Homework Set #2 Solutions
IE 336
Spring 2011
1.
(a) Let
X
equal number of good chips that appear.
Then,
X
∼
Binomial
(
n, p
), where
n >
1000
and
p
= 2
/
5. Since
n
is sufficiently large, the binomial distribution can be approximated using a
normal distribution, i.e.
X
∼
N
(
np, np
(1

p
))
.
P
(4
< X <
8)
≈
P
4

np
p
np
(1

p
)
< Z <
8

np
p
np
(1

p
)
!
≈
P
Z <
8

np
p
np
(1

p
)
!

P
Z <
4

np
p
np
(1

p
)
!
≈
Φ
8

np
p
np
(1

p
)
!

Φ
4

np
p
np
(1

p
)
!
≈
0
Since
n
is very large, Φ(
·
) evaluates very close to 0.
(b)
E
(
X
) =
np
=
2
5
n
(c) Let
Y
equal the number of bad chips selected.
Then, the probability that exactly 3 of the 4
selected chips are bad is equal to:
P
(
Y
= 3)
=
(1

p
)
3
·
p
=
3
5
3
2
5
2.
X
∼
U
(
a, b
). The expected value and the variance are equal to:
E
(
X
) =
a
+
b
2
= 2
V ar
(
X
) =
(
b

a
)
2
12
=
1
3
Solving for
a
and
b
yields
a
= 1 and
b
= 3.
1
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E
(
X
n
)
=
Z
∞
∞
x
n
f
(
x
)
dx
=
Z
b
a
1
b

a
x
n
dx
=
1
b

a
1
n
+ 1
x
n
+1
b
a
=
1
b

a
1
n
+ 1
(
b
n
+1

a
n
+1
)
10
=
1
3

1
1
n
+ 1
(
3
n
+1

1
)
20
=
3
n
+1

1
n
+ 1
20
n
+ 21
=
3
n
+1
There is no easy way to solve for
n
, therefore this answer will be accepted.
However, a numerical
computation (e.g. Goal Seek in Excel) yields
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 Spring '08
 Bruce,S
 Probability theory, np np, CE CE

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