Question 1088691: A survey found that women's heights are normally distributed with mean 63.5 in and standard deviation 2.2 in. A branch of the military requires women's heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z value for 58 is (58-63.5/2.2=-5.5/2.2=-2.5
z value for 80 is (80-63.5)/2.2=16.5/2.2=7.5
The probability of z being between these is 0.994
More women will be denied because they are too short. Virtually all will be within the high height limit.
for the shortest 1% the z value is -2.326
for the tallest 2%, the z-value is +2.055
multiply this by the sd to get -5.12, which makes the lower limit 63.5-5.12=58.4 inches lower limit
and +4.52 above this to get 68.0 inches upper limit (58.4, 68.0)
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