A survey found that women's heights are normally distributed with mean $63.6 \mathrm{in}$. and standard deviation 3.9 in. The survey also found that men's heights are normally distributed with mean $67.3 \mathrm{in}$. and standard deviation 3.3 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 55 in. and a maximum of 62 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 $\%$. (Round to two decimal places as needed.)

Step 1 ：Given that the mean height for men is 67.3 inches with a standard deviation of 3.3 inches, we need to find the percentage of men whose height falls within the range of 55 to 62 inches.

Step 2 ：We calculate the z-scores for these heights using the formula \(z = \frac{X - \mu}{\sigma}\), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：For the minimum height of 55 inches, the z-score is calculated as \(z_{min} = \frac{55 - 67.3}{3.3} = -3.73\).

Step 4 ：For the maximum height of 62 inches, the z-score is calculated as \(z_{max} = \frac{62 - 67.3}{3.3} = -1.61\).

Step 5 ：We then find the area under the normal distribution curve between these z-scores using the cumulative distribution function (CDF) for a normal distribution. This gives us the percentage of men who meet the height requirement.

Step 6 ：The percentage is calculated to be approximately 5.40%.

Step 7 ：\(\boxed{5.40\%}\) of men meet the height requirement. This suggests that a small percentage of men meet the height requirements to be employed as characters at the amusement park, implying that the majority of the characters are likely to be women.