6.3.24 Question Help * Men\'s heights are normally distributed with mean 70.2 in
ID: 3066733 • Letter: 6
Question
6.3.24 Question Help * Men's heights are normally distributed with mean 70.2 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?Explanation / Answer
Man
Mean= 70.2
S.D =2.8
Woman
Mean= 64
S.D.= 2.5
X= 80
A)
For man
Z = (X - ?) / ?
Z = (80 - 70.2) / 2.8
Z = 3.5
FROM Z-SCORE TABLE
P(X>80) = 0.0002
Z = (X - ?) / ?
Z = (80 - 64) / 2.5
Z = 6.4
FROM Z-SCORE TABLE
From Z table
P(X>80)=0.00
b) given =
P(x>?)=0.05
Z value at P value of 0.05
Z = 1.64485
1.645 = (X - ?) / ?
1.645= (X - 70.2) / 2.8
X=70.2+4.606=74.806
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