1. Suppose you are in charge of ordering caps and gowns for senior graduation in
ID: 2959183 • Letter: 1
Question
1. Suppose you are in charge of ordering caps and gowns for senior graduation in May. You know the heights of the population of students at your college are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. The gown manufacturer wants to know how many students will need to special order their gowns because they are very tall.A) Find the percentage of students who are above 72 inches tall.
B)Suppose the school administrator believes that the company should only consider a gown as a special order if only 2% of the population is tall enough to have to "special order". What is the height such that only 2% of the population is taller?
C)Suppose there are 22 seniors graduating. What is the probability that the average height of the 22 seniors is between 66 and 70 inches?
Explanation / Answer
Given X~Normal(=68, s=3)
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A) Find the percentage of students who are above 72 inches tall.
P(X>72) = P((X-)/s > (72-68)/3) = P(Z>1.33) =0.0917 (9.17%) (check standard normal table)
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What is the height such that only 2% of the population is taller?
P(X>c)=0.02
--> P(X<c) = 1-0.02 =0.98
--> P(Z<(c-68)/3) = 0.98
--> (c-68)/3 = 2.05(check standard normal table)
--> c = 68 + 3*2.05 =74.15
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What is the probability that the average height of the 22 seniors is between 66 and 70 inches?
P(66<xbar<70) = P((66-68)/(3/(sqrt(22))) <Z< (70-68)/(3/(sqrt(22))))
=P( -3.13<Z<3.13)
=0.9982 (check standard normal table)
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