APPM 4/5570
Solutions to Problems from Section 4.4
60. The pdf for
X
is
f
(
x
) = 0
.
01386
e

0
.
01386
x
,
x >
0
and zero otherwise.
It will be convenient to find the cdf once and use it to answer all questions.
F
(
x
) =
Z
x
0
0
.
01386
e

0
.
01386
u
du
= 1

e

0
.
01386
x
.
(This holds for
x >
0, for
x <
0,
F
(
x
) = 0.)
a.
P
(
X
≤
100) =
F
(100) = 1

e

0
.
01386(100)
≈
0
.
7499
P
(
X
≤
200) =
F
(200) = 1

e

0
.
01386(200)
≈
0
.
9375
P
(100
< X <
200) =
F
(200)

F
(100)
≈
0
.
9375

0
.
7499 = 0
.
1875
(I used “unrounded” values to compute 0
.
1875.
It is okay to give the
answer as 0
.
1876.)
b. The mean for an exponential random variable with rate
λ
is 1
/λ
.
The
variance is 1
/λ
2
. So, in this case,
μ
=
1
λ
=
1
0
.
01386
≈
72
.
15
σ
1
=
1
λ
2
=
1
0
.
01386
2
≈
5205
.
64
We want
P
(
X > μ
+2
σ
) =
P
(
X >
216
.
45) = 1

F
(216
.
45) = 1

h
1

e

0
.
01386(216
.
45)
i
≈
0
.
05
66.
a. We know that for
X
∼
Γ(
α, β
),
E
[
X
] =
αβ
and
V ar
[
X
] =
αβ
2
. So, we
equate
αβ
= 20
and
αβ
2
= 80
Solving for
α
and
β
gives
α
= 5
and
β
= 4
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b. We want to transform the problem from one involving a Γ(5
,
4) distribu
tion to one involving a Γ(5
,
1) distribution since the latter has cdf values
tabulated in Table A.4.
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 '08
 staff
 Probability theory, Exponential distribution, γ, Table A.4

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