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. Executive Jet Doorway the Gulfstream 100 is an executive jet that seats six, and it has a doorway height of 51.6 in.

a. What percentage of adult men can fit through the door without bending?

b. Does the door design with a height of 51.6 in. appear to be adequate? Why didn’t the engineers design a larger door?

c. What doorway height would allow 40% of men to fit without bending?

The height requirements Men’s heights are normally distributed with mean 68.6 in., and standard deviation 2.8 in.

Doorway height is 51.6 in.

Let X be the random variable for height of men.

Then,

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

a.

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

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

Z-score associated to height 51.6 in is,

$$\begin{gathered} z=\frac{51.6-68.6}{2.8} \\ =-6.0714 \end{gathered}$$

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

In standard normal table, the cumulative probability is obtained for z-score -6.07 corresponding to row -3.5 and less as 0.0001.

Thus,

$P\left(Z<-6.07\right)=0.0001$

The percentage of adult men can fit through the door without bending is $0.0001\times 100=0.01\%$.

b.

The door design fits very less men and hence does not seem adequate to fit most men. The engineers may have not designed a larger door due only 6 seats in the jet and to maintain the efficiency of the built.

c.

Let x be the maximum height of men for shorted 40%, and z be the corresponding z-score.

Then,

$$\begin{gathered} P\left(X<x\right)=0.40 \\ P\left(Z<z\right)=0.40 \end{gathered}$$

From the standard normal table, the cumulative probability of 0.40 corresponds to row -0.2 and column 0.05, which implies the z-score of -0.25.

Thus, the required height is,

$$\begin{gathered} -0.25=\frac{x-68.6}{2.8} \\ x=67.89\text{in} \end{gathered}$$

Thus the doorway height of 68.9 in would fit 40% of men without bending.