A survey found that women\'s heights are normally distributed with mean 62.9 in.

Question

A survey found that women's heights are normally distributed with mean 62.9 in. and standard deviation 2.4 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 3 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is? 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

Answer to the question is as follows:

Women:
Men = 62.9
Stdev = 2.4

Men:
Mean = 68.9
Stdev = 2.8

a. %age of women meeting 4ft9in to 6ft3in =?
4ft9in = 57 inch and 6ft3in = 75 inch since 1ft = 12in

So, %women = P(57<X<75)
= P((57-62.9)/2.4<Z< (75-63.9)/2.4)
= P(-2.46<Z<4.3)
= 1-.0069 = .9931

So, 99.31% of women will clear

b.%men meeting the 4ft9in to 6ft3in requirement:

So, %women = P(57<X<75)
= P((57-68.9)/2.8<Z< (75-68.9)/2.8)
= P(-4.25<Z<2.18)
= .9853-0
= .9853

c.

Tallest 5% of mean are excluded
So, Upper boundry = 1.645*2.8+68.9 = 73.506
Shortest 5% of women are excluded
So, Lower boundry = -1.645*2.4+62.9 = 58.952

The new height requiredments instead of 57 to 75 is 58.952 to 73.506 inches

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