Question

use z scores to compare the given values. the tallest living man at one time had a height of 227 cm. the shortest living man at that time had a height of 95.2 cm. heights of men at that time had a mean of 171.33 cm and a standard deviation of 5.63 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 the z - score formula

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

Step2: Calculate the z - score for the tallest man

Given \(x = 227\), \(\mu=171.33\), \(\sigma = 5.63\).

$$ z=\frac{227 - 171.33}{5.63}=\frac{55.67}{5.63}\approx9.89 $$

Step3: Calculate the z - score for the shortest man

Given \(x = 95.2\), \(\mu = 171.33\), \(\sigma=5.63\).

$$ z=\frac{95.2-171.33}{5.63}=\frac{- 76.13}{5.63}\approx - 13.52 $$

Answer:

Since the z - score for the tallest man is \(z = 9.89\) and the z - score for the shortest man is \(z=-13.52\), the shortest man had the height that was more extreme.