A survey found that women\'s heights are normally distributed with mean 62.5 in.

Question

A survey found that women's heights are normally distributed with mean 62.5 in. and standard deviation 2.8 in. The survey also found that men's heights are normlly distributed with a mean 67.1 in. and standard deviation 2.9. Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft7 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 96. Round to two decimal places as needed.)

Explanation / Answer

Answer in detail with calc below:

For women
Mean = 62.5
Stdev = 2.8

For men
Mean = 67.1
Stdev = 2.9

a. P(4ft7in <X< 6ft3in) = P(55<Z<75) = P((55-62.5) / 2.8 <Z< (75-62.5)/2.8) = P(-2.68<Z<4.464) = 1-.00368= .99632
So, 99.63% of women clear these requirements

b. P(4ft7in <X< 6ft3in) = P(55<Z<75) = P((55-67.1)/2.9 <Z< (75-67.1)/2.9) = P(-4.17<Z<2.72) = .99674-0 = 99.67%
So, 99.67% of mean clear these requirement

c. To calcualte this:

5% of tallest men: Z*Sigma+Mean = 1.645*2.9+67.1 = 71.87
5% of shortest women:Z*Sigma+Mean = -1.645*2.8+62.5 = 57.894
So, it should be alteast 57.89 inches and atmost 71.87 inches

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