Math331, Spring 2008
Instructor: David Anderson
Section 4.4 Homework Answers
Homework:
pgs. 173  174, #’s 2, 3, 7, 11, 12.
2. Suppose the person chooses to park illegally. Let
X
be the amount of money he will pay
on a given day. We have
E
[
X
] = 25
*
.
6 + 0
*
.
4 = 15
.
Therefore he can expect to pay 15 per day on average. It is better to park legally.
3.
Let
X
be the winnings from a single ticket in dollars.
Then, based on what we were
given,
p
(1
,
200
,
000) = 1
/
(2
*
10
6
),
p
(800) = 500
/
(2
*
10
6
),
p
(30) = 4000
/
(2
*
10
6
) and
p
(0) = 1

1
/
(2
*
10
6
)

500
/
(2
*
10
6
)

4000
/
(2
*
10
6
). Thus,
E
[
X
] = 1
,
200
,
000
1
2
*
10
6
+ 800
500
2
*
10
6
+ 30
4000
2
*
10
6
+ 0 =
.
86
.
7.
We are given that if
X
is the demand for a magazine at a specific newsstand, then
R
(
X
) =
{
4
,
5
,
6
,
7
}
. The magazine sells for $
a
and costs $2
a/
3 to the owner, and the owner
can not return unsold magazines. So, how many magazines should be ordered?
For each
i
∈
R
(
X
), let
Z
i
represent the profit in a given week for the newsstand if the owner purchases
i
magazines. Then, we need
E
[
Z
i
] for each
i
∈ {
4
,
5
,
6
,
7
}
. The maximum value will give us
the number of magazines we should order.
Therefore,
E
[
Z
4
] = (4
a

4
*
(2
/
3)
a
)(
p
(4) +
p
(5) +
p
(6) +
p
(7) = (4
/
3)
a
*
1 = (4
/
3)
a.
E
[
Z
5
] = (4
a

5
*
(2
/
3)
a
)
p
(4) + (5
a

5
*
2
/
3)(
p
(5) +
p
(6) +
p
(7))
= (2
/
3)
a
6
18
+ (5
/
3)
a
*
12
18
=
4
3
a.
E
[
Z
6
] = (4
a

6
*
(2
/
3)
a
)
p
(4) + (5
a

6
*
(2
/
3)
a
)
p
(5) + (6
a

6
*
2
/
3)(
p
(6) +
p
(7))
= 0
6
18
+
a
5
18
+ (6
/
3)
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 Spring '08
 Anderson
 Math, Probability, 11:11, Probability mass function

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