Questions: A survey found that women's heights are normally distributed with mean 63.5 in. and standard deviation 3.5 in. The survey also found that men's heights are normally distributed with mean 67.5 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in. and a maximum of 63 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.)

A survey found that women's heights are normally distributed with mean 63.5 in. and standard deviation 3.5 in. The survey also found that men's heights are normally distributed with mean 67.5 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in. and a maximum of 63 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.)

Transcript text: A survey found that women's heights are normally distributed with mean 63.5 in. and standard deviation 3.5 in. The survey also found that men's heights are normally distributed with mean 67.5 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in. and a maximum of 63 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.)

Solution

Solution Steps

Step 1: Define the Problem

We need to find the percentage of men who meet the height requirement of being between $56$ in. and $63$ in. The heights of men are normally distributed with a mean ($\mu$) of $67.5$ in. and a standard deviation ($\sigma$) of $3.1$ in.

Step 2: Calculate Z-scores

To find the probability that a man meets the height requirement, we first calculate the Z-scores for the lower and upper bounds of the height requirement.

The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \]

For the lower bound ($X = 56$): \[ Z_{start} = \frac{56 - 67.5}{3.1} \approx -3.7097 \]

For the upper bound ($X = 63$): \[ Z_{end} = \frac{63 - 67.5}{3.1} \approx -1.4516 \]

Step 3: Calculate the Probability

Using the Z-scores, we can find the probability that a man meets the height requirement: \[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(-1.4516) - \Phi(-3.7097) \approx 0.0732 \]

Step 4: Convert Probability to Percentage

To express the probability as a percentage, we multiply by $100$: \[ \text{Percentage} = P \times 100 \approx 0.0732 \times 100 = 7.32\% \]