Question

the heights of adult men in america are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.69 inches.
the heights of adult women in america are also normally distributed, but with a mean of 64.7 inches and a standard deviation of 2.59 inches.

a) if a man is 6 feet 3 inches tall, what is his z-score, rounded to two decimal places? ( z = )

b) what percentage (%) of men are shorter than 6 feet 3 inches? round to nearest tenth of a percent.
%

c) if a woman is 5 feet 11 inches tall, what is her z-score, rounded to two decimal places? ( z = )

d) what percentage (%) of women are taller than 5 feet 11 inches? round to nearest tenth of a percent.
%

Explanation:

Step1: Convert man's height to inches

6 feet 3 inches = \(6 \times 12 + 3 = 75\) inches

Step2: Calculate z-score for man

\(z = \frac{X - \mu}{\sigma} = \frac{75 - 69.3}{2.69} \approx 2.12\)

Step3: Find percentage of men shorter than 75 inches

Use z=2.12: cumulative probability ≈ 0.9830 → 98.3%

Step4: Convert woman's height to inches

5 feet 11 inches = \(5 \times 12 + 11 = 71\) inches

Step5: Calculate z-score for woman

\(z = \frac{71 - 64.7}{2.59} \approx 2.43\)

Step6: Find percentage of women taller than 71 inches

Use z=2.43: cumulative probability ≈ 0.9925 → 1 - 0.9925 = 0.0075 → 0.8%

Answer:

a) 2.12
b) 98.3
c) 2.43
d) 0.8