A survey found that woman\'s heights are normally distributed with mean 63.4 in.

Question

A survey found that woman's heights are normally distributed with mean 63.4 in. and standard deviation 2.1 in. the survey also found that men's heights are normally distributed with a mean 67.5 and standard deviation 2.9 in. most of the live characters at an amusement park have height requirements with a minimum of 4ft 9in an a maximum of 6ft 2in.

A.) find the percentage of women meeting the height requirement

B.) find the percentage of men meeting the height requirement

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?

Explanation / Answer

a) z1 = (57 - 63.4)/2.1 = - 3.05

z2 = (74 - 63.4)/2.1 = 5.05

P = 0.999

So, around 99.9%

b) z1 = (57 - 67.5/2.9 = - 3.62

z2 = (74 - 67.5)/2.9 = 2.24

P = 98.73%

c) Tallest 5% of men height:

(x - 67.5)/2.9 = 1.64

x = 72.26

shortest 5% of women height:

(x - 63.4)/2.1 = -1.64

x = 59.956

So, new requirements = {59.956, 72.26}

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