ts are normally distributed with mean 69.0 inn and standard deviation 2.8 in. Wo
ID: 3073685 • Letter: T
Question
ts are normally distributed with mean 69.0 inn and standard deviation 2.8 in. Women's hei ights are normally distributed with mean 63.6 in. and standard deviation 2.5 in. A. Tal Il Clubs International is at least 74 in. tal a. What percentage of men meet that requirement? b. a social organization for tall people. It has a requirement that men m ust be ll, and women must be at least 70 in. tall. What percentage of women meet that requirement? a. Find the percentage of men who meet the height requirements b, If the height requirements are changed so that all men are eligible except the shortest 3% and 8. The U.S. Marine Corps requires that men have heights between 64 in. and 80 in. the tallest 4%, what are the new height requirements?Explanation / Answer
P(X < A) = P(Z < (A - mean)/standard deviation)
A. a. P(X > 74) = 1 - P(X < 74)
= 1 - P(Z < (74 - 69)/2.8)
= 1 - P(Z < 1.79)
= 1 - 0.9633
= 0.0367
b. P(X > 70) = 1 - P(X < 70)
= 1 - P(Z < (70 - 63.6)/2.5)
= 1 - P(Z < 2.56)
= 1 - 0.9948
= 0.0052
B. P(64 < X < 80) = P(X < 80) - P(X < 64)
= P(Z < (80 - 69)/2.8) - P(Z < (64 - 69)/2.8)
= P(Z < (3.93) - P(Z < -1.79)
= 1 - 0.0357
= 0.9633
Let S denote the shortest 3% and T denote the tallest 4%
P(X < S) = 0.03
P(Z < (S - 69)/2.8) = 0.03
(S - 69)/2.8 = -1.88
S = 63.74 in
P(X > T) = 0.04
P(Z < (T - 69)/2.8) = 0.96
(T - 69)/2.8 = 1.75
T = 73.9 in
New heigh requirement is between 63.74 in and 73.9 in
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