A survey found that women's heights are normally distributed with mean 62.2 in. and standard deviation 2.9 in. The survey also found that men's heights are normally distributed with mean $69.5 \mathrm{in}$. and standard deviation 3.6 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 64 in. Complete parts (a) and (b) below.
a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park?
The percentage of men who meet the height requirement is $\square \%$. (Round to two decimal places as needed.)
Since most men
the height requirement, it is likely that most of the characters are

Step 1 ：Given that the mean height for men is 69.5 inches and the standard deviation is 3.6 inches, we need to find the percentage of men whose height falls within the range of 57 to 64 inches.

Step 2 ：To do this, we calculate the z-scores for these heights. The z-score is calculated as \((X - \mu) / \sigma\), where X is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Calculating the z-scores for the minimum and maximum heights gives us -3.4722222222222223 and -1.5277777777777777 respectively.

Step 4 ：We then find the area under the normal distribution curve between these z-scores, which gives us the percentage of men who meet the height requirement.

Step 5 ：The percentage of men who meet the height requirement is found to be 6.302577629916504.

Step 6 ：Rounding this to two decimal places, we get \(\boxed{6.30\%}\).

Step 7 ：This suggests that most of the characters at the amusement park are likely not men, as only a small percentage of men meet the height requirement.