# 7.2 Z-scores

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If $X$ is a normally distributed random variable and $X$ ~ $\mathrm{N\left(\mu , \sigma \right)}$ , then the z-score is:

$z=\frac{x-\mu }{\sigma }$

The z-score tells you how many standard deviations that 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 $0$ .

Suppose $X$ ~ $\mathrm{N\left(5, 6\right)}$ . This says that $X$ is a normally distributed random variable with mean $\mathrm{\mu = 5}$ and standard deviation $\mathrm{\sigma = 6}$ . Suppose $\mathrm{x = 17}$ . Then:

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

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

Notice that:

$5+2\cdot 6=17\phantom{\rule{20pt}{0ex}}\text{(The pattern is}\phantom{\rule{5pt}{0ex}}\mu +z\sigma =x.\text{)}$

Now suppose $\mathrm{x=1}$ . Then:

$z=\frac{x-\mu }{\sigma }=\frac{1-5}{6}=-0.67\phantom{\rule{20pt}{0ex}}\text{(rounded to two decimal places)}$

This means that $\mathrm{x = 1}$ is 0.67 standard deviations $\mathrm{\left(- 0.67\sigma \right)}$ below or to the left of the mean $\mathrm{\mu = 5}$ . Notice that:

$5+\left(-0.67\right)\left(6\right)$ is approximately equal to 1 $\phantom{\rule{20pt}{0ex}}$ (This has the pattern $\mu +\left(-0.67\right)\sigma =1$ )

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 $\mu$ .

Some doctors believe that a person can lose 5 pounds, on the average, in a month by reducing his/her fat intake and by exercising consistently. Suppose weight loss has anormal distribution. Let $X$ = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of 2 pounds. $X$ ~ $\mathrm{N\left(5, 2\right)}$ . Fill in the blanks.

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

This z-score tells you that $\mathrm{x = 10}$ is 2.5 standard deviations to the right of the mean 5 .

Suppose a person gained 3 pounds (a negative weight loss). Then $z$ = __________. This z-score tells you that $\mathrm{x = -3}$ is ________ standard deviations to the __________ (right or left) of the mean.

$z$ = -4 . This z-score tells you that $\mathrm{x = -3}$ is 4 standard deviations to the left of the mean.

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

$z=\frac{y-\mu }{\sigma }=\frac{4-2}{1}=2\phantom{\rule{20pt}{0ex}}\text{where μ=2 and σ=1.}$

The z-score for $\mathrm{y = 4}$ is $\mathrm{z = 2}$ . This means that 4 is $\mathrm{z = 2}$ standard deviations to the right of the mean. Therefore, $\mathrm{x = 17}$ and $\mathrm{y = 4}$ are both 2 (of their ) 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 $X$ ~ $\mathrm{N\left(5, 6\right)}$ represents weight gains for one group of people who are trying to gain weight in a 6 week period and $Y$ ~ $\mathrm{N\left(2, 1\right)}$ measures the same weight gain for a second group of people. A negative weight gain would be a weight loss.Since $\mathrm{x = 17}$ and $\mathrm{y = 4}$ are each 2 standard deviations to the right of their means, they represent the same weight gain relative to their means .

## The empirical rule

If $X$ is a random variable and has a normal distribution with mean $µ$ and standard deviation $\sigma$ then the Empirical Rule says (See the figure below)
• About 68.27% of the $x$ values lie between -1 $\sigma$ and +1 $\sigma$ of the mean $µ$ (within 1 standard deviation of the mean).
• About 95.45% of the $x$ values lie between -2 $\sigma$ and +2 $\sigma$ of the mean $µ$ (within 2 standard deviations of the mean).
• About 99.73% of the $x$ values lie between -3 $\sigma$ and +3 $\sigma$ of the mean $µ$ (within 3 standard deviations of the mean). Notice that almost all the $x$ values lie within 3 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.

Suppose $X$ has a normal distribution with mean 50 and standard deviation 6.

• About 68.27% of the $x$ values lie between -1 $\sigma$ = (-1)(6) = -6 and 1 $\sigma$ = (1)(6) = 6 of the mean 50. The values 50 - 6 = 44 and 50 + 6 = 56 are within 1 standard deviation of the mean 50. The z-scores are -1 and +1 for 44 and 56, respectively.
• About 95.45% of the $x$ values lie between -2 $\sigma$ = (-2)(6) = -12 and 2 $\sigma$ = (2)(6) = 12 of the mean 50. The values 50 - 12 = 38 and 50 + 12 = 62 are within 2 standard deviations of the mean 50. The z-scores are -2 and 2 for 38 and 62, respectively.
• About 99.73% of the $x$ values lie between -3 $\sigma$ = (-3)(6) = -18 and 3 $\sigma$ = (3)(6) = 18 of the mean 50. The values 50 - 18 = 32 and 50 + 18 = 68 are within 3 standard deviations of the mean 50. The z-scores are -3 and +3 for 32 and 68, respectively.

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