A survey found that women's heights are normally distributed with mean 62.7 in. and standard deviation 3.7 in. The survey also found that men's heights are normally distributed with mean 68.3 in deviation 3.3 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 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 \( 2.81 \% \) (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 \( \square \) in. and a maximum of \( \square \) in (Round to one decimal place as needed)

We begin by recalling that men's heights follow a normal distribution with mean μ = 68.3 inches and standard deviation σ = 3.3 inches. ────────────────────────────── Part (a): The original height requirement is for a height between 56 and 62 inches. To find the percentage of men in that range, we first convert these heights to z‐scores. For x = 56 inches:   z₁ = (56 − 68.3) / 3.3 ≈ (−12.3) / 3.3 ≈ −3.73 For x = 62 inches:   z₂ = (62 − 68.3) / 3.3 ≈ (−6.3) / 3.3 ≈ −1.91 The probability that a man’s height is between 56 and 62 inches is:   P(56 ≤ x ≤ 62) = P(z₁ ≤ z ≤ z₂) = Φ(−1.91) − Φ(−3.73) From the standard normal table (or using technology):   Φ(−1.91) ≈ 0.0281   Φ(−3.73) is very close to 0 (approximately 0.0001). Thus,   P(56 ≤ x ≤ 62) ≈ 0.0281 − 0.0001 = 0.0280, or about 2.81%. This very low percentage indicates that very few men meet the park’s original height requirement. Hence, it is likely that most of the live characters employed (who must meet the height requirement) are women. ────────────────────────────── Part (b): Now, the amusement park intends to change the requirements so that they exclude only the tallest 50% of men and the shortest 5% of men. This means that:  • The new maximum height is set at the 50th percentile (the median) of the men’s distribution.  • The new minimum height is set at the 5th percentile of the men’s distribution. For a normal distribution, the median is the same as the mean. Therefore:  New maximum = 68.3 inches Next, we find the 5th percentile. The z‐score corresponding to the 5th percentile is approximately z = –1.645. Now compute the corresponding height:  New minimum = μ + z·σ   = 68.3 + (–1.645)(3.3)   = 68.3 − 5.4285   ≈ 62.9 inches (rounded to one decimal place) ────────────────────────────── Final Answers: a. Approximately 2.81% of men meet the original height requirement, suggesting that most of the characters (who must meet this height criterion) are likely to be women. b. The new height requirements are a minimum of 62.9 inches and a maximum of 68.3 inches.