Question

In Exercises 21–24, use these parameters (based on Data Set 1 “Body Data” in Appendix B):• Men’s heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in.• Women’s heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.Mickey Mouse Disney World requires that people employed as a Mickey Mouse character must have a height between 56 in. and 62 in.

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 Mickey Mouse characters?

b. If the height requirements are changed to exclude the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

Short answer

a. 0.90% of men meet the height requirement. As very few men meet the requirement, it is highly likely that the women are employed for the character.

b. The height requirements are changed to exclude the tallest 50% of the men and the shortest 5% of the men. Then, the new height requirements are between 64.0 in and 68.6 in.

Step-by-step solution

Given information 

The men’s heights are normally distributed with a mean of 68.6 in and a standard deviation of 2.8 in.

The Mickey Mouse character must have a height between 56 in and 62 in.

Describe the random variable

Let X be the height of the men.

Then,

$$\begin{gathered} X~\text{N}\left(\mu ,{\sigma }^2\right) \\ \text{∼N}\left(68.6,{2.8}^2\right) \end{gathered}$$

Compute the z-scores 

a.

The z-score is the standardized score for a specific value. It is computed as follows.

$z=\frac{x-\mu }{\sigma }$

The z-score associated with the heights 56 in and 62 in are as follows.

$$\begin{gathered} z_1=\frac{56-68.6}{2.8} \\ =-4.5 \\ {z}_2=\frac{62-68.6}{2.8} \\ =-2.3571 \end{gathered}$$

Compute the probability

From the standard normal table, find the cumulative probabilities associated with each z-score.

In the standard normal table, the cumulative probability is obtained.

For the z-score -4.5 corresponding to row -4.5 and column 0.00, it is 0.0001.

For the z-score -2.36 corresponding to row -2.3 and column 0.06, it is 0.0091.

Thus,

$$\begin{gathered} P\left(Z<-4.5\right)=0.0001 \\ P\left(Z<-2.36\right)=0.0091 \end{gathered}$$

The probability that the height of a randomly selected man lies between 56 in and 62 in is

$$\begin{gathered} P\left(56<X<62\right)=P\left(-4.5<z<-2.36\right) \\ =P\left(Z<-2.36\right)-P\left(Z<-4.5\right) \\ =0.0091-0.0001 \\ =-0.0090 \end{gathered}$$

Then, the percentage of the men meeting the height requirement is$-0.009\times 100=-0.90\%$.

Therefore, only of the men meet the height requirement, which is quite less. Therefore, it is more likely that most Mickey Mouse characters are women.

Obtain the measure for the z-scores

b.

The new height requirement excludes the tallest 50% and the shortest 5% men.

Let the minimum height of the tallest 50% men be $x_1\left(z_1\right)$, and the maximum height of the shortest 5% men be $x_2\left(z_2\right)$.

Thus,

$$\begin{gathered} P\left(X<x_2\right)=0.05 \\ P\left(Z<z_2\right)=0.05 \\ P\left(X>x_1\right)=0.5 \\ P\left(Z>z_1\right)=0.5 \end{gathered}$$

From the standard normal table, the values of the z-scores are obtained as follows.

The cumulative area of 0.05 corresponds to the row to -1.6 and 0.045, which implies the z-score of -1.645.

The cumulative area of 0.5 corresponds to the row to 0.0 and 0.00, which implies the z-score of 0.

Compute the associated values of the heights

The minimum height for the shortest 5% of the men is computed as follows.

$$\begin{gathered} z_1=\frac{x_1-68.6}{2.8} \\ -1.645=\frac{x_1-68.6}{2.8} \\ {x}_1=63.994\text{in} \end{gathered}$$

The minimum height for the tallest 50% of the men is computed as follows.

$$\begin{gathered} z_2=\frac{x_2-68.6}{2.8} \\ 0=\frac{x_2-68.6}{2.8} \\ {x}_2=68.6\text{in} \end{gathered}$$

Thus, the shortest 5% of the men are shorter than 64.0 in, and the tallest 50% of the men are taller than 68.6 in. The new height requirements are between 64.0 in and 68.6 in.