Men\'s heights are normally distributed with Mean: 68.8 in standard deviation: 2
ID: 3053053 • Letter: M
Question
Men's heights are normally distributed with Mean: 68.8 in standard deviation: 2.8 in. Women's heights are normally distributed with Mean: 63.5 in standard deviation: 2.5 in The standard doorway height is 80 in. a. What percentage of men are too tall to fit through a standard doorway withoutbending, 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? Please show steps so I can work remaining problems. Thank you so very much!!Explanation / Answer
Ans:
a)For men:
mean=68.8
standard deviation=2.8
z=(80-68.8)/2.8=4
P(z<4)=0.99997 i.e. 99.997 % men can pass through without bending.
For women:
mean=63.5,standard deviation=2.5
z=(80-63.5)/2.5=6.6
P(z<6.6)=1 i.e. 100% of the women can pass through without bending.
b)P(Z<=z)=0.95
(i.e. 95% can pass through except the tallest 5%)
z=1.645
x=68.8+1.645*2.8=73.41,height of the doorway should be 73.41 inch