Question

use z - scores to compare the given values. the tallest living man at one time had a height of 229 cm. the shortest living man at that time had a height of 97.8 cm. heights of men at that time had a mean of 173.91 cm and a standard deviation of 5.43 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 $x = 229$ cm, $\mu=173.91$ cm and $\sigma = 5.43$ cm. Substitute into the formula: $z_1=\frac{229 - 173.91}{5.43}=\frac{55.09}{5.43}\approx10.15$.

Step3: Calculate z - score for the shortest man

Given $x = 97.8$ cm, $\mu = 173.91$ cm and $\sigma=5.43$ cm. Substitute into the formula: $z_2=\frac{97.8 - 173.91}{5.43}=\frac{- 76.11}{5.43}\approx - 14.02$.

Step4: Compare the absolute values of z - scores

The absolute value of $z_1$ is $|z_1| = 10.15$ and the absolute value of $z_2$ is $|z_2|=14.02$. Since $14.02>10.15$, the shortest man has a more extreme height.

Answer:

Since the z - score for the tallest man is $z\approx10.15$ and the z - score for the shortest man is $z\approx - 14.02$, the shortest man had the height that was more extreme.