Men’s heights are normally distributed with mean 69.5 inches and a standard devi

ID: 3180742 • Letter: M

Question

Men’s heights are normally distributed with mean 69.5 inches and a standard deviation of 2.4 inches.

Women’s heights are normally distributed with mean 63.8 inches and a standard deviation of 2.6 inches

The Gulfstream 100 is an executive jet that seats six and it has a doorway height of 51.6 inches.

a. What percentage of adult men can fit through the door without bending?

b. what percentage of adult women can fit through the door without bending?

c. Does the door design with a height of 51.6 inches appear to be adequate? Why didn’t engineers design a larger door?

d. What doorway height would allow 60% of men to fit without bending?

***Can you show the calculator shortcut for this?

The problem is essentially asking what percentage of men&women have heights less than 51.6 inches.

a)z-statistic=(51.6-69.5)/2.4=-7.4583

as you asked for shortcut ill give a function in excel which gives directly the area-> NORMSDIST(Z)

substituting we will get a very low value of 4.3823E-14

so very few men can fit through door approx 0%.

b.)for women z=(51.6-63.8)/2.6=-4.692.

keeping in the same function we get 1.35274E-06

c)NO it is not adequate.May be the airplane is small and they needed of that size.

d)use function INV.NORM(Probabilty,Mean,S.D) in excel

INV.NORM(0.6,69.5,2.4)=70.10803

SO for 60% door height is 70.10803 inches.

1.35274E-06