Question

use z - scores to compare the given values. the tallest living man at one time had a height of 249 cm. the shortest living man at that time had a height of 55.3 cm. heights of men at that time had a mean of 172.98 cm and a standard deviation of 7.30 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 $x = 249$, $\mu=172.98$, $\sigma = 7.30$. Then $z_1=\frac{249 - 172.98}{7.30}=\frac{76.02}{7.30}\approx10.41$.

Step3: Calculate z - score for the shortest man

Given $x = 55.3$, $\mu=172.98$, $\sigma = 7.30$. Then $z_2=\frac{55.3 - 172.98}{7.30}=\frac{- 117.68}{7.30}\approx - 16.12$.

Answer:

Since the z - score for the tallest man is $z = 10.41$ and the z - score for the shortest man is $z=-16.12$, the shortest man had the height that was more extreme.