Question

Men's heights are normally distributed with mean 68.9 in and standard deviation of 2.8 in. ​Women's heights are normally distributed with mean 63.6 in and standard deviation of 2.5 in. The standard doorway height is 80 in. a. What percentage of men are too tall to fit through a standard doorway without​ bending, and what percentage of women are too tall to fit through a standard doorway without​ bending? b. If a statistician designs a house so that all of the doorways have heights that are sufficient for all men except the tallest​ 5%, what doorway height would be​ used?

A. The percentage of men who are too tall to fit through a standard door without bending is _____ %. (Round two decimal places)

B. The percentage of women who are too tall to fit through a standard door without bending is ______ %. (ROund two decimals)

C. The statistician would design a house with doorway height ____ in. (round to nearest 10th).

Answer 1

For men

For women

X = standard height = 80

For men :

Zm = 3.96     ( rounded into 2 decimal)

If height is less than 80 then men no need to bend

P( x < 80 )= P( Z < 3.96)

Using Z table for 3.96

P( Z < 3.96)= 0.9999 or 1

100%*1 = 100

The percentage of men who are too tall to fit through a standard door without bending is 100% or 99.99%

For women :

Zw= 6.56

P( x < 80 )= P( Z < 6.56) = 1

Using Z table for 6.56

P( Z < 6.56)= 1

100% *1 = 100 %

The percentage of women who are too tall to fit through a standard door without bending is 100%

Part C.

tallest = 5%

smaller height = 100-5% = 95%

Area = 0.95

Using Z table for 0.95 area score will be

Z = 1.645

The statistician would design a house with doorway height is 73.5 inch