ISyE 3232
Stochastic Manufacturing and Service Systems
Spring 2011
H.Ayhan
Solutions to Homework 1
1. (a) Poisson (b) exponential (c) geometric (d) Bernoulli (e) binomial (f) normal
2. Since
E
[
X
] = 4 and Var(
X
) = 100,
(
a
)
E
[6

4
X
] = 6

4
E
[
X
] =

10 and Var(6

4
X
) = 16Var(
X
) = 1
,
600.
(
b
)
E
[(
X

3)
/
5] =
1
5
E
[
X
]

3
5
= 1
/
5 and Var((
X

3)
/
5) =
1
25
Var(
X
) = 4.
3. (
a
)
1 =
3
X
k
=1
P
(
X
= 2
k

1) =
3
X
k
=1
(2
k

1)
c
= 9
c
so it follows that c = 1/9.
(
b
)
E
[
X
] =
∑
3
i
=1
(2
k

1)
P
(
X
= 2
k

1) = (1
/
9)(1 + 9 + 25) = 35
/
9
(
c
)
E
[
X
2
] =
∑
3
i
=1
(2
k

1)
2
P
(
X
= 2
k

1) = (1
/
9)(1 + 27 + 125) = 17
(
d
) Var(
X
) =
E
[
X
2
]

(
E
[
X
])
2
= 17

(35
/
9)
2
= 1
.
8765
(
e
) Note that
(
X

2)
+
=
0 with probability 1
/
9
1 with probability 3
/
9
3 with probability 5
/
9
Hence,
E
[(
X

2)
+
] = 0
·
(1
/
9) + 3
/
9 + 3
×
5
/
9 = 2
4. (
a
)
p
k
=
P
(
X
=
k
) =
e

5 5
k
k
!
,
k
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 Fall '07
 Billings
 Central Limit Theorem, Normal Distribution, Probability theory, var, Stochastic Manufacturing, processing time

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