Module 6: Normal Distribution

# The Standard Normal Distribution

Barbara Illowsky & OpenStax et al.

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, and 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.

### Example

Suppose *X* ~ *N(5, 6)*. This says that *x* is a normally distributed random variable with mean *μ* = 5 and standard deviation *σ* = 6. Suppose *x* = 17. Then:

[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** (2*σ*) above or to the right of the mean *μ* = 5. The standard deviation is *σ* = 6.

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

Now suppose *x* = 1. Then: [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 one (This has the pattern

*μ*+ (–0.67)σ = 1)

Summarizing, when *z* is positive, *x* is above or to the right of *μ* and when *z*is negative, *x* is to the left of or below *μ*. Or, when *z* is positive, *x* is greater than *μ*, and when *z* is negative *x* is less than *μ*.

### try it

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

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

### Example

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

Solution:

- This
*z*-score tells you that*x*= 10 is**2.5**standard deviations to the**right**of the mean**five**. *z*= –4. This*z*-score tells you that*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 *z* = 2. This means that four is *z* = 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?).

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 –1*σ*and +1*σ*of the mean*µ*(within one standard deviation of the mean). - About 95% of the
*x*values lie between –2*σ*and +2*σ*of the mean*µ*(within two standard deviations of the mean). - About 99.7% of the
*x*values lie between –3*σ*and +3*σ*of the mean*µ*(within three standard deviations of the mean). Notice that almost all the*x*values lie within three standard deviations of the mean. - The
*z*-scores for +1*σ*and –1*σ*are +1 and –1, respectively. - The
*z*-scores for +2*σ*and –2*σ*are +2 and –2, respectively. - The
*z*-scores for +3*σ*and –3*σ*are +3 and –3 respectively.

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

### Example

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 _____ (What is the mean?).

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.

Solution:

a. –0.32, 0.32, left, 170

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.

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]displaystylefrac{{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

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?

Solution:

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?

The *z*-score for *x*1 = 325 is *z*1 = –1.14.

The *z*-score for *x*2 = 366.21 is *z*2 = –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

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

- About 68% of the
*x*values lie between –1*σ*= (–1)(6) = –6 and 1*σ*= (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 –2*σ*= (–2)(6) = –12 and 2*σ*= (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 –3*σ*= (–3)(6) = –18 and 3*σ*= (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?

Solution:

Between 20 and 30.

### Example

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. Then *Y* ~ *N*(172.36, 6.34).

- 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 .7% of the values lie between 153.34 and 191.38. The*z*-scores are –3 and 3.- About 68% of the
*y*values lie between what two values? These values are ________________. The*z*-scores are ________________, respectively.

- About 68% of the

### 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.

.7% of the values lie between 153.34 and 191.38. The *z*-scores are –3 and 3.

- 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.

Solution:

- About 68% of the values lie between the values 41 and 63. The
*z*-scores are –1 and 1, respectively. - About 95% of the values lie between the values 30 and 74. The
*z*-scores are –2 and 2, respectively. - About 99.7% of the values lie between the values 19 and 85. The
*z*-scores are –3 and 3, respectively.

## 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 find the *K*th percentile of *X* when the *z*-scores is known:

*k* = *μ* + (*z*)*σ*

*z*-score:[latex]displaystyle{z}=frac{{x - mu}}{{sigma}}[/latex]

*Z* = the random variable for *z*-scores

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