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* = *Î¼* + (*z*)(*Ïƒ*) = 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* = $\frac{x\xe2\u02c6\u2019\mathrm{\xce\xbc}}{\mathrm{\xcf\u0192}}$ produces the distribution *Z* ~ *N*(0, 1). The value *x* in the given equation comes from a normal distribution with mean *Î¼* and standard deviation *Ïƒ*.

*Z*-Scores

If *X* is a normally distributed random variable and *X* ~ *N(Î¼, Ïƒ)*, then the *z*-score is:

**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, Î¼.** 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 *Î¼* = 5 and standard deviation *Ïƒ* = 6. Suppose *x* = 17. Then:

This means that *x* = 17 is **two standard deviations** (2*Ïƒ*) above or to the right of the mean *Î¼* = 5.

Notice that: 5 + (2)(6) = 17 (The pattern is *Î¼* + *zÏƒ* = *x*)

Now suppose *x* = 1. Then: *z* = $\frac{x\xe2\u20ac\u201c\mathrm{\xce\xbc}}{\mathrm{\xcf\u0192}}$ = $\frac{1\xe2\u20ac\u201c5}{6}$ = â€“0.67 (rounded to two decimal places)

**This means that x = 1 is 0.67 standard deviations (â€“0.67Ïƒ) below or to the left of the mean Î¼ = 5. Notice that:** 5 + (â€“0.67)(6) is approximately equal to one (This has the pattern

*Î¼*+ (â€“0.67)Ïƒ = 1)

Summarizing, when *z* is positive, *x* is above or to the right of *Î¼* and when *z* is negative, *x* is to the left of or below *Î¼*. Or, when *z* is positive, *x* is greater than *Î¼*, and when *z* is negative *x* is less than *Î¼*.

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 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.

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?).

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

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*?

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**.

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*â€“score when *x* = 10 is â€“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 *Âµ* and standard deviation *Ïƒ*, then the Empirical Rule states the following:

- About 68% of the
*x*values lie between â€“1*Ïƒ*and +1*Ïƒ*of the mean*Âµ*(within one standard deviation of the mean). - About 95% of the
*x*values lie between â€“2*Ïƒ*and +2*Ïƒ*of the mean*Âµ*(within two standard deviations of the mean). - About 99.7% of the
*x*values lie between â€“3*Ïƒ*and +3*Ïƒ*of the mean*Âµ*(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*Ïƒ*and â€“1*Ïƒ*are +1 and â€“1, respectively. - The
*z*-scores for +2*Ïƒ*and â€“2*Ïƒ*are +2 and â€“2, respectively. - The
*z*-scores for +3*Ïƒ*and â€“3*Ïƒ*are +3 and â€“3 respectively.

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

### 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).

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

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*= â€“2. What is the maleâ€™s height? The*z*-score (*z*= â€“2) tells you that the maleâ€™s height is ________ standard deviations to the __________ (right or left) of the mean.

### Example 6.4

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 as they compare to their respective means and standard deviations?

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean *Âµ* = 496 and a standard deviation *Ïƒ* = 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 â€“1*Ïƒ*= (â€“1)(6) = â€“6 and 1*Ïƒ*= (1)(6) = 6 of the mean 50. The values 50 â€“ 6 = 44 and 50 + 6 = 56 are within one standard deviation from the mean 50. The*z*-scores are â€“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 â€“2*Ïƒ*= (â€“2)(6) = â€“12 and 2*Ïƒ*= (2)(6) = 12. The values 50 â€“ 12 = 38 and 50 + 12 = 62 are within two standard deviations from the mean 50. The*z*-scores are â€“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 â€“3*Ïƒ*= (â€“3)(6) = â€“18 and 3*Ïƒ*= (3)(6) = 18 from the mean 50. The values 50 â€“ 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The*z*-scores are â€“3 and +3 for 32 and 68, respectively.

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

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.

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