Question

use z scores to compare the given values. the tallest living man at one time had a height of 248 cm. the shortest living man at that time had a height of 53.5 cm. heights of men at that time had a mean of 171.33 cm and a standard deviation of 7.14 cm. which of these two men had the height that was more extreme? since the z score for the tallest man is z = and the z score for the shortest man is z =, the man had the height that was more extreme. (round to two decimal places.)

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Calculate z - score for the tallest man

Given $x = 248$ cm, $\mu=171.33$ cm, and $\sigma = 7.14$ cm.
$z_1=\frac{248 - 171.33}{7.14}=\frac{76.67}{7.14}\approx10.74$

Step3: Calculate z - score for the shortest man

Given $x = 53.5$ cm, $\mu = 171.33$ cm, and $\sigma=7.14$ cm.
$z_2=\frac{53.5 - 171.33}{7.14}=\frac{- 117.83}{7.14}\approx - 16.50$

Answer:

Since the z - score for the tallest man is $z = 10.74$ and the z - score for the shortest man is $z=-16.50$, the shortest man had the height that was more extreme.