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home / study / math / statistics and probability / statistics and probability questions and answers / a survey found that​ women's heights are normally distributed with mean 62.3 in. and standard ... Your question has been answered Let us know if you got a helpful answer. Rate this answer Question: A survey found that​ women's heights are normally distributed with mean 62.3 in. and standard dev... A survey found that​ women's heights are normally distributed with mean 62.3 in. and standard deviation 2.9 in. The survey also found that​ men's heights are normally distributed with a mean 67.7 in. and standard deviation 2.9. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 8 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is . Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is c. 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? The new height requirements are at least nothing in. and at most nothing in.

Answer 1

A survey found that women's heights are normally distributed with mean 62.3 in. and standard deviation 2.9 in.
The survey also found that; men's heights are normally distributed with a mean 67.7 in. and standard deviation 2.9.
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Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 8 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the height requirement.
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z(4'8") = z(56") = (56-62.3)/2.9 = -2.1724
z(6'3") = z(75") = (75-62.3)/2.9 = 4.3793
P(56<= x <=75) = normalcdf(-2.1724,4.3793)" =0.985
The percentage of women who meet the height requirement is 98.5%.

The percentage of men who meet the height requirement is %

z(4'8") = z(56") = (56-67.7)/2.9 = -4.034
z(6'3") = z(75") = (75-67.7)/2.9 = 2.517
P(56<= x <=75) = normalcdf(-4.034,2.517)" =0.99396
The percentage of women who meet the height requirement is 99.3%.


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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*2.9 + 67.7 = 72.47"