Question

A survey found that​ women's heights are normally distributed with mean

62.7

in. and standard deviation

3.3

in. The survey also found that​ men's heights are normally distributed with mean

68.2

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

64

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 ___​%.

Answer 1

Answer 2

Part (a): Percentage of Men Meeting Height Requirements

Given:

• Men's heights: Normally distributed with

• Mean (μ) = 68.2 in

• Standard deviation (σ) = 3.3 in

• Height requirement: Between 55 in and 64 in.

• 
  

Step 1: Calculate Z-scores
Convert the height limits to Z-scores using:

$$Z=\frac{X-\mu }{\sigma }$$

• Lower limit (55 in):

  $$Z_{\text{lower}}=\frac{55-68.2}{3.3}=\frac{-13.2}{3.3}=-4.00$$

• Upper limit (64 in):

  $$Z_{\text{upper}}=\frac{64-68.2}{3.3}=\frac{-4.2}{3.3}\approx -1.27$$

• 
  

Step 2: Find Probabilities Using Z-Table

• $P(Z<-4.00)\approx 0$ (extremely rare, ≈ 0% chance).

• $P(Z<-1.27)\approx 0.1020$ (10.20%).

• 
  

Step 3: Calculate Percentage Meeting Requirement

$$P(-4.00<Z<-1.27)=P(Z<-1.27)-P(Z<-4.00)=10.20\mathrm{\%}-0\mathrm{\%}=10.20\mathrm{\%}$$

Interpretation:
Only 10.2% of men meet the height requirement, suggesting that most employed characters are likely women, as their distribution (mean 62.7 in) aligns better with the 55–64 in range.

Answer:
The percentage of men who meet the height requirement is $\boxed{{\displaystyle 10.2\mathrm{\%}}}$