Math331, Spring 2008
Instructor: David Anderson
Section 10.1 Homework Answers
Homework:
pgs. 412  414, #’s 3, 5, 11, 13.
3. We have that
V ar
(
X
) =
V ar
(
Y
) =
V ar
(
Z
) = 1 and
E
[
X
] =
E
[
Y
] =
E
[
Z
] = 0, therefore,
E
[
X
2
] =
E
[
Y
2
] =
E
[
Z
2
] = 1. Using this and independence gives
E
[
X
2
(
Y
+ 5
Z
)
2
] =
E
[
X
2
(
Y
2
+ 10
Y Z
+ 25
Z
2
)] =
E
[
X
2
Y
2
] + 10
E
[
X
2
Y Z
] + 25
E
[
X
2
Z
2
]
=
E
[
X
2
]
E
[
Y
2
] + 10
E
[
X
2
]
E
[
Y
]
E
[
Z
] + 25
E
[
X
2
]
E
[
Z
2
]
= 1
∗
1 + 10
∗
1
∗
0
∗
0 + 25
∗
1
∗
1
= 26
.
5.
Let
X
be the number of cereal boxes the customer must open until all five prizes are
found.
Let
X
1
be the number of cereal boxes needed to be opened before the customer
finds the first unique prize,
X
2
be the number of cereal boxes needed to be opened after
the customer finds the first prize and until he/she finds a second unique prize,
X
3
be the
number of cereal boxes needed to be opened after the customer finds the second unique
prize and until he/she finds a third unique prize, etc. for
X
4
and
X
5
. Therefore, counting
the number of unique prizes available between each time a new prize is found, we see that
X
1
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 Spring '08
 Anderson
 Math, Probability, cereal boxes, Booby prize, unique prize

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