91影视

The height requirements Men鈥檚 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,

$$\begin{gathered} \text{齿鈭糔}\left(渭,{蟽}^2\right) \\ \text{鈭糔}\left(68.6,{2.8}^2\right) \end{gathered}$$

a.

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

$z=\frac{x-渭}{蟽}$

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 $0.0001脳100=0.01\%$.

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.