A survey found that women's heights are normally distributed with mean 63.9 and standard deviation 3.1 The survey also found that men's heights are normally distributed with mean 68.1 and standard deviation 3.4 Consider an executive jet that seats six with a doorway height of 56.1 Complete parts (a) through (c) below.
What percentage of adult men can fit through the door without bending?
The percentage of men who can fit without bending is

enter your response here%.
(Round to two decimal places as needed.)
Part 2
b. Does the door design with a height of 56.1 appear to be adequate? Why didn't the engineers design a larger door?
A.
The door design is inadequate, but because the jet is relatively small and seats only six people, a much higher door would require major changes in the design and cost of the jet, making a larger height not practical.
B.
The door design is inadequate, because every person needs to be able to get into the aircraft without bending. There is no reason why this should not be implemented.
C.
The door design is adequate, because the majority of people will be able to fit without bending. Thus, a larger door is not needed.
D.
The door design is adequate, because although many men will not be able to fit without bending, most women will be able to fit without bending. Thus, a larger door is not needed.
What doorway height would allow 40% of men to fit without bending?
The doorway height that would allow 40% of men to fit without bending is

enter your response here in.
(Round to one decimal place as needed.)

b) A. The door design is inadequate, but because the jet is relatively small and seats only six people, a much higher door would require major changes in the design and cost of the jet, making a larger height not practical.

c) The doorway height that would allow 40% of men to fit without bending is 67.2 inches.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

For men, the mean and the standard deviation are given by:

[tex]\mu = 68.1, \sigma = 3.4[/tex]

The proportion that can fit without bending is the p-value of Z when X = 56.1, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (56.1 - 68.1)/3.4

Z = -3.53.

Z = -3.53 has a p-value of 0.0002.

The percentage of men who can fit without bending is 0.02%.

This is a very small percentage. For women, more than half of them will also have to bend, as their mean height is above 56.1. However, since the capacity of the airplane is small, the correct option for item b is given by:

A. The door design is inadequate, but because the jet is relatively small and seats only six people, a much higher door would require major changes in the design and cost of the jet, making a larger height not practical.

The doorway height that would allow 40% of men to fit without bending is X when Z has a p-value of 0.4, so X when Z = -0.253.

Then:

-0.253 = (X - 68.1)/3.4

X - 68.1 = -0.253 x 3.4

X = 67.2.

The doorway height that would allow 40% of men to fit without bending is 67.2 inches.

More can be learned about the normal distribution at https://brainly.com/question/4079902

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