15 2.7 Measures of the Spread of Data
An 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, that
7 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 integer
- One 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: [latex]\displaystyle{x}=\overline{{x}}+[/latex](# of STDEV)[latex]{({s})}[/latex]
- Population: [latex]\displaystyle{x}=\mu+[/latex](# of STDEV)[latex]{(\sigma)}[/latex]
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 [latex]\displaystyle\overline{{x}}[/latex] is the sample mean and the Greek symbol μ is the population mean.
Calculating the Standard Deviation
If 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 [latex]\displaystyle{x}-\overline{{x}}[/latex].
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 – [latex]\overline{{x}}[/latex] 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 Deviation[latex]\displaystyle{s}=\sqrt{{\frac{{\sum{({x}-\overline{{x}})}^{{2}}}}{{{n}-{1}}}}}{\quad\text{or}\quad}{s}=\sqrt{{\frac{{\sum{(f){{({x}-\overline{{x}})}}}^{{2}}}}{{{n}-{1}}}}}[/latex] For the sample standard deviation, the denominator is n – 1, that is the sample size – 1. Formulas for the Population Standard Deviation[latex]\displaystyle\sigma=\sqrt{{\frac{{\sum{({x}-\mu)}^{{2}}}}{{{N}}}}}{\quad\text{or}\quad}\sigma=\sqrt{{\frac{{\sum{(f){{({x}-\mu)}}}^{{2}}}}{{{N}}}}}[/latex] For 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.
Sampling Variability of a Statistic
How 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 [latex]\frac{{\sigma}}{{\sqrt{n}}}[/latex] 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 1
The 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; 11.5
1. The teacher was interested in the average age and the sample standard deviation of the ages of her students.
Show Answer
The 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.
| Data | Freq. | Deviations | Deviations2 | (Freq.)( Deviations2) |
|---|---|---|---|---|
| x | f | ( x – [latex]\overline{x}[/latex]) | ( x –[latex]\overline{x}[/latex])2 | (f)(x –[latex]\overline{x}[/latex])2 |
| 9 | 1 | 9 – 10.525 = –1.525 | (–1.525) 2 = 2.325625 | 1 × 2.325625 = 2.325625 |
| 9.5 | 2 | 9.5 – 10.525 = –1.025 | (–1.025) 2 = 1.050625 | 2 × 1.050625 = 2.101250 |
| 10 | 4 | 10 – 10.525 = –0.525 | (–0.525) 2 = 0.275625 | 4 × 0.275625 = 1.1025 |
| 10.5 | 4 | 10.5 – 10.525 = –0.025 | (–0.025) 2 = 0.000625 | 4 × 0.000625 = 0.0025 |
| 11 | 6 | 11 – 10.525 = 0.475 | (0.475) 2 = 0.225625 | 6 × 0.225625 = 1.35375 |
| 11.5 | 3 | 11.5 – 10.525 = 0.975 | (0.975) 2 = 0.950625 | 3 × 0.950625 = 2.851875 |
| The total is 9.7375 |
The sample variance, [latex]\displaystyle{s}^{2}[/latex], is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1): s2 = [latex]\frac{9.7375}{20-1}[/latex] = 0.5125
The sample standard deviation s is equal to the square root of the sample variance: which is rounded to two decimal places, s = [latex]\sqrt{0.5125}[/latex] = 0.72.
[latex]\overline{x} = {9+9.5(2)+10(4)+10.5(4)+11(6)+11.5(3)20}={10.525}[/latex]
2. Find the value that is one standard deviation above the mean. (Find [latex]\overline{x}[/latex] + 1s. )
Show Answer
[latex]\overline{x}[/latex]+ 1s = 10.53 + (1)(0.72) = 11.25
3. Find the value that is two standard deviations below the mean. (Find [latex]\overline{x}[/latex] – 2s. )
Show Answer
[latex]\overline{x}[/latex] – 2s = 10.53 – (2)(0.72) = 9.09
4. Find the values that are 1.5 standard deviations from(below and above) the mean.
Show Answer
([latex]\displaystyle\overline{x}[/latex] – 1.5s) = 10.53 – (1.5)(0.72) = 9.45
([latex]\displaystyle\overline{x}[/latex] + 1.5s) = 10.53 + (1.5)(0.72) = 11.61
Typically, 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 = [latex]\overline{x}[/latex] + (# of STDEV)(s)
For a population: x = μ + (# of STDEV)(σ)
For this example, use x =[latex]\overline{x}[/latex] + (# of STDEV)(s) because the data is from a sample.
Verify the mean and standard deviation on your calculator or computer:
|
Try It
On a baseball team, the ages of each of the players are as follows:
[latex]\displaystyle {21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40}[/latex]
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 Answer
μ = 30.68
s = 6.09
[latex]\displaystyle\overline{x}[/latex] + 2s = 30.68 + (2)(6.09) = 42.86.
Explanation of the standard deviation calculation shown in the table
The 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 2
Use 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; 100
- Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
Show Answer
Data Frequency Relative Frequency Cumulative Relative Frequency 33 1 0.032 0.032 42 1 0.032 0.064 49 2 0.065 0.129 53 1 0.032 0.161 55 2 0.065 0.226 61 1 0.032 0.258 63 1 0.032 0.29 67 1 0.032 0.322 68 2 0.065 0.387 69 2 0.065 0.452 72 1 0.032 0.484 73 1 0.032 0.516 74 1 0.032 0.548 78 1 0.032 0.580 80 1 0.032 0.612 83 1 0.032 0.644 88 3 0.097 0.741 90 1 0.032 0.773 92 1 0.032 0.805 94 4 0.129 0.934 96 1 0.032 0.966 100 1 0.032 0.998 (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 Answer
The sample mean = 73.5
- The sample standard deviation
Show Answer
The sample standard deviation = 17.9
- The median
Show Answer
The median = 73
- The first quartile
Show Answer
The first quartile = 61
- The third quartile
Show Answer
The third quartile = 90
- IQR
Show Answer
IQR = 90 – 61 = 29
- The sample mean
- 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 Answer
The 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 It
The following data show the different types of pet food stores in the area carry.
[latex]\displaystyle {6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;}[/latex]
Calculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.
Show Answer
μ = 9.3
s = 2.2
Standard Deviation of Grouped Frequency Tables
Recall 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, [latex]\overline{x}[/latex] = [latex]\frac{{\sum(fm)}}{{\sum(f)}}[/latex]
Standard Deviation of Frequency Table, [latex]{s}_{x}=\sqrt{{\frac{{f{(m-\overline{x})}^{2}}}{{n-1}}}}[/latex]
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 3
| Class | Frequency |
| 0 – 2 | 1 |
| 3 – 5 | 6 |
| 6 – 8 | 10 |
| 9 – 11 | 7 |
| 12 – 14 | 0 |
| 15 – 17 | 2 |
Find the standard deviation for the data in the table.
Show Answer
Steps 1: Find the mean.
| Class | Frequency, f | Midpoint, m |
|---|---|---|
| 0–2 | 1 | 1 |
| 3–5 | 6 | 4 |
| 6–8 | 10 | 7 |
| 9–11 | 7 | 10 |
| 12–14 | 0 | 13 |
| 15–17 | 2 | 16 |
For this data set, we have the mean, [latex]\overline{x}[/latex] = [latex]\frac{\sum{fm}}{n}[/latex] = [latex]\frac{(1)(1) + (6)(4)+(10)(7)+(7)(10)+(0)(13)+(2)(16)}{26}[/latex] = 7.58.
Step 2: Use the mean to find the standard deviation.
| Class | Frequency, f | Midpoint, m | [latex]m-\overline{x}[/latex] | [latex](m-\overline{x})^2[/latex] | [latex](f)(m-\overline{x})^2[/latex] |
|---|---|---|---|---|---|
| 0–2 | 1 | 1 | 1 – 7.58 = -6.58 | [latex](-6.58)^2[/latex] = 43.2964 | (1)(43.296) = 43.2964 |
| 3–5 | 6 | 4 | 4 – 7.58 = -3.58 | [latex](-3.58)^2[/latex] = 12.8164 | (6)(12.816) = 76.8984 |
| 6–8 | 10 | 7 | 7 – 7.58 = -0.58 | [latex](-0.58)^2[/latex] = 0.3364 | (10)(0.3364) = 3.364 |
| 9–11 | 7 | 10 | 10 – 7.58 = 2.42 | [latex](2.42)^2[/latex] = 5.8564 | (7)(5.8564) = 40.9948 |
| 12–14 | 0 | 13 | 13 – 7.58 = 5.42 | [latex](5.42)^2[/latex] = 29.3764 | (0)(29.376) = 0 |
| 15–17 | 2 | 16 | 16 – 7.58 = 8.42 | [latex](8.42)^2[/latex] = 70.8964 | (2)(70.896) = 141.7928 |
The standard deviation, [latex]{s}_{x} = \sqrt{\frac{\sum{(f)(m-\overline{x})^2}}{n-1}} = \sqrt{\frac{43.2964 + 76.8984 + 3.364 + 40.9948 + 0 + 141.7928}{26-1}} = \sqrt{\frac{306.3464}{25}} = \sqrt{12.2539} = 3.5[/latex].
|
Mean of Frequency Table, [latex]\overline{x}[/latex] = [latex]\frac{{\sum(fm)}}{{\sum(f)}}[/latex] Standard Deviation of Frequency Table, [latex]{s}_{x}=\sqrt{{\frac{{f{(m-\overline{x})}^{2}}}{{n-1}}}}[/latex] where f = interval frequencies and m = interval midpoints. The calculations are tedious. It is usually best to use technology when performing the calculations. |
Try It
Find the standard deviation for the data from the previous example
| Class | Frequency, f |
|---|---|
| 0–2 | 1 |
| 3–5 | 6 |
| 6–8 | 10 |
| 9–11 | 7 |
| 12–14 | 0 |
| 15–17 | 2 |
Use TI-Calculators to find the standard deviation:
Input the midpoint values into L1 and the frequencies into L2
Select STAT, CALC, and 1: 1-Var Stats
Select 2nd then 1 then , 2nd then 2 Enter
You will see displayed both a population standard deviation, σx, and the sample standard deviation, sx.
Comparing Values from Different Data Sets
The 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 deviation
- Compare 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, x | Z-Score of data value | |
| Sample | x = [latex]\overline{x}+(z)(s)[/latex] | z = [latex]\frac{x-\overline{x}}{s}[/latex] |
| Population | x = [latex]\mu+(z)(\sigma)[/latex] | z = [latex]\frac{x-\mu}{\sigma}[/latex] |
Example 4
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school.
| Student | GPA | School Mean GPA | School Standard Deviation |
|---|---|---|---|
| John | 2.85 | 3.0 | 0.7 |
| Ali | 77 | 80 | 10 |
Which student had the highest GPA when compared to his school?
Show Answer
For 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 = [latex]\frac{x-μ}{σ}[/latex]
| John | Ali |
| z =[latex]\frac{{2.85 - 3.00}}{{0.7}}[/latex] = −0.21 | z = [latex]\frac{{77 - 80}}{{10}}[/latex]= −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 It
Two 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?
| Swimmer | Time (seconds) | Team Mean Time | Team Standard Deviation |
|---|---|---|---|
| Angie | 26.2 | 27.2 | 0.8 |
| Beth | 27.3 | 30.1 | 1.4 |
Show Answer
For Angie: z = [latex]\frac{26.2-27.2}{0.8}[/latex] = [latex]\frac{-1}{0.8}[/latex] = –1.25
For Beth: z = [latex]\frac{27.3-30.1}{1.4}[/latex] = [latex]\frac{-2.8}{1.4}[/latex] = –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.
References
Data 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 Review
The 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.
- [latex]\displaystyle{s}_{x}=\sqrt{{\frac{{f{(m-\overline{x})}^{2}}}{{n-1}}}}[/latex] 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 [latex]\displaystyle{\sigma}=\sqrt{{\frac{{f{(x-\mu)}^{2}}}{{N}}}}[/latex]
Formula Review
[latex]\displaystyle{s}_{x}=\sqrt{{\frac{{\sum{fm}^{2}}}{{n}} - {x}^{2}}}[/latex]
where [latex]\displaystyle{s}_{x}[/latex]=sample standard deviation, [latex]\displaystyle\overline{x}[/latex] = sample mean
