Question 1078672: A survey found that women's heights are normally distributed with mean 62.6 in. and standard deviation 2.7 in. The survey also found that; men's heights are normally distributed with a mean 68.9 in. and standard deviation 2.8. Complete parts a through c below.. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 4 in. Find the percentage of women meeting the height requirement.
The percentage of women who meet the height requirement is %.
&(Round to two decimal places as; needed.)
The percentage of men who meet the height requirement is %
If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A survey found that women's heights are normally distributed with mean 62.6 in. and standard deviation 2.7 in.
The survey also found that; men's heights are normally distributed with a mean 68.9 in. and standard deviation 2.8.
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Complete parts a through c below..
Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 4 in. Find the percentage of women meeting the height requirement.
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z(4'9") = z(57") = (57-62.6)/2.7 = -2.0741
z(6'4") = z(76") = (76-62.6)/2.7 = 4.96
P(47<= x <=76) = P(-2.0741
The percentage of women who meet the height requirement is 98%.
(Round to two decimal places as; needed.)
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The percentage of men who meet the height requirement is %
Comment:: Follow the same procedure as used for the women.
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If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?
For men::
Find the z-value with a right-tail of 5%
invNorm(0.95) = 1.645
Find the corresponding men's height:: x = 1.645*2.8 + 68.9 = 73.5"
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Comment:: Follow the same procedure as used for the men.
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Cheers,
Stan H.
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