A survey found that women's heights are normally distributed with mean 63.6 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 $5.40 \%$. (Round to two decimal places as needed.) Since most men do not meet the height requirement, it is likely that most of the characters are women. b. If the height requirements are changed to exclude only the tallest $50 \%$ of men and the shortest $5 \%$ of men, what are the new height requirements? The new height requirements are a minimum of in. and a maximum of in. (Round to one decimal place as needed.)
Solution
Step 1 ：Given that the mean height of men is 67.3 inches with a standard deviation of 3.3 inches, we need to find the percentage of men who meet the height requirement of between 55 and 62 inches.
Step 2 ：We calculate the z-scores for the lower and upper height limits. 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 ：For the lower limit of 55 inches, the z-score is \((-3.7272727272727266)\).
Step 4 ：For the upper limit of 62 inches, the z-score is \((-1.6060606060606053)\).
Step 5 ：We find the area under the normal distribution curve between these two z-scores using the cumulative distribution function (CDF) of the normal distribution.
Step 6 ：The percentage of men who meet the height requirement is approximately \(5.403351568847136\) percent.
Step 7 ：Final Answer: The percentage of men who meet the height requirement is approximately \(\boxed{5.40\%}\).
