Men\'s heights are normally distributed with mean 70 1 in and standard deviation
ID: 3245107 • Letter: M
Question
Men's heights are normally distributed with mean 70 1 in and standard deviation of 2.8 in. Women's heights are normally distributed with mean 63.3 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.)Explanation / Answer
a) P(Men too tall to fit)
= P(X > 80)
= P(z > (80 - 70.1)/2.8)
= P(z > 3.54)
= 0.0002
Hence,
Percentage of men too tall to fit = 0.02 %
P(Women too tall to fit)
= P(X > 80)
= P(z > (80 - 63.3)/2.5)
= P(z > 6.68)
= 0.0000
Hence,
Percentage of women too tall to fit = 0 %
b) From z table,
P(z > 1.645) = 0.05
Hence,
Standard doorway size should be = 70.1 + 1.645*2.8 = 74.7 in