Question

# Men's heights are normally distributed with mean 68.8 in and standard deviation of 2.8 in. ​Women's...

Men's heights are normally distributed with mean 68.8 in and standard deviation of 2.8 in. ​Women's heights are normally distributed with mean 64 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 to two decimal places as​ needed.)

The percentage of women who are too tall to fit through a standard door without bending is __​%. ​(Round to two decimal places as​ needed.)

b. The statistician would design a house with doorway height __ in. ​(Round to the nearest tenth as​ needed.)

(a)

For men:

The z-score for X= 80 is

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

P(X > 80) = P(z > 4) =1 -P(z <= 4) =  0.0000

For women:

The z-score for X= 80 is

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

P(X > 80) = P(z > 6.4) =1 -P(z <= 6.4) =  0.0000

(b)

Here we need z-score that has 0.05 area to its right. The z-score 1.645 has 0.05 area to its right.

The required height is