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Homework Set #2
Solutions
Fall 2008
1. (a) Let
p
=
P
(
Chip
=
Good
) =
3
4
, and let
q
=
P
(
Chip
=
Bad
)
.
Then,
q
= 1

P
(
Chip
=
Good
) = 1

p
= 1

3
4
=
1
4
(b) Let X = Number of bad chips in group of n
X
∼
Binomial
(
n,p
)
Then,
P
(
X
≤
x
) =
±
n
0
²
p
0
(1

p
)
n
+
±
n
1
²
p
(1

p
)
n

1
+
±
n
2
²
p
2
(1

p
)
n

2
+
±
n
3
²
p
3
(1

p
)
n

3
(c) Since ”the number of bad chips” follows binomial distribution,
E
(
x
) =
nq
=
1
4
n
2. The mean and variance of the uniform distribution U(a,b) are,
E
(
X
) =
a
+
b
2
,
V ar
(
X
) =
(
b

a
)
2
12
.
So,
E
(
X
) =
a
+
b
2
= 2
and,
V ar
(
X
) =
(
b

a
)
2
12
=
1
3
. Solving for a and b we obtain a = 1 and b = 3.
E
[
X
n
]
=
R
x
x
n
f
(
x
)
dx
=
R
b
a
x
n
1
b

a
dx
=
1
b

a
h
1
n
+1
x
n
+1
i
b
a
=
1
b

a
h
b
n
+1

a
n
+1
n
+1
i
10
=
1
3

1
h
3
n
+1

1
n
+1
i
20
=
3
n
+1

1
n
+1
20
n
+ 21
=
3
n
+1
There is no easy way to solve for n, therefore we accept this as the answer. However, knowledge of any
technique be it theoretical or practical (i.e. Goal Seek in Excel) will yield an answer of n = 3.
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This note was uploaded on 03/29/2010 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue University.
 Spring '08
 Leyla,O
 Operations Research

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