Disney Monorail The Mark VI monorail used at Disney World has doors with a height of 72 in. Heights of men are normally distributed with a mean of in. and a standard deviation of in. (based on Data Set 1 "Body Data" in Appendix B).

a. What percentage of adult men can fit through the doors without bending? Does the door design with a height of 72 in. appear to be adequate? Explain.

b. What doorway height would allow of adult men to fit without bending?

The population of heights of men is normally distributed with a mean value equal to 68.6 inches and a standard deviation equal to 2.8 inches.

The Mark VI monorail used at Disney World has doors with a height of 72 inches.

Here, the population mean value is equal to $\mu =68.6$.

The population standard deviation is equal to $\sigma =2.8$.

The sample value given is equal to x=72 inches.

The following formula is used to convert a given sample value (x=72) to a z-score:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{72-68.6}{2.8} \\ =1.21 \end{gathered}$$

The required probability value can be computed using the value of z-score.

a.

In order to fit through doors of height equal to 72 inches, the following probability value is computed using the standard normal table:

$$\begin{gathered} P\left(x⩽72\right)=P\left(z⩽1.21\right) \\ =0.8869 \end{gathered}$$

By converting the probability value to a percentage, the following value is obtained:

$$\begin{gathered} \text{Percentage}=0.8869\times 100\% \\ =88.69\% \end{gathered}$$

Therefore, the percentage of men who will fit through doors of height 72 inches is equal to 88.69%.

This implies that approximately 11% of the men cannot fit through doors of height 72 inches which is a significantly large value.

Thus, the height of doors equal to 72 inches appears to be inadequate.

b.

Let X denote the height of men and Z denote the corresponding z-value.

Thus, the value of doorway height that would allow 99% of adult men to fit without bending has the following expression:

$P\left(Z⩽z\right)=0.99$

The corresponding z-score for the left-tailed probability value equal to 0.99 is seen from the table and is approximately equal to 2.33.

Thus, $P\left(z⩽2.33\right)=0.99$.

The value that will allow 99% of adult men to fit without bending is computed below:

$$\begin{gathered} x=\mu +z\sigma \\ =68.6+2.33(2.8) \\ =75.124 \\ \approx 75.1 \end{gathered}$$

Therefore, the height of men that will allow 99% of men to fit through the doors without bending is equal to 75.1 inches.