Question

the heights of adult men in america are normally distributed, with a mean of 69.2 inches and a standard deviation of 2.66 inches. the heights of adult women in america are also normally distributed, but with a mean of 64.1 inches and a standard deviation of 2.52 inches.

a. if a man is 6 feet 3 inches tall, what is his z-score (to 4 decimal places)?
$z = \square$

b. if a woman is 5 feet 11 inches tall, what is her z-score (to 4 decimal places)?
$z = \square$

c. who is relatively taller?
\\(\bigcirc\\) the 6 foot 3 inch american man
\\(\bigcirc\\) the 5 foot 11 inch american woman

Explanation:

Part a

Step1: Convert height to inches

6 feet 3 inches = \(6\times12 + 3 = 75\) inches.

Step2: Use z - score formula \(z=\frac{x-\mu}{\sigma}\)

Here, \(x = 75\), \(\mu = 69.2\), \(\sigma = 2.66\).
\(z=\frac{75 - 69.2}{2.66}=\frac{5.8}{2.66}\approx2.1804\)

Step1: Convert height to inches

5 feet 11 inches = \(5\times12+11 = 71\) inches.

Step2: Use z - score formula \(z=\frac{x - \mu}{\sigma}\)

Here, \(x = 71\), \(\mu = 64.1\), \(\sigma = 2.52\).
\(z=\frac{71 - 64.1}{2.52}=\frac{6.9}{2.52}\approx2.7381\)

To determine who is relatively taller, we compare the z - scores. The z - score of the man is approximately \(2.1804\) and the z - score of the woman is approximately \(2.7381\). A higher z - score means the height is more extreme (taller relative to the respective population). Since \(2.7381>2.1804\), the woman's height is more extreme relative to her population.

Answer:

\(2.1804\)

Part b