Question

use z scores to compare the given values.
the tallest living man at one time had a height of 262 cm. the shortest living man at that time had a height of 114.1 cm. heights of men at that time had a mean of 174.24 cm and a standard deviation of 8.29 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: Calculate the z - score for the tallest man

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For the tallest man, \(x = 262\), \(\mu=174.24\), \(\sigma = 8.29\)
\(z=\frac{262 - 174.24}{8.29}=\frac{87.76}{8.29}\approx10.59\)

Step2: Calculate the z - score for the shortest man

For the shortest man, \(x = 114.1\), \(\mu = 174.24\), \(\sigma=8.29\)
\(z=\frac{114.1-174.24}{8.29}=\frac{- 60.14}{8.29}\approx - 7.25\)

Answer:

Since the z - score for the tallest man is \(z = 10.59\) and the z - score for the shortest man is \(z=-7.25\), the tallest man had the height that was more extreme.