Men\'s heights are normally distributed with mean 68.6 in and standard deviation
ID: 3293989 • Letter: M
Question
Men's heights are normally distributed with mean 68.6 in and standard deviation of 2.8 in. Women's heights are normally distributed with mean 63.9 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? The percentage o 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.) The statistician would design a house with doorway height in. (Round to two decimal places as needed.)Explanation / Answer
P(X < A) = P(Z < (A - mean)/standard deviation)
a) Mean = 68.8 in
Standard deviation = 2.8 in
P(X >80) = 1 - P(X < 80)
= 1 - P(Z < (80-68.8)/2.8)
= 1 - P(Z < 4)
= 1 - 0.99998
= 0.00002
The percentage of men who are too tall to fit through a standard door without bending is 0.00%
Since the mean and standard deviation of heights of women are smaller than that of men,
The percentage of women who are too tall to fit through a standard door without bending is 0.00%
b) Let A be the doorway height that is sufficient for 95% of men.
P(X < A) = 0.95
P(Z < (A - 68.6)/2.8) = 0.95
(A - 68.6)/2.8 = 1.645 (from standard normal distribution table)
A = 73.2 inch