Questions: Use z scores to compare the given values. The tallest living man at one time had a height of 255 cm. The shortest living man at that time had a height of 110.6 cm. Heights of men at that time had a mean of 174.08 cm and a standard deviation of 7.77 cm. Which of these two men had the height that was more extreme? Since the z score for the tallest man is z= find the z score for the shortest man is z=, the man had the height that was more extreme. (Round to two decimal places)

Solution Steps

To find the Z-score for the tallest man, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 255 \) cm (height of the tallest man)
• \( \mu = 174.08 \) cm (mean height)
• \( \sigma = 7.77 \) cm (standard deviation)

Substituting the values:

\[ z = \frac{255 - 174.08}{7.77} = \frac{80.92}{7.77} \approx 10.41 \]

Thus, the Z-score for the tallest man is \( z \approx 10.41 \).

Next, we calculate the Z-score for the shortest man using the same formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 110.6 \) cm (height of the shortest man)

Substituting the values:

\[ z = \frac{110.6 - 174.08}{7.77} = \frac{-63.48}{7.77} \approx -8.17 \]

Thus, the Z-score for the shortest man is \( z \approx -8.17 \).

To determine which man had the height that was more extreme, we compare the absolute values of the Z-scores:

• For the tallest man: \( |z| \approx 10.41 \)
• For the shortest man: \( |z| \approx 8.17 \)

Since \( 10.41 > 8.17 \), the tallest man had the height that was more extreme.

Final Answer