The standard normal distribution is a normal distribution of **standardized values called** *z*-scores. **A z-score is measured in units of the standard deviation.** 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:

*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 *z* = $\frac{x-\mu}{\sigma}$ produces the distribution *Z* ~ *N*(0, 1). The value *x* in the given equation comes from a normal distribution with mean *μ* and standard deviation *σ*.

*Z*-Scores

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

$$z=\frac{x\text{}\u2013\text{}\mu}{\sigma}$$

**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 6.1

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

$$z=\frac{x\u2013\mu}{\sigma}=\frac{17\u20135}{6}=2$$

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

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

Now suppose *x* = 1. Then: *z* = $\frac{x\u2013\mu}{\sigma}$ = $\frac{1\u20135}{6}$ = –0.67 (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 6.1

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

## Example 6.2

Some doctors believe that a person can lose five pounds, on the average, in a month by reducing their 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.

### Problem

a. 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?).

### Solution

a. This *z*-score tells you that *x* = 10 is **2.5** standard deviations to the **right** of the mean **five**.

### Problem

b. 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

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

### Problem

c. 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*?

### Solution

c. *z* = $\frac{y-\mu}{\sigma}$ = $\frac{4-2}{1}$ = 2 where *µ* = 2 and *σ* = 1.

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 6.2

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?).

The Empirical RuleIf *X* is a random variable and has a normal distribution with mean *µ* and standard deviation *σ*, then the Empirical Rule states 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.

Figure 6.3

## Example 6.3

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).

### Problem

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 cm, 1.27, right

## Try It 6.3

Use the information in Example 6.3 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.

## Example 6.4

### Problem

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 as they compare to their respective means and standard deviations?

### 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 6.4

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 as they compare to their respective means and standard deviations?

## Example 6.5

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

- About 68% of the
*x*values lie within one standard deviation of the mean. Therefore, 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 from the mean 50. The*z*-scores are –1 and +1 for 44 and 56, respectively. - About 95% of the
*x*values lie within two standard deviations of the mean. Therefore, 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 from the mean 50. The*z*-scores are –2 and +2 for 38 and 62, respectively. - About 99.7% of the
*x*values lie within three standard deviations of the mean. Therefore, about 99.7% of the*x*values lie between –3*σ*= (–3)(6) = –18 and 3*σ*= (3)(6) = 18 from 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 6.5

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

## Example 6.6

### Problem

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

### Solution

- About 68% of the values lie between 166.02 cm and 178.7 cm. The
*z*-scores are –1 and 1. - About 95% of the values lie between 159.68 cm and 185.04 cm. The
*z*-scores are –2 and 2. - About 99.7% of the values lie between 153.34 cm and 191.38 cm. The
*z*-scores are –3 and 3.

## Try It 6.6

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

- 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.