Homework12,13
61. Assume that the heights of adult males residing in the US (’males’ for short) are approximately
normally distributed with mean 69.0 inches and standard deviation 2.8 inches.
a. Find the proportion of males whose heights lie between 65.5 inches and 74.6 inches.
b. If a male’s height is 0.4 standard deviations below the mean, how tall is he and what is
the probability that a male, chosen at random, is within two inches of his height?
c. In a random sample of 6 males, what is the probability that exactly 2 of them will be over
six feet tall?
SOLUTION:
Let X indicate the height of randomly selected adult male residing in the US,
i.e. X
∼
N(69
,
2
.
8
2
)
a.
P(65
.
5
≤
X
≤
74
.
6) = P(
65
.
5
−
69
2
.
8
≤
Z
≤
74
.
6
−
69
2
.
8
)
= P(Z
≤
2)
−
P(Z
≤ −
1
.
25) = 0
.
9772
−
0
.
1056 = 0
.
8716.
b. Height=69(0.4)(2.8)=67.88.
P(67
.
88
−
2
≤
X
≤
67
.
88 + 2) = P(
65
.
88
−
69
2
.
8
≤
Z
≤
69
.
88
−
69
2
.
8
)
= P(Z
≤
0
.
31)
−
P(Z
≤ −
1
.
11) = 0
.
6217
−
0
.
1335 = 0
.
4882.
c. Let Y indicate the number of males taller than 6 feet in the sample (of size 6). Y
∼
Bin(6
, p
)
where
p
= P(X
>
6(12)) = P(Z
>
72
−
69
2
.
8
) = 1
−
P(Z
≤
1
.
07) = 1
−
0
.
8577 = 0
.
1423.
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 Fall '11
 DANIELCONUS
 Statistics, Normal Distribution, Standard Deviation, average net weight

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