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Solutions to Homework 10, ISyE 2027 Fall 2006
Problem 1
(a) geometric (b) Poisson (c) binomial (d) normal (e) exponential
Problem 2
(a)
P
(
Z
≤
1
.
44) = 0
.
9251.
(b) We see that
P
(

Z
 ≤
1
.
44)
=
P
(

1
.
44
≤
Z
≤
1
.
44)
=
P
(
Z
≤
1
.
44)

P
(
Z
≤ 
1
.
44)
=
2
P
(
Z
≤
1
.
44)

1
=
0
.
8502
.
(c)
P
(
Z >
1
.
44) = 1

P
(
Z
≤
1
.
44) = 0
.
0749.
(d)
P
(
Z
= 1
.
44) = 0 (
Z
is continuous).
(e) We see that
0
.
05 =
P
(

Z

> x
)
=
P
(
Z
≥
x
) +
P
(
Z
≤ 
x
)
=
2
P
(
Z
≥
x
)
=
2(1

P
(
Z
≤
x
))
.
This implies that
P
(
Z
≤
x
) = 0
.
975. After using a normal table, we see that
x
= 1
.
96.
Problem 3
Now suppose
X
∼
N
(2
,
100). Then
(a)
P
(
X
≤
9) =
P
(10
Z
+ 2
≤
9) =
P
(
Z
≤
.
7) = 0
.
7580.
(b)
P
(

X

>
9)
=
1

P
(

X
 ≤
9)
=
1

P
(

9
≤
X
≤
9)
=
1

P
(

9
≤
10
Z
+ 2
≤
9)
=
1

P
(

1
.
1
≤
Z
≤
0
.
7)
=
2

P
(
Z
≤
0
.
7)

P
(
Z
≤
1
.
1) = 0
.
3777
.
(c) Now,
0
.
05
=
P
(

X

2
 ≥
3
x
)
=
P
(
X

2
≥
3
x
) +
P
(
X

2
≤ 
3
x
)
=
P
(10
Z
≥
3
x
) +
P
(10
Z
≤
3
x
)
=
2
P
(
Z
≥
0
.
3
x
)
so now we ﬁnd that
P
(
Z
≤
0
.
3
x
) = 0
.
975
1
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which implies that 0
.
3
x
= 1
.
96, or
x
= 6
.
533.
Problem 4
Here
P
(
μ

kσ < Y < μ
+
kσ
)
=
P
(
μ

kσ < μ
+
σZ < μ
+
kσ
)
=
P
(

k < Z < k
)
=
2
P
(
Z
≤
k
)

1
.
If
k
= 0, the probability is 0. If
k
= 1, the probability is 0.6826. If
k
= 2, the
probability is 0
.
9544. Finally, if
k
= 3, the probability is 0
.
9974.
Problem 5
Using the formula given for the even moments of a standard normal r.v., we see
that
E
[
Z
2
] = 1 and
E
[
Z
4
] = 3 (we’ll use these later).
(a) The mean of
X
n
is just
E
[
X
n
]
=
E
[
n
X
k
=1
Z
2
k
]
=
n
X
k
=1
E
[
Z
2
k
]
=
nE
[
Z
2
1
] =
n.
The variance of
X
n
is
V ar
(
X
n
)
=
V ar
(
n
X
k
=1
Z
2
k
)
=
n
X
k
=1
V ar
(
Z
2
k
)
=
nV ar
(
Z
2
1
)
.
But
V ar
(
Z
2
1
) =
E
[
Z
4
1
]

E
[
Z
2
1
]
2
= 3

1 = 2, so
V ar
(
X
n
) = 2
n
.
(b) Now
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This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn

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