3. (20 pts) A survey found that women\'s heights are normally distributed with m

Question

3. (20 pts) A survey found that women's heights are normally distributed with mean 62.5 in and standard deviation 2.9 in. The survey also found that men's heights are normally distributed with a mean 67.3 in' and standard deviation 2.8 in. Complete parts (a) through (c) below. a) Most of the life characters at an amusement park have height requirements with a minimum of 4ft 3 in. and a maximum of 6ft 6in. Find the percentage of women meeting the height requirements. Round to four decimal places. The percentage of women that meet the requirements is

Explanation / Answer

For women, mean = 62.5 in

Standard deviation = 2.9 in

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

A) P(women between 4ft 3 in and 6ft 6 in)

= P(51 < X < 78)

= P(X < 78) - P(X < 51)

= P(Z < (78 - 62.5)/2.9) - P(Z < (51-62.5)/2.9)

= P(Z < 5.34) - P(Z < -3.84)

= 1 - 0.0001

= 0.9999

= 99.99%

B) For men, mean = 67.3 in

Standard deviation = 2.8 in

Probability that a man meets requirement

= P(X < 78) - P(X < 51)

= P(Z < (78 - 67.3)/2.8) - 0

= P(Z < 3.82)

= 0.9999

= 99.99%

C) Let the minimum height be E and maximum height be F

P(X < E) = 0.04, for women

So, P(Z < (E - 62.5)/2.9) = 0.04

(E - 62.5)/2.9 = -1.75

E = 57.425 in

For men, P(X > F) = 0.04

P(X < F) = 0.96

P(Z < (F - 67.3)/2.8) = 0.96

(F - 67.3)/2.8) = 1.75

F = 72.2 in

So, the new height requirement is at least 57.425 in and at most 72.2 in

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