SOLUTION: Men’s highs are normally distributed with 69.0 in. and standard deviation 2.8 in. Women’s highs are normally distributed with mean 63.6 in. and standard deviation 2.5 in. a. If

Algebra ->  Probability-and-statistics -> SOLUTION: Men’s highs are normally distributed with 69.0 in. and standard deviation 2.8 in. Women’s highs are normally distributed with mean 63.6 in. and standard deviation 2.5 in. a. If       Log On


   



Question 1050306: Men’s highs are normally distributed with 69.0 in. and standard deviation 2.8 in.
Women’s highs are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
a. If the requirements are changed so that the tallest 4% of men are eligible, what is the new minimum height for men?
b. If the requirements are changed so that the tallest 4% of women are eligible, what is the new minimum height for women?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Men’s highs are normally distributed with 69.0 in. and standard deviation 2.8 in.
Women’s highs are normally distributed with mean 63.6 in. and standard deviation 2.5 in.
a. If the requirements are changed so that the tallest 4% of men are eligible, what is the new minimum height for men?
Find the z-value with a left tail of 0.96:: invNorm(0.96) = 1.7507
Find the correspond height which is 1.7507 std above the mean::
x = 1.7507*2.8 + 69 = 73.9 inches
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b. If the requirements are changed so that the tallest 4% of women are eligible, what is the new minimum height for women?
x = 1.7507*2.5+63.6 = 68 inches
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Cheers,
Stan H.
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