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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 $渭=68.6$.

The population standard deviation is equal to $蟽=2.8$.

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:

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

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:

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

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

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

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:

$P\left(Z猢�z\right)=0.99$

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, $P\left(z猢�2.33\right)=0.99$.

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

$$\begin{gathered} x=渭+z蟽 \\ =68.6+2.33(2.8) \\ =75.124 \\ 鈮�75.1 \end{gathered}$$

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