A survey found that women\'s heights are normally distributed with mean 63.7 in.

Question

A survey found that women's heights are normally distributed with mean 63.7 in. and standard deviation 2.4 in. The survey also found that men's heights are normally distributed with a mean 68.1 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 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? (round to two decimal places as needed) 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) For women, Mean = 63.7 in

Standard deviation = 2.4 in

P(X < A) = P(Z < (A -mean)/standard deviation)

Height requirements = 57 in to 75 in

Percenrtage of women meeting height requirement = P(57 < X < 75)

= P(X < 75) - P(X < 57)

= P(Z < (75 - 63.7)/2.4) - P(Z < (57 - 63.7)/2.4)

= P(Z < 4.7) - P(Z < -2.79)

= 1 - 0.0026

= 0.9974

b) For men, Mean = 68.1 in

Standard deviation = 2.9 in

Percenrtage of men meeting height requirement = P(57 < X < 75)

= P(X < 75) - P(X < 57)

= P(Z < (75 - 68.1)/2.9) - P(Z < (57 - 68.1)/2.9)

= P(Z < 2.38) - P(Z < -3.83)

= 0.9938 - 0.0001

= 0.9937

c) Let M be the height that restricks tallest 5% men

Then, P(X < M) = P(Z < (M - 68.1)/2.9) = 0.95

(M - 68.1)/2.9 = 1.645

M = 72.87 in

Let W be the height that restricts shortest 5% women

P(X < W) = P(Z < (W - 63.7)/2.4) =0.05

= (W - 63.7)/2.4 = -1.645

W = 59.75

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