Question

The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.65 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 2.54 inches.

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

z =


b) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?

z =


c) Who is relatively taller?

The 5 foot 11 inch American woman
The 6 foot 3 inch American man

Answer 1

To calculate the z-scores for the heights provided, we can use the z-score formula:

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

where:
- \(X\) is the value for which we are calculating the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.

### a) For the man who is 6 feet 3 inches tall

First, convert 6 feet 3 inches to inches:

\[
6 \text{ feet} = 6 \times 12 = 72 \text{ inches}
\]
\[
6 \text{ feet 3 inches} = 72 + 3 = 75 \text{ inches}
\]

Now, using the mean (\(\mu\) = 69.7 inches) and the standard deviation (\(\sigma\) = 2.65 inches):

\[
z = \frac{(75 - 69.7)}{2.65} = \frac{5.3}{2.65} \approx 2.00
\]

So, the man's z-score is:

\[
z = 2.00
\]

### b) For the woman who is 5 feet 11 inches tall

Convert 5 feet 11 inches to inches:

\[
5 \text{ feet} = 5 \times 12 = 60 \text{ inches}
\]
\[
5 \text{ feet 11 inches} = 60 + 11 = 71 \text{ inches}
\]

Using the mean (\(\mu\) = 64.7 inches) and the standard deviation (\(\sigma\) = 2.54 inches):

\[
z = \frac{(71 - 64.7)}{2.54} = \frac{6.3}{2.54} \approx 2.48
\]

So, the woman's z-score is:

\[
z = 2.48
\]

### c) Who is relatively taller?

To determine who is relatively taller, we compare the z-scores:

- The man's z-score is 2.00
- The woman's z-score is 2.48

The woman has a higher z-score, indicating that she is relatively taller when compared to her gender's height distribution.

Thus, the answer is:

**The 5 foot 11 inch American woman**.