen makes the first shot is 0.75. The probability that Helen makes the second shot is 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?Show AnswerLet C be the event that Helen makes the first shot. Let D be the event that Helen makes the second shot. P(C) = 0.75, P(D) = 0.75, P(D|C) = 0.85 P(C AND D) = P(D AND C) = P(D|C)P(C) = (0.85)(0.75) = 0.6375 Helen makes the first and second free throws with probability 0.6375. Example 3A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.What is the probability that the member is a novice swimmer? Show Answer What is the probability that the member practices four times a week? Show Answer What is the probability that the member is an advanced swimmer and practices four times a week? Show Answer What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? Show AnswerA swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time, so P(advanced AND intermediate) = 0. These are mutually exclusive events. Are being a novice swimmer and practicing four times a week independent events? Why or why not? Show AnswerNo, these are not independent events. P(novice AND practices four times per week) = 0.0667 P(novice) * P(practices four times per week) = 0.09960. P(novice AND practices four times per week) P(novice) * P(practices four times per week) Try ItA school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?Show AnswerThe probability that a senior is taking a gap year = Example 4Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.Let M = math class, S = speech class, M|S = math given speechWhat is the probability that Felicity enrolls in math and speech?What is the probability that Felicity enrolls in math or speech classes?Are M and S independent?Are M and S mutually exclusive?SolutionShow AnswerP(M AND S) = P(M|S)P(S) = 0.25 * 0.65 = 0.1625.Show AnswerP(M OR S) = P(M) + P(S) – P(M AND S) = 0.2 + 0.65 – 0.1625 = 0.6875Show AnswerP(M AND S) = 0.1625 P(M) * P(S) = 0.2 * 0.65 = 0.13 Since P(M AND S) P(M) * P(S), M and S are not independent. Show AnswerSince P(M AND S) = 0.1625 0, M and S are not mutually exclusie. Try ItA student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.Find P(B AND D). Show AnswerP(B AND D) = P(D|B)P(B) = (0.5)(0.4) = 0.20. Find P(B OR D). Show AnswerP(B OR D) = P(B) + P(D) − P(B AND D) = 0.40 + 0.30 − 0.20 = 0.50 Example 5Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.What is the probability that the woman develops breast cancer? Show AnswerP(B) = 0.143 What is the probability that woman tests negative? Show AnswerP(N) = 0.85 Given that the woman has breast cancer, what is the probability that she tests negative? Show AnswerP(N|B) = 0.02 What is the probability that the woman has breast cancer AND tests negative? Show AnswerP(B AND N) = P(B)P(N|B) = (0.143)(0.02) = 0.0029 What is the probability that the woman has breast cancer or tests negative? Show AnswerP(B OR N) = P(B) + P(N) – P(B AND N) = 0.143 + 0.85 – 0.0029 = 0.9901 Are having breast cancer and testing negative independent events? Show AnswerNo. P(N) = 0.85; P(N|B) = 0.02. So, P(N|B) does not equal P(N). Are having breast cancer and testing negative mutually exclusive? Show AnswerNo. P(B AND N) = 0.0029. For B and N to be mutually exclusive, P(B AND N) must be zero. Try ItA school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?Show AnswerLet A = student is a senior going to college. Let B = student plays sports. Show AnswerP(A and B) = = = Example 6Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer, let N = tests negative, and let P = tests positive. Suppose one woman is selected at random.Given that a woman develops breast cancer, what is the probability that she tests positive? Show AnswerP(P|B) = 1 – P(N|B) = 1 – 0.02 = 0.98. What is the probability that a woman develops breast cancer and tests positive? Show AnswerP(B AND P) = P(P|B)P(B) = 0.98 * 0.143 = 0.1401. What is the probability that a woman does not develop breast cancer? Show AnswerP(B′) = 1 – P(B) = 1 – 0.143 = 0.857. What is the probability that a woman tests positive for breast cancer? Show AnswerP(P) = 1 – P(N) = 1 – 0.85 = 0.15. Try ItA student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D|B) = 0.5.Find P(B′).Find P(D AND B).Find P(B|D).Find P(D AND B′).Find P(D|B′).Show AnswerP(B′) = 0.60P(D AND B) = P(D|B) * P(B) = 0.20P(B|D)=\frac{{{P}{({B}\text{ AND } {D})}}}{{{P}{({D})}}}=\frac{{{0.20}}}{{{0.30}}}={0.66}P(D AND B′) = P(D) – P(D AND B) = 0.30 – 0.20 = 0.10P(D|B′) = P(D AND B′) * P(B′) = (P(D) – P(D AND B))(0.60) = (0.10)(0.60) = 0.06 ReferencesDiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013).Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013).“Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).“Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013).Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013).Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013).Data from U.S. Census Bureau.Data from the Wall Street Journal.Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013).Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013).Concept ReviewThe multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events A and B are mutually exclusive events when they do not have any outcomes in common.Formula ReviewThe multiplication rule: P(A AND B) = P(A|B)P(B)The addition rule: P(A OR B) = P(A) + P(B) – P(A AND B)Chapter 213.4 Contingency TablesA contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.The following video shows and example of finding the probability of an event from a table.Thumbnail for the embedded element "Ex: Basic Example of Finding Probability From a Table"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=40 Example 1Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:Speeding violation in the last yearNo speeding violation in the last yearTotalCell phone user25280305Not a cell phone user45405450Total70685755The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.Calculate the following probabilities using the table.Find P(Person is a car phone user). Show Answer Find P(person had no violation in the last year). Show Answer Find P(Person had no violation in the last year AND was a car phone user). Show Answer Find P(Person is a car phone user OR person had no violation in the last year). Show Answer Find P(Person is a car phone user GIVEN person had a violation in the last year). Show Answer(The sample space is reduced to the number of persons who had a violation.) Find P(Person had no violation last year GIVEN person was not a car phone user) Show Answer (The sample space is reduced to the number of persons who were not car phone users.) This video shows an example of how to determine the probability of an AND event using a contingency table.Thumbnail for the embedded element "Ex: Determine a Probability with AND using a Table"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=40 Try ItThis table shows the number of athletes who stretch before exercising and how many had injuries within the past year.Injury in last yearNo injury in last yearTotalStretches55295350Does not stretch231219450Total286514800What is P(athlete stretches before exercising)? Show AnswerP(athlete stretches before exercising) = = 0.4375 What is P(athlete stretches before exercising|no injury in the last year)? Show AnswerP(athlete stretches before exercising|no injury in the last year) = = 0.5739 Example 2This table shows a random sample of 100 hikers and the areas of hiking they prefer.Hiking Area PreferenceSexThe CoastlineNear Lakes and StreamsOn Mountain PeaksTotalFemale1816___45Male______1455Total___41______Complete the table. Show AnswerHiking Area PreferenceSexThe CoastlineNear Lakes and StreamsOn Mountain PeaksTotalFemale18161145Male16251455Total344125100 Are the events “being female” and “preferring the coastline” independent events? Hint:Let F = being female and let C = preferring the coastline.Check if P(F AND C) = P(F) * P(C).If P(F AND C) = P(F) * P(C), then F and C are independent. If P(F AND C) P(F) * P(C), then F and C are not independent. Show AnswerP(F AND C) = = 0.18 P(F) * P(C) = = (0.45)(0.34) = 0.153 P(F AND C) ≠ P(F) * P(C), so the events F and C are not independent. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Hint:Let M = being male, and let L = prefers hiking near lakes and streams.What word tells you this is a conditional?Fill in the blanks and calculate the probability: P(___|___) = ___.Is the sample space for this problem all 100 hikers? If not, what is it? Show AnswerThe word given tells you that this is a conditional. No, the sample space for this problem is the 41 hikers who prefer lakes and streams. Find the probability that a person is female or prefers hiking on mountain peaks. Hint:Let F = being female, and let P = prefers mountain peaks.Find P(F).Find P(P).Find P(F AND P).Find P(F OR P). Show AnswerThe probability that a person is female or prefers hiking on mountain peaks = P(F) = \displaystyle\frac{{45}}{{100}}P(P) = \displaystyle\frac{{25}}{{100}}P(F AND P) = \displaystyle\frac{{11}}{{100}}P(F OR P) = \displaystyle\frac{{45}}{{100}} + \frac{{25}}{{100}} - \frac{{11}}{{100}} = \frac{{59}}{{100}} Try ItThis table shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.GenderLake PathHilly PathWooded PathTotalFemale453827110Male26521290Total719039200Out of the males, what is the probability that the cyclist prefers a hilly path? Show AnswerP(H|M) = = 0.5778 Are the events “being male” and “preferring the hilly path” independent events? Show AnswerFor M and H to be independent, show P(H|M) = P(H) P(H|M) = 0.5778, P(H) = = 0.45 P(H|M) P(H), so M and H are not independent. Example 3Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is and the probability he is not caught is . If he goes out the second door, the probability he gets caught by Alissa is and the probability he is not caught is . The probability that Alissa catches Muddy coming out of the third door is and the probability she does not catch Muddy is . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is .Door ChoiceCaught or NotDoor OneDoor TwoDoor ThreeTotalCaught\displaystyle\frac{{1}}{{15}}\displaystyle\frac{{1}}{{12}}\displaystyle\frac{{1}}{{6}}____Not Caught\displaystyle\frac{{4}}{{15}}\displaystyle\frac{{3}}{{12}}\displaystyle\frac{{1}}{{6}}____Total____________1The first entry \displaystyle\frac{{1}}{{15}}={(\frac{{1}}{{5}})}{(\frac{{1}}{{3}})} is P(Door One AND Caught)The entry \displaystyle\frac{{4}}{{15}}={(\frac{{4}}{{5}})}{(\frac{{1}}{{3}})} is P(Door One AND Not Caught)Verify the remaining entries.Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1. Show AnswerDoor ChoiceCaught or NotDoor OneDoor TwoDoor ThreeTotalCaught\displaystyle\frac{{1}}{{15}}\displaystyle\frac{{1}}{{12}}\displaystyle\frac{{1}}{{6}}\displaystyle\frac{{19}}{{60}}Not Caught\displaystyle\frac{{4}}{{15}}\displaystyle\frac{{3}}{{12}}\displaystyle\frac{{1}}{{6}}\displaystyle\frac{{41}}{{60}}Total\displaystyle\frac{{5}}{{15}}\displaystyle\frac{{4}}{{12}}\displaystyle\frac{{2}}{{16}}1 What is the probability that Alissa does not catch Muddy? Show Answer What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa? Show Answer Example 4This table contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.United States Crime Index Rates Per 100,000 Inhabitants 2008–2011YearRobberyBurglaryRapeVehicleTotal2008145.7732.129.7314.72009133.1717.729.1259.22010119.370127.7239.12011113.7702.226.8229.6TotalTOTAL each column and each row. Total data = 4,520.7Find P(2009 AND Robbery). Show Answer0.0294 Find P(2010 AND Burglary). Show Answer0.1551 Find P(2010 OR Burglary). Show Answer0.7165 Find P(2011|Rape). Show Answer0.2365 Find P(Vehicle|2008). Show Answer0.2575 This video gives and example of determining an “OR” probability given a table.Thumbnail for the embedded element "Ex: Determine the Probability of a Union Using a Table - Not Mutually Exclusive"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=40 Try ItThis table relates the weights and heights of a group of individuals participating in an observational study.Weight/HeightTallMediumShortTotalsObese182814Normal205128Underweight12259TotalsFind the total for each row and column. Show AnswerWeight/HeightTallMediumShortTotalsObese18281460Normal20512899Underweight1225946Totals5010451205 Find the probability that a randomly chosen individual from this group is Tall. Show AnswerP(Tall) = = 0.244 Find the probability that a randomly chosen individual from this group is Obese and Tall. Show AnswerP(Obese AND Tall) = = 0.088 Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese. Show AnswerP(Tall|Obese) = = 0.3 Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall. Show AnswerP(Obese|Tall) = = 0.36 Find the probability a randomly chosen individual from this group is Tall and Underweight. Show AnswerP(Tall AND Underweight = = 0.0585 Are the events Obese and Tall independent? Show AnswerNo. P(Tall) (Tall|Obese). References“Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (accessed May 3, 2013).Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services.Data from United States Senate. Available online at www.senate.gov (accessed May 2, 2013).Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013).“Human Blood Types.” Unite Blood Services, 2011. Available online at http://www.unitedbloodservices.org/learnMore.aspx (accessed May 2, 2013).Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013).“United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013).Concept ReviewThere are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables.Chapter 223.5 Tree and Venn DiagramsSometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.Tree DiagramsA tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of “branches” that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.Example 1In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball.a. Use tree diagram to show all possible outcomes.Show AnswerTotal = 64 + 24 + 24 + 9 = 121The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:R1R1 R1R2 R1R3 R2R1 R2R2 R2R3 R3R1 R3R2 R3R3The other outcomes are similar.There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.b. List the 24 BR outcomes.Show AnswerB1R1, B1R2, B1R3, B2R1, B2R2, B2R3, B3R1, B3R2, B3R3, B4R1, B4R2, B4R3, B5R1, B5R2, B5R3, B6R1, B6R2, B6R3, B7R1, B7R2, B7R3, B8R1, B8R2, B8R3c. Using the tree diagram, calculate P(RR).Show AnswerP(RR) = = d. Using the tree diagram, calculate P(RB OR BR).Show AnswerP(RB OR BR) = ()() + ()() = e. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).Show AnswerP(R on 1st draw AND B on 2nd draw) =()() = f. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).Show AnswerP(R on 2nd B on 1st) = = g. Using the tree diagram, calculate P(BB).Show AnswerP(BB) = h. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).Show Answer]P(B on 2nd R on 1st) = There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. Therefore, P(B on the 2nd draw given R on the first draw) = = Try ItIn a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).Show AnswerTotal number of outcomes is 144 + 480 + 480 + 1600 = 2,704.P(FF) = “Without replacement” means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches.Example 2An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn.This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.Total = \displaystyle\frac{{56+24+24+6}}{{110}}=\frac{{110}}{{110}}=1Note:If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.a. P(RR) = ________Show AnswerP(RR) = b. Fill in the blanks:P(RB or BR) = ____________Show AnswerP(RB or BR) = c. P(R on 2nd|B on 1st) = ___________Show AnswerP(R on 2nd|B on 1st) = d. Fill in the blanks.P(R on 1st AND B on 2nd) = P(RB) = (___)(___) = \frac{24}{100}Show AnswerP(R on 1st AND B on 2nd) = P(RB) = ()() = e. Find P(BB).Show AnswerP(BB) = ()() = f. Find P(B on 2nd|R on 1st).Show AnswerP(B on 2nd|R on 1st) = If we are using probabilities, we can label the tree in the following general way.This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).P(R|R) here means P(R on 2nd|R on 1st)P(B|R) here means P(B on 2nd|R on 1st)P(R|B) here means P(R on 2nd|B on 1st)P(B|B) here means P(B on 2nd|B on 1st)Thumbnail for the embedded element "Count outcomes using tree diagram | Statistics and probability | 7th grade | Khan Academy"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=41 Try ItA litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.What is the probability that both kittens are tabby?Show AnswerP(both kittens are tabby) = What is the probability that one kitten of each coloring is selected?Show AnswerP(one kitten of each coloring is selected) = P(TB) + P(BT) = What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?Show AnswerP(a tabby is chosen as the second kitten when a black kitten was chosen as the first) = What is the probability of choosing two kittens of the same color?Show AnswerP(choosing two kittens of the same color) = P(TT) + P(BB) = Example 3Suppose there are 4 red balls and 3 yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected? Show AnswerP(one of each coloring is selected) = P(RY) + P(YR) = Venn Diagram A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events. Example 4Suppose an experiment has the outcomes 1, 2, 3, … , 12 where each outcome has an equal chance of occurring. Let event A = {1, 2, 3, 4, 5, 6} and event B = {6, 7, 8, 9}. Then A AND B = {6} and A OR B = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:A Venn diagram. An oval representing set A contains the values 1, 2, 3, 4, 5, and 6. An oval representing set B also contains the 6, along with 7, 8, and 9. The values 10, 11, and 12 are present but not contained in either set.Try ItSuppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.Show Answer Example 5Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore,A AND B = {TT}. A OR B = {TH, TT, HT}.The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. The Venn diagram is as follows:This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.Try ItRoll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.Show Answer Example 640% of the students at a local college belong to a club and 50% work part time. 5% of the students work part time and belong to a club.a. Draw a Venn diagram showing the relationships. Show AnswerLet C = student belongs to a club and PT = student works part time.b. If a student is selected at random, find the probability that the student belongs to a club.Show AnswerP(the student belongs to a club) = 0.40c. If a student is selected at random, find the probability that the student works part time.Show AnswerP(the student works part time) = 0.50d. If a student is selected at random, find the probability that the student belongs to a club AND works part time.Show AnswerP(the student belongs to a club AND works part time) = 0.05e. If a student is selected at random, find the probability that the student belongs to a club given that the student works part time.Show AnswerP(the student belongs to a club given that the student works part time) = AND the student works part time)}{\text{P(the student works part time)} = \frac{0.05}{0.50}= 0.1" title="\frac{\text{P(the student belongs to a club AND the student works part time)}{\text{P(the student works part time)} = \frac{0.05}{0.50}= 0.1" class="latex mathjax">f. If a student is selected at random, find the probability that the student belongs to a club OR works part time.Show AnswerP(the student belongs to a club OR works part time) = P(the student belongs to a club) + P(the student works part time) – P(the student belongs to a club AND works part time) = 0.40 + 0.50 – 0.05 = 0.85Try ItFifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works.Show AnswerExample 7In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.Draw a Venn diagram representing the situation. Show AnswerIn the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction. Find the probability that the customer buys either a novel or anon-fiction book. Show AnswerP(novel or non-fiction) = P(Blue OR Red) = P(Blue) + P(Red) – P(Blue AND Red) = 0.6 + 0.4 – 0.2 = 0.8. In the Venn diagram, describe the overlapping area using a complete sentence. Show AnswerThe overlapping area of the blue oval and red oval represents the customers buying both a novel and a nonfiction book. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event. Show AnswerIn the following Venn diagram below, the blue oval represent customers buying a novel, the red oval represents customer buying non-fiction, and the yellow oval customer who buy compact disks. https://youtu.be/MassxXy8iko Glossary Tree Diagramthe useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies) Venn Diagramthe visual representation of a sample space and events in the form of circles or ovals showing their intersections Solutions to Try These 1:a. B1R1 B1R2 B1R3 B2R1 B2R2 B2R3 B3R1 B3R2 B3R3 B4R1 B4R2 B4R3 B5R1 B5R2 B5R3 B6R1 B6R2 B6R3 B7R1B7R2 B7R3 B8R1 B8R2 B8R3b. P(RR) = (311)(311) = 9121c. P(RB OR BR) = (311)(811) + (811)(311) = 48121d. P(R on 1st draw AND B on 2nd draw) = P(RB) = (311)(811) = 24121e. P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = 2488 = 311This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. 2488 = 311.f. P(BB) = 64121g. P(B on 2nd draw|R on 1st draw) = 811There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then 2433.Solutions to Try These 2:a. P(RR) = (311)(210)=6110b. P(RB OR BR) = (311)(810) + (811)(310) = 48110c. P(R on 2nd|B on 1st) = 310d. P(R on 1st AND B on 2nd) = P(RB) = (311)(810) = 24100e. P(BB) = (811)(710)f. Using the tree diagram, P(B on 2nd | R on 1st) = P(R | B) = 810.PART IVDiscrete Random VariablesChapter 23Introduction: Discrete Random VariablesThis photo shows branch lightening coming from a dark cloud and hitting the ground.You can use probability and discrete random variables to calculate the likelihood of lightning striking the ground five times during a half-hour thunderstorm. (Credit: Leszek Leszczynski)Learning ObjectivesBy the end of this chapter, the student should be able to:Recognize and understand discrete probability distribution functions, in general.Calculate and interpret expected values.Recognize the binomial probability distribution and apply it appropriately.Recognize the Poisson probability distribution and apply it appropriately.Recognize the geometric probability distribution and apply it appropriately.Recognize the hypergeometric probability distribution and apply it appropriately.Classify discrete word problems by their distributions. A student takes a ten-question, true-false quiz. Because the student had such a busy schedule, he or she could not study and guesses randomly at each answer. What is the probability of the student passing the test with at least a 70%?Small companies might be interested in the number of long-distance phone calls their employees make during the peak time of the day. Suppose the average is 20 calls. What is the probability that the employees make more than 20 long-distance phone calls during the peak time?These two examples illustrate two different types of probability problems involving discrete random variables. Recall that discrete data are data that you can count. Arandom variable describes the outcomes of a statistical experiment in words. The values of a random variable can vary with each repetition of an experiment.Random Variable NotationUpper case letters such as X or Y denote a random variable. Lower case letters like x or y denote the value of a random variable. If X is a random variable, then X is written in words, and x is quantitative.For example, let X = the number of heads you get when you toss three fair coins. The sample space for the toss of three fair coins is TTT; THH; HTH; HHT; HTT; THT; TTH;HHH. Then, x = 0, 1, 2, 3. X is in words and x is a number. Notice that for this example, the x values are countable outcomes. Because you can count the possible values that X can take on and the outcomes are random (the x values 0, 1, 2, 3), X is a discrete random variable.GlossaryRandom Variable (RV)a characteristic of interest in a population being studied; common notation for variables are upper case Latin letters X, Y, Z,…; common notation for a specific value from the domain (set of all possible values of a variable) are lower case Latin letters x, y, and z. For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,…. Variables in statistics differ from variables in intermediate algebra in the two following ways. The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color then the domain is {black, blond, gray, green, orange}.We can tell what specific value x the random variable X takes only after performing the experiment. Chapter 244.1 Probability Distribution Function (PDF) for a Discrete Random VariableThe idea of a random variable can be confusing. In this video we help you learn what a random variable is, and the difference between discrete and continuous random variables.Thumbnail for the embedded element "Random Variables and Probability Distributions"A YouTube element has been excluded from this version of the text. You can view it online here: https://library.achievingthedream.org/odessastatistics/?p=44 A discrete probability distribution function has two characteristics:Each probability is between zero and one, inclusive.The sum of the probabilities is one.Example 1:A child psychologist is interested in the number of times a newborn baby’s crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby’s crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.P(x) = probability that X takes on a value x.Probability distribution table for Example 1xP(x)0P(x = 0) = \frac{2}{50}1P(x = 1) = \frac{11}{50}2P(x = 2) = \frac{23}{50}3P(x = 3) = \frac{9}{50}4P(x = 4) = \frac{4}{50}5P(x = 5) = \frac{1}{50}X takes on the values 0, 1, 2, 3, 4, 5. This is a discrete PDF because:Each P(x) is between zero and one, inclusive. P(x = 0) = \frac{2}{50} > 0 P(x = 1) = \frac{11}{50}> 0 P(x = 2) = \frac{23}{50} > 0 P(x = 3) = \frac{9}{50} > 0 P(x = 4) = \frac{4}{50} > 0 P(x = 5) = \frac{1}{50} > 0The sum of the probabilities is one, that is, P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) =\frac{2}{50} + \frac{11}{50} + \frac{23}{50} + \frac{9}{50} + \frac{4}{50} + \frac{1}{50} = 1 Try ItSuppose Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected. Let X = the number of days Nancy has classes in a week.a. X takes on what values? Show Answer0 day, 1 day, 2 days, 3 days b. Suppose one week is randomly chosen. Construct a probability distribution table (called a PDF table). The table should have two columns labeled x and P(x). Show AnswerXP(X)01% = 0.0114% = 0.04215% = 0.15380% = 0.80 c. What does the P(x) column sum to? Show AnswerP(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) = 0.01 + 0.04 + 0.15 + 0.80 = 1. Example 2Jeremiah has basketball practice two days a week. Ninety percent of the time, he attends both practices. Eight percent of the time, he attends one practice. Two percent of the time, he does not attend either practice. What is X and what values does it take on?Show AnswerX is the number of days Jeremiah attends basketball practice per week.X takes on the values 0, 1, and 2.Number of days Jeremiah attends basketball practice per week, XP(X)02% = 0.0218% = 0.08290% = 0.90Concept ReviewThe characteristics of a probability distribution function (PDF) for a discrete random variable are as follows:Each probability is between zero and one, inclusive (inclusive means to include zero and one).The sum of the probabilities is one. Solutions to Try These:a. Let X = the number of days Nancy attends class per week.b. 0, 1, 2, and 3c.xP(x)00.0110.0420.1530.80Chapter 254.2 Mean / Expected Value and Standard Deviation of Discrete Random VariableThe expected value is often referred to as the “long-term” average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.You toss a coin and record the result. What is the probability that the result is heads? If you flip a coin two times, does probability tell you that these flips will result in one heads and one tail? You might toss a fair coin ten times and record nine heads. Probability does not describe the short-term results of an experiment. It gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times. In his experiment, Pearson illustrated the Law of Large Numbers.The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero (the theoretical probability and the relative frequency get closer and closer together). When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment and is denoted by the Greek letter μ. In other words, after conducting many trials of an experiment, you would expect this average value.Expected Value, \mu = {x}_{1}*P({x}_{1}) + {x}_{2}*P({x}_{2}) + {x}_{3}*P({x}_{3}) + …. = sum of all (x * P(x) )Variance, \sigma2 = sum of all ((x – μ)2 ⋅ P(x))Standard deviation, \sigma = \sqrt{{\sigma}^{2}}Example 1A men’s soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, μ, of the number of days per week the men’s soccer team plays soccer.Solution:To do the problem, first let the random variable X = the number of days the men’s soccer team plays soccer per week. X takes on the values 0, 1, 2. Construct a PDF table adding a column x ⋅ P(x). In this column, you will multiply each x value by its probability.Expected Value Table. This table is called an expected value table. The table helps you calculate the expected value or long-term average.xP(x)x⋅ P(x)00.2(0)(0.2) = 010.5(1)(0.5) = 0.520.3(2)(0.3) = 0.6Add the last column x ⋅ P(x) to find the long term average or expected value: (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1.The expected value is 1.1. The men’s soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the men’s soccer team plays soccer week after week after week. We say μ = 1.1.Example 2 The probability that a newborn baby does not cry after midnight is . The probability that a newborn baby cries once after midnight is . The probability that a newborn baby cries twice after midnight is . The probability that a newborn baby cries thrice after midnight is . The probability that a newborn baby cries for 4 times after midnight is . The probability that a newborn baby cries for 5 times after midnight is .Find the expected value of the number of times a newborn baby’s crying wakes its mother after midnight. (The expected value is the expected number of times per week a newborn baby’s crying wakes its mother after midnight. )Calculate the standard deviation of the variable as well.Solution:You expect a newborn to wake its mother after midnight 2.1 times per week, on the average.x (Number of times a newborn baby crying after midnight)P(x)x⋅ P(x)(x – μ)2 ⋅ P(x)0\displaystyle{P}{({x}={0})}=\frac{{2}}{{50}}\displaystyle{({0})}{(\frac{{2}}{{50}})}=\frac{{0}}{{50}}\displaystyle { ( { 0 } - { 2.1 } ) } ^{ { 2 } } \cdot { 0.04}={0.1764}1\displaystyle{P}{({x}={1})}=\frac{{11}}{{50}}\displaystyle{({1})}{(\frac{{11}}{{50}})}=\frac{{11}}{{50}}\displaystyle{({1}-{2.1})}^{{2}}\cdot{0.22}={0.2662}2\displaystyle{P}{({x}={2})}=\frac{{23}}{{50}}\displaystyle{({2})}{(\frac{{23}}{{50}})}=\frac{{46}}{{50}}\displaystyle{({2}-{2.1})}^{{2}}\cdot{0.46}={0.0046}3\displaystyle{P}{({x}={3})}=\frac{{9}}{{50}}\displaystyle{({3})}{(\frac{{9}}{{50}})}=\frac{{27}}{{50}}\displaystyle{({3}-{2.1})}^{{2}}\cdot{0.18}={0.1458}4\displaystyle{P}{({x}={4})}=\frac{{4}}{{50}}\displaystyle{({4})}{(\frac{{4}}{{50}})}=\frac{{16}}{{50}}\displaystyle{({4}-{2.1})}^{{2}}\cdot{0.08}={0.2888}5\displaystyle{P}{({x}={5})}=\frac{{1}}{{50}}\displaystyle{({5})}{(\frac{{1}}{{50}})}=\frac{{5}}{{50}}\displaystyle{({5}-{2.1})}^{{2}}\cdot{0.02}={0.1682}sum of all (x ⋅ P(x) ) = 2.1 sum of all ( (x – μ)2 ⋅ P(x) ) = 1.05Add the values in the third column of the table to find the expected value of X:μ = Expected Value = + + + + + = = 2.12. Use μ to complete the table. The fourth column of this table will provide the values you need to calculate the standard deviation. For each value x, multiply the square of its deviation by its probability. (Each deviation has the format x – μ).Add the values in the fourth column of the table:variance of X = 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05The standard deviation of X is the square root of this sum: Try ItA hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?[practice-area rows=”1″][/practice-area]Click here to show answer: xP(x)0\displaystyle{P}{({x}={0})}=\frac{{4}}{{50}}1\displaystyle{P}{({x}={1})}=\frac{{8}}{{50}}2\displaystyle{P}{({x}={2})}=\frac{{16}}{{50}}3\displaystyle{P}{({x}={3})}=\frac{{14}}{{50}}4\displaystyle{P}{({x}={4})}=\frac{{6}}{{50}}5\displaystyle{P}{({x}={5})}=\frac{{2}}{{50}}(The expected value is 2.32. Example 3Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game?Solution:To do this problem, set up an expected value table for the amount of money you can profit.Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.If you purchase a ticket, you will either make a profit of $100,000 or lose $2. Since you are interested in your profit, the values of x are 100,000 dollars and −2 dollars.To win, you must get all five numbers correct, in order. The probability of choosing one correct number is because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order isTherefore, the probability of winning is 0.00001 and the probability of losing is The expected value table is as follows:xP(x)x⋅ P(x)Loss–20.99999(–2)(0.99999) = –1.99998Profit100,0000.00001(100000)(0.00001) = 1Αdd the value of x ⋅ P(x). Expected value, = –1.99998 + 1 = –0.99998 1Since –0.99998 is closed to –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over.Try ItYou are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term?Click here to show answer: Let X = the amount of money you profit. The x-values are –$1 and $256.The probability of guessing the right suit each time isThe probability of losing is Expected profit of playing the game over the long term = (0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023 or 0.23 cents. Example 4Suppose you play a game with a biased coin. You play each game by tossing the coin once. P(heads) = If you toss a head, you pay $6. If you toss a tail, you win $10.Define a random variable X.Complete the following expected value table. x________WIN10\displaystyle \frac{{1}}{{3}}____LOSE________\displaystyle \frac{{-12}}{{3}} What is the expected value, μ? Do you come out ahead?Show AnswerX = amount of profitxP(x)x⋅ P(x)WIN10\displaystyle \frac{{1}}{{3}} \\\displaystyle \frac{{10}}{{3}} \\LOSE–6\displaystyle \frac{{2}}{{3}} \\\displaystyle \frac{{-12}}{{3}} \\The expected value μ = sum of all (x ⋅ P(x)) = \frac{10}{3} + \frac{-12}{3} = \displaystyle \frac{{-2}}{{3}}. You lose, on average, about 67 cents each time you play the game so you do not come out ahead. Example 5Suppose you play a game with a spinner. You play each game by spinning the spinner once. P(red) = . If you land on red, you pay $10. If you land on blue, you don’t pay or win anything. If you land on green, you win $10.Complete the following expected value table.Let X be your profit.xP(x)x ⋅ P(x)Red– \frac{{20}}{{5}}Blue\frac{2}{5}Green10Show AnswerxP(x)x⋅ P(x)Red–10\frac{{2}}{{5}}-\frac{{20}}{{5}}Blue0\frac{{2}}{{5}}\frac{{0}}{{5}}Green10\frac{{1}}{{5}}\frac{{10}}{{5}} Like data, probability distributions have standard deviations. To calculate the standard deviation (σ) of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root. To understand how to do the calculation, look at the table for the number of days per week a men’s soccer team plays soccer. To find the standard deviation, add the entries in the column labeled (x – μ)2P(x) and take the square root.xP(x)x⋅ P(x)(x – μ)2⋅ P(x)00.2(0)(0.2) = 0(0 – 1.1 )2⋅ (0.2) = 0.24210.5(1)(0.5) = 0.5(1 – 1.1 )2⋅ (0.5) = 0.00520.3