ST 314
HW #3
Solutions
1. Text, p. 133, exercise 3.40.
The r.v. Y is distributed exponential with
λ
= 0.01.
a.
P(Y < 30) = 1 – e

λ
(30)
= 1 – e
.01(30)
= 1 e
.30
= .2592
b.
P(Y > 15) = 1 – [1 – e

λ
(15)
] = e
.15
= .8607
c.
P(Y = 100) = 0
d.
P(50
≤
Y
≤
150) =
[1 – e

λ
(150)
] – [1 – e

λ
(50)
] = e
1.5
+ e
.5
= .3834
2. Text, p. 141, exercise 3.47.
Let Y = thickness of bolts.
Then Y is normally distributed with mean
μ
= 10.0 mm and
standard deviation
σ
= 1.6 mm.
a.
P(9.2
≤
Y
≤
10.8) = P( (9.210)/1.6
≤
(Y
μ
)/
σ
≤
(10.810)/1.6 ) = P (.5
≤
Z
≤
.5) = P(Z < .5) – P(Z < .5) = .6915 – .3085 = .3830
b.
P(Y
≤
9.2) = P ((Y
μ
)/
σ
≤
(9.210)/1.6) = P(Z
≤
.5) = .3085
3. Text, p. 142, exercise 3.53.
Let Y = volume percent.
Then Y is normally distributed with
μ
= 70 and
σ
= 2.
a.
P(Y > 75) = P((Y
μ
)/
σ
> (7570)/2) = P(Z > 2.5) = 1  .9938 = .0062
b.
P(67 < Y < 73) = P((6770)/2 < Z < (7370)/2) = P(1.5 < Z < 1.5) = P(Z < 1.5) –
P(Z < 1.5) = .9332 – 0.668 = 0.8664
c.
We want to find
μ
so that P(Y < 70) < .02.
But P(Y < 70) = P((Y
μ
)/
σ
< (70
μ
)/
σ
))
= P(Z < (70
μ
)/2) and from Table 1 we see
(70
μ
)/2 = 2.06 so that
μ
= 70 – 2(2.06) = 74.12
4. Text, p. 152, exercise 3.56.
Let Y = concentricity of an engine oil seal groove.
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 Spring '08
 Arthur
 Normal Distribution, Standard Deviation, a. b. c., engine oil seal

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