# 6.2: The Standard Normal Distribution

## Z-Scores

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.

Definition: Z-Score

If $$X$$ is a normally distributed random variable and $$X \sim N(\mu, \sigma)$$, then the z-score is:

$z = \dfrac{x - \mu}{\sigma} \label{zscore}$

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, $$\mu$$. 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. 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:

\begin{align*} x &= \mu + (z)(\sigma) \\[5pt] &= 5 + (3)(2) = 11 \end{align*}

The z-score is three.

Since the mean for the standard normal distribution is zero and the standard deviation is one, then the transformation in Equation \ref{zscore} produces the distribution $$Z \sim N(0, 1)$$. The value $$x$$ comes from a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$.

A z-score is measured in units of the standard deviation.

Example $$\PageIndex{1}$$

Suppose $$X \sim N(5, 6)$$. This says that $$x$$ is a normally distributed random variable with mean $$\mu = 5$$ and standard deviation $$\sigma = 6$$. Suppose $$x = 17$$. Then (via Equation \ref{zscore}):

$z = \dfrac{x-\mu}{\sigma} = \dfrac{17-5}{6} = 2 \nonumber$

This means that $$x = 17$$ is two standard deviations (2$$\sigma$$) above or to the right of the mean $$\mu = 5$$. The standard deviation is $$\sigma = 6$$.

Notice that: $$5 + (2)(6) = 17$$ (The pattern is $$\mu + z \sigma = x$$)

Now suppose $$x = 1$$. Then:

$z = \dfrac{x-\mu}{\sigma} = \dfrac{1-5}{6} = -0.67 \nonumber$

(rounded to two decimal places)

This means that $$x = 1$$ is $$0.67$$ standard deviations ($$–0.67\sigma$$) below or to the left of the mean $$\mu = 5$$. Notice that: $$5 + (–0.67)(6)$$ is approximately equal to one (This has the pattern $$\mu + (–0.67)\sigma = 1$$)

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

Exercise $$\PageIndex{1}$$

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

$$z = \dfrac{1-12}{3} \approx -3.67$$

Example $$\PageIndex{2}$$

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. $$X \sim N(5, 2)$$. Fill in the blanks.

1. Suppose a person lost ten pounds in a month. The $$z$$-score when $$x = 10$$ pounds is $$x = 2.5$$ (verify). This $$z$$-score tells you that $$x = 10$$ is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. 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.

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

b. Suppose the random variables $$X$$ and $$Y$$ have the following normal distributions: $$X \sim N(5, 6)$$ and $$Y \sim N(2, 1)$$. If $$x = 17$$, then $$z = 2$$. (This was previously shown.) If $$y = 4$$, what is $$z$$?

$z = \dfrac{y-\mu}{\sigma} = \dfrac{4-2}{1} = 2 \nonumber$

where $$\mu = 2$$ and $$\sigma = 1$$.

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 $$X \sim N(5, 6)$$ represents weight gains for one group of people who are trying to gain weight in a six week period and $$Y \sim N(2, 1)$$ 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.

Exercise $$\PageIndex{2}$$

Fill in the blanks.

Jerome averages 16 points a game with a standard deviation of four points. $$X \sim N(16, 4)$$. 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 $$\mu$$ and standard deviation $$\sigma$$, then the Empirical Rule says the following:

• About 68% of the $$x$$ values lie between –1$$\sigma$$ and +1$$\sigma$$ of the mean $$\mu$$ (within one standard deviation of the mean).
• About 95% of the $$x$$ values lie between –2$$\sigma$$ and +2$$\sigma$$ of the mean $$\mu$$ (within two standard deviations of the mean).
• About 99.7% of the $$x$$ values lie between –3$$\sigma$$ and +3$$\sigma$$ of the mean $$\mu$$ (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$$\sigma$$ and –1$$\sigma$$ are +1 and –1, respectively.
• The $$z$$-scores for +2$$\sigma$$ and –2$$\sigma$$ are +2 and –2, respectively.
• The $$z$$-scores for +3$$\sigma$$ and –3$$\sigma$$ are +3 and –3 respectively.

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

Example $$\PageIndex{3}$$

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 $$X \sim N(170, 6.28)$$.

1. 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?).
2. 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.

1. –0.32, 0.32, left, 170
2. 177.98, 1.27, right

Exercise $$\PageIndex{3}$$

Use the information in Example $$\PageIndex{3}$$ to answer the following questions.

1. 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?).
2. 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 $$z = \dfrac{x-\mu}{\sigma}$$ for $$z$$. $$x = \mu+ (z)(\sigma)$$

$$z = \dfrac{176-170}{6.28}$$, This z-score tells you that $$x = 176$$ cm is 0.96 standard deviations to the right of the mean 170 cm.

Solve the equation $$z = \dfrac{x-\mu}{\sigma}$$ for $$z$$. $$x = \mu+ (z)(\sigma)$$

$$X = 157.44$$ cm, The $$z$$-score($$z = –2$$) tells you that the male’s height is two standard deviations to the left of the mean.

Example $$\PageIndex{4}$$

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 $$Y \sim N(172.36, 6.34)$$.

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 $$X \sim N(170, 6.28)$$.

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?

• The $$z$$-score (Equation \ref{zscore}) 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.

Exercise $$\PageIndex{4}$$

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean $$\mu = 496$$ and a standard deviation $$\sigma = 114$$. Let $$X =$$ a SAT exam verbal section score in 2012. Then $$X \sim N(496, 114)$$.

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 (Equation \ref{zscore}) for $$x_{1} = 325$$ is $$z_{1} = –1.14$$.

The z-score (Equation \ref{zscore}) 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 $$\PageIndex{5}$$

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 –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 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 –2σ = (–2)(6) = –12 and 2σ = (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 –3σ = (–3)(6) = –18 and 3σ = (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.

Exercise $$\PageIndex{5}$$

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

between 20 and 30.

Example $$\PageIndex{6}$$

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 $$Y \sim N(172.36, 6.34)$$.

1. About 68% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________, respectively.
2. About 95% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________ respectively.
3. About 99.7% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________, respectively.

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

Exercise $$\PageIndex{6}$$

The scores on a college entrance exam have an approximate normal distribution with mean, $$\mu = 52$$ points and a standard deviation, $$\sigma = 11$$ points.

1. About 68% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________, respectively.
2. About 95% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________, respectively.
3. About 99.7% of the $$y$$ values lie between what two values? These values are ________________. The $$z$$-scores are ________________, respectively.

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.

## Summary

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

## Formula Review

$$Z \sim N(0, 1)$$

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

mean = 0; standard deviation = 1

To find the $$K$$th percentile of $$X$$ when the $$z$$-scores is known:

$$k = \mu + (z)\sigma$$

$$z$$-score: $$z = \dfrac{x-\mu}{\sigma}$$

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

$$Z \sim N(0, 1)$$

## Glossary

Standard Normal Distribution
a continuous random variable (RV) $$X \sim N(0, 1)$$; when $$X$$ follows the standard normal distribution, it is often noted as $$Z \sim N(0, 1)\. \(z$$-score
the linear transformation of the form $$z = \dfrac{x-\mu}{\sigma}$$; if this transformation is applied to any normal distribution $$X \sim N(\mu, \sigma$$ the result is the standard normal distribution $$Z \sim N(0,1)$$. If this transformation is applied to any specific value $$x$$ of the RV with mean $$\mu$$ and standard deviation $$\sigma$$, the result is called the $$z$$-score of $$x$$. The $$z$$-score allows us to compare data that are normally distributed but scaled differently.

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