A survey found that women\'s heights are normally distributed with mean 62.6 in.

Question

A survey found that women's heights are normally distributed with mean 62.6 in. and standard deviation 2.9 in. The survey also found that men's heights are normally distributed with a mean 68.3 in. and standard deviation 2.9. Complete parts a through c below. 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 (Round to two decimal places as needed.) Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is (Round to two decimal places as needed.) 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 in. and at most in. (Round to one decimal place as needed.)

Explanation / Answer

a) From information given, Xbar=62.6, sd=2.9, compute z score corresponding to Xi=56 in and 75 in (values obtained after converting feet into inches, 1foot=12 inches)

Z1=(Xi-Xbar)/sd=(56-62.6)/2.9=-2.28 Z2=(75-62.6)/2.9=4.28

P(4 ft 8 in <X < 6 ft 3 in)=P(-2.28<Z<4.28)=0.4887+0.4999=0.9886

b) Xbar=68.3, sd=2.9

Z1=(56-68.3)/2.9=-4.24 Z2=(75-68.3)/2.9=2.31

P(4 ft 8 in <X < 6 ft 3 in)=P(-4.24<Z<2.31)=0.4999+0.4896=0.9895

c) Z score corresponding to tallest 5% is 1.60, find Xi

1.60=(Xi-68.3)/2.9

Xi=72.94

Z score correspond to -1.60

-1.60=(Xi-62.6)/2.9

Xi=57.96

72.94 to 57.96

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