Question

A survey found that​ women's heights are normally distributed with mean 63.7 . And standard deviation 3.7 . The survey also found that​ men's heights are normally distributed with mean 67.5. And standard deviation 3.7. Consider an executive jet that seats six with a doorway height of 56.4. Complete parts​ (a) through​ (c) below.

Answer 1

Complete question :

A survey found that women's heights are normally distributed with mean 63.6 in and standard deviation 3.7 in. The survey also found that men's heights are normally distributed with mean 69.3 in and standard deviation 3.7 in. Consider an executive jet that seats six with a doorway height of 55.7 in.

1-What percentage of adult men can fit through the door without bending?

2-) Does the door design with a height of 56.35in. appear to be adequate? Why didn't the engineers design a larger door? Choose answer below

A. 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.

B. The door design is adequate, because the majority of people will be able to fit without bending. Thus, a larger door is not needed.

C. 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.

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.

Answer:

A.) 0.135%

B.) C. 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.)

Step-by-step explanation:

Percentage of adult men that can fit the door without bending :

Obtain the Zscore :

Mean height of men, m = 67.5

Standard deviation, s = 3.7

Zscore = (x - m) / s ; where x = 56.4

Zscore = (56.4 - 67.5) / 3.7

Zscore = - 11.1 / 3.7

Zscore = - 3

P(Z < - 3) = 0.0013499 (Z probability calculator)

P(Z < - 3) = 0.0013499 * 100% = 0.13499%