# 30 6.1 The Standard Normal Distribution

The **standard normal distribution** is a normal distribution of **standardized values called z-scores**.

**A**. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:

*z*-score is measured in units of the standard deviation*x* = *μ* + (*z*)(*σ*) = 5 + (3)(2) = 11

The *z*-score is three.

The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation [latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex] produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean *μ* and standard deviation *σ*.

The following two videos give a description of what it means to have a data set that is “normally” distributed.

*Z*-Scores

If *X* is a normally distributed random variable and *X* ~ *N(μ, σ)*, then the *z*-score is:

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]

**The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ.**

- Values of
*x*that are larger than the mean have positive*z*-scores. - Values of
*x*that are smaller than the mean have negative*z*-scores. - If
*x*equals the mean, then*x*has a*z*-score of zero.

**Note: **

- The
*z*-scores for*µ*+1*σ*and*µ*–1*σ*are +1 and –1, respectively. - The
*z*-scores for*µ*+2*σ*and*µ*–2*σ*are +2 and –2, respectively. - The
*z*-scores for*µ*+3*σ*and*µ*–3*σ*are +3 and –3 respectively.

### Example 1

Suppose *X* ~ *N(5, 6)*.

This says that *x* is a normally distributed random variable with mean *μ* = 5 and standard deviation *σ* = 6.

**1.** Suppose there is a raw data, *x* of 17. What is the z-score?

## Show Answer

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]

[latex]\displaystyle{z}=\frac{{17-5}}{{6}}={2}[/latex]

This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5.

(given that the standard deviation is *σ* = 6. )

*** Notice that: 5 + (2)(6) = 17 (The pattern is *μ* + *zσ* = *x*) ***

**2.** What is the z-score for raw data x = 1?

## Show Answer

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]

[latex]\displaystyle {z}=\frac{{1-5}}{{6}} = -{0.67}[/latex]

(rounded to two decimal places)

This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5.

Notice that: 5 + (–0.67)(6) is approximately equal to 1. (This has the pattern μ + zσ = raw data.) |

Summarizing,

- When
*z*is positive,*x*is greater than or to the right of*μ.*

(when x is greater than μ, the corresponding z-score is positive,. ) *W*hen*z*is negative,*x*is less than or to the left of*μ*.

(When*x*is less than*μ, the corresponding z-score is negative*. )

### Try It

What is the *z*-score of *x*, when *x* = 1 and *X* ~ *N*(12,3)?

## Show Answer

[latex]\displaystyle {z}=\frac{{1-12}}{{3}} = -{3.67}[/latex]

### Example 2

Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let *X* = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. *X* ~ *N*(5, 2). Fill in the blanks.

- Suppose a person lost ten pounds in a month. The
*z*-score when*x*= 10 pounds is*z*= 2.5 (verify). This*z*-score tells you that*x*= 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

## Show Answer

This z-score tells you that data 10 is 2.5 standard deviations to the

**right**of the mean**five**. - Suppose a person gained three pounds (a negative weight loss). Then
*z*= __________. This*z*-score tells you that data*x*= –3 is ________ standard deviations to the __________ (right or left) of the mean.

## Show Answer

z=

**–4**. This z-score tells you that data x = –3 is**4**standard deviations to the**left**of the mean.

Suppose the random variables *X* and *Y* have the following normal distributions: *X* ~ *N*(5, 6) and *Y* ~ *N*(2, 1).

If *x* = 17, then *z* = 2. (This was previously shown.)

If *y* = 4, what is *z*?

[latex]\displaystyle {z}=\frac{{y - \mu}}{{\sigma}} = \frac{{4-2}}{{1}}[/latex].

The *z*-score for *y* = 4 is 2.

This means that raw data 4 is 2 standard deviations to the right of the mean.

Therefore, *x* = 17 and *y* = 4 are both two (**of their own**) standard deviations to the right of their respective means.

**The z-score allows us to compare data that are scaled differently.**

To understand the concept, suppose

*X*~

*N*(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and

*Y*~

*N*(2, 1) measures the same weight gain for a second group of people. (A negative weight gain would be a weight loss. )

Since

*x*= 17 and

*y*= 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.

### Try It

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. *X* ~*N*(16,4). Suppose Jerome scores ten points in a game. The *z*–score when *x* = 10 is –1.5. This score tells you that *x* = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).

[practice-area rows=”1″][/practice-area]

## Show Answer

1.5, left, 16

## The Empirical Rule

If *X* is a random variable and has a **normal distribution** with mean *µ* and standard deviation *σ*,

then the **Empirical Rule** says the following:

- About 68% of the
*x*values lie between the range between*µ – σ and µ + σ*(within one standard deviation of the mean). - About 95% of the
*x*values lie between the range between*µ – 2σ and µ + 2σ*(within two standard deviations of the mean). - About 99.7% of the
*x*values lie between the range between*µ – 3σ and µ + 3σ*(within three standard deviations of the mean).

Notice that almost all the*x-*values/data lie within three standard deviations of the mean.

The empirical rule is also known as the 68-95-99.7 rule.

### Example 3

The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm.

Male heights are known to follow a **normal distribution**.

Let *X* = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then *X* ~ *N*(170, 6.28).

a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010.

The *z*-score when *x* = 168 cm is *z* = _______.

This *z*-score tells you that *x* = 168 is ________ standard deviations to the ________ (right or left) of the mean _____.

## Show Answer

a. –0.32, 0.32, left, 170

b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a *z*-score of *z* = 1.27.

What is the male’s height?

The *z*-score (*z* = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

## Show Answer

b. 177.98, 1.27, right

### Try It

Use the information in Example 3 to answer the following questions.

- Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The
*z*-score when*x*= 176 cm is*z*= _______. This*z*-score tells you that*x*= 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?). - Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a
*z*-score of*z*= –2. What is the male’s height? The*z*-score (*z*= –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

[practice-area rows=”3″][/practice-area]

## Show Answer

Solve the equation [latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex] for x. x = μ + (z)(σ) for x. x = μ + (z)(σ) z<=[latex]\displaystyle\frac{{176-170}}{{0.96}}[/latex], This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm. X = 157.44 cm, The z-score(z = –2) tells you that the male’s height is two standard deviations to the left of the mean.

### Example 4

From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let *Y* = the height of 15 to 18-year-old males from 1984 to 1985. Then *Y* ~ *N*(172.36, 6.34).

The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let *X* = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then *X* ~ *N*(170, 6.28).

- Find the
*z*-scores for*x*= 160.58 cm and*y*= 162.85 cm. - Interpret each
*z*-score. What can you say about*x*= 160.58 cm and*y*= 162.85 cm?

## Show Answer

The z-score for x = 160.58 is z = –1.5.

The z-score for y = 162.85 is z = –1.5.

Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.

### Try It

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean*µ* = 496 and a standard deviation *σ* = 114. Let *X* = a SAT exam verbal section score in 2012. Then *X* ~ *N*(496, 114).

Find the *z*-scores for *x*1 = 325 and *x*2 = 366.21. Interpret each *z*-score. What can you say about *x*1 = 325 and *x*2 = 366.21?

[practice-area rows=”3″][/practice-area]

## Show Answer

The z-score for x1 = 325 is z1 = –1.14. The z-score for x2 = 366.21 is z2 = –1.14. Student 2 scored closer to the mean than Student 1 and, since they both had negative z-scores, Student 2 had the better score.

**Example 5**

Suppose *x* has a normal distribution with mean 50 and standard deviation 6.

Use empirical rule to interpret the data distribution.

## Show Answer

- About 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50.

The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation of the mean 50.

The z-scores are –1 and +1 for 44 and 56, respectively. - About 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12.

The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations of the mean 50.

The z-scores are –2 and +2 for 38 and 62,respectively. - About 99.7% of the x values lie between –3σ = (–3)(6) = –18 and 3σ= (3)(6) = 18 of the mean 50.

The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50.

The z-scores are –3 and +3 for 32 and 68, respectively.

### Try It

Suppose *X* has a normal distribution with mean 25 and standard deviation five. Between what values of *x* do 68% of the values lie?

## Show Answer

Between 20 and 30.

### Example 6

From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm.

Let *Y* = the height of 15 to 18-year-old males in 1984 to 1985, *Y* ~ *N*(172.36, 6.34).

(The variable Y is normally distributed. )

- About 68% of the
*y*values lie between what two values?

These values are ________________. The*z*-scores are ________________, respectively. - About 95% of the
*y*values lie between what two values?

These values are ________________. The*z*-scores are ________________ respectively. - About 99.7% of the
*y*values lie between what two values?

These values are ________________. The*z*-scores are ________________ respectively.

## Show Answer

Since the variable Y is normally distributed, we can apply the Empirical Rule.

*µ*–1*σ = 172.36 – 6.34 = 166.02 ; µ+1σ = 172.36 + 6.34 = 178.7*

About 68% of the*y*values lie between 166.02 cm and 178.7 cm.

The*z*-scores are -1 and 1, respectively.*µ*–2*σ = 172.36 – 2(6.34) = 159.68 ; µ+2σ = 172.36 + 2(6.34) = 185.04*

About 95% of the*y*values lie between 159.68 cm and 185.04 cm.

The*z*-scores are -2 and 2, respectively.*µ*–3*σ = 172.36 – 3(6.34) = 153.34 ; µ+3σ = 172.36 + 3(6.34) = 191.38*

About 99.7% of the*y*values lie between 153.34 cm and 191.38 cm.

The*z*-scores are -3 and 3, respectively.

### Try It

The scores on a college entrance exam have an approximate normal distribution with mean, *µ* = 52 points and a standard deviation, *σ* = 11 points.

[practice-area rows=”4″][/practice-area]

## Show Answer

About 68% of the values lie between the values 41 and 63.

About 95% of the values lie between the values 30 and 74.

About 99.7% of the values lie between the values 19 and 85.

.

## References

“Blood Pressure of Males and Females.” StatCruch, 2013. Available online at http://www.statcrunch.com/5.0/viewreport.php?reportid=11960 (accessed May 14, 2013).

“The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.” London School of Hygiene and Tropical Medicine, 2009. Available online at http://conflict.lshtm.ac.uk/page_125.htm (accessed May 14, 2013).

“2012 College-Bound Seniors Total Group Profile Report.” CollegeBoard, 2012. Available online at http://media.collegeboard.com/digitalServices/pdf/research/TotalGroup-2012.pdf (accessed May 14, 2013).

“Digest of Education Statistics: ACT score average and standard deviations by sex and race/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.” National Center for Education Statistics. Available online at http://nces.ed.gov/programs/digest/d09/tables/dt09_147.asp (accessed May 14, 2013).

Data from the *San Jose Mercury News*.

Data from *The World Almanac and Book of Facts*.

“List of stadiums by capacity.” Wikipedia. Available online at https://en.wikipedia.org/wiki/List_of_stadiums_by_capacity (accessed May 14, 2013).

Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).

# Concept Review

A *z*-score is a standardized value. Its distribution is the standard normal, *Z* ~*N*(0, 1). The mean of the *z*-scores is zero and the standard deviation is one. If *z*is the *z*-score for a value *x* from the normal distribution *N*(*µ*, *σ*) then *z* tells you how many standard deviations *x* is above (greater than) or below (less than) *µ*.

# Formula Review

*Z* ~ *N*(0, 1)

*z* = a standardized value (*z*-score)

mean = 0; standard deviation = 1

**To convert z-score into raw data:** *raw data* = *μ* + (*z*)*σ
*

**To convert data into z-score:** * *[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]