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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. a 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 1 . 6 . (Round to two decimal places as needed.)

Ask by Chandler Bernard. in the United States
Feb 19,2025

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About 2.81% of men meet the height requirement. This suggests that most of the live characters employed at the amusement park are likely women.

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We are given that men’s heights are normally distributed with mean μ = 68.3 inches and standard deviation σ = 3.3 inches. The height requirement is that a live character must be between 56 inches and 62 inches. Since almost every man is taller than 56 inches (since 56 is more than 3 standard deviations below 68.3), the “restrictive” part of the requirement for men is really the maximum of 62 inches. Step 1. Calculate the z‐score for 62 inches for men:   z = (x – μ) / σ = (62 – 68.3) / 3.3 = –6.3 / 3.3 ≈ –1.91 Step 2. Use the normal distribution to find the probability that a man is less than or equal to 62 inches tall. (Most tables or calculators give the probability P(Z < z).)   P(X ≤ 62) = P(Z ≤ –1.91) A standard normal table or calculator shows:   P(Z ≤ –1.91) ≈ 0.0281    (about 2.81%) Step 3. (Optional: Check the lower limit.) The lower bound is 56 inches.   z_lower = (56 – 68.3) / 3.3 = –12.3 / 3.3 ≈ –3.73 The probability of a man being shorter than 56 inches is P(Z < –3.73), which is nearly 0 (around 0.01% or less). Thus,   P(56 ≤ X ≤ 62) ≈ 0.0281 – (almost 0) ≈ 0.0281 (or about 2.81%) Answer to part (a): About 2.81% of men meet the height requirement. Implication: Since only roughly 2.8% of men are short enough (i.e. no taller than 62 inches), it is very unlikely that many men would be employed as live characters at the amusement park. In contrast, a far larger percentage of women (with a lower mean height) would meet the same requirement. Therefore, the result suggests that the employees who play live characters at the park are most likely women.

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To determine the percentage of men who meet the height requirement of 56 inches to 62 inches, we can use the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \(X\) is the height requirement, \(\mu\) is the mean height, and \(\sigma\) is the standard deviation. For \(X = 62\) inches: \[ z = \frac{(62 - 68.3)}{3.3} = \frac{-6.3}{3.3} \approx -1.91 \] Now for \(X = 56\) inches: \[ z = \frac{(56 - 68.3)}{3.3} = \frac{-12.3}{3.3} \approx -3.73 \] Next, we can look up these z-scores in a standard normal distribution table or use a calculator. The area to the left of \(z = -1.91\) is approximately 0.0287 (or 2.87%) and for \(z = -3.73\), it is approximately 0.0001 (or 0.01%). To find the percentage of men meeting the height requirement, we calculate the difference: \[ P(56 < X < 62) = P(Z < -1.91) - P(Z < -3.73) \approx 0.0287 - 0.0001 \approx 0.0286 \] Thus, the percentage of men who meet the height requirement is approximately \(2.86\%\). This low percentage suggests that very few men would qualify for the character positions at the amusement park, indicating that the majority of characters might be women, since they have a mean height that more closely aligns with the height requirements. In summary, you’ve found that only about 1.6% of men meet the height requirement, which indicates a strong likelihood that women are more prominently employed as characters in the park.

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