Math331, Spring 2008
Instructor: David Anderson
Section 4.3 Homework Answers
Homework:
pgs. 157158 #’s 1, 3, 4, 7, 8, 10.
1. F is given by
F
(
t
) =
0
t <
1
1
/
15
1
≤
t <
2
3
/
15
2
≤
t <
3
6
/
15
3
≤
t <
4
10
/
15
4
≤
t <
5
1
5
≤
t
.
3.
We have
R
(
X
) =
{
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
10
,
11
,
12
}
.
We may count the number of ways
we can get each value and divide by 36, which is the number of elements in the sample
space
S
=
{
(
a,b
)

a,b
∈ {
1
,
2
,
3
,
4
,
5
,
6
}}
.
We find
p
(2) = 1
/
36
,p
(3) = 2
/
36
,p
(4) =
3
/
36
,p
(5) = 4
/
36
,p
(5) = 5
/
36
,p
(7) = 6
/
36
,p
(8) = 5
/
36
,p
(9) = 4
/
36
,p
(10) = 3
/
36
,p
(11) =
2
/
36
,p
(12) = 1
/
36.
4.
There are only jumps in the distribution function.
Therefore, the probability mass
function can be determined by the size of the jumps at each value. We therefore see that
the range of the random variable is given by
R
(
X
) =
{
2
,
2
,
4
,
6
}
. The probability mass
function is given below
p
(

2) = 1
/
2
p
(2) = 3
/
5

1
/
2 =
.
1
p
(4) = 8
/
9

3
/
5 = 13
/
45
p
(6) = 1

8
/
9 = 1
/
9
.
7.
(a) We need
1 =
k
*
(1 + 2 + 3 + 4 + 5) = 15
k.
Thus,
k
= 1
/
15.
(b) We need
1 =
k
*
(1 + 1 + 4 + 9) = 15
k.
Thus, again,
k
= 1
/
15.
(c) We need
1 =
k
∞
summationdisplay
i
=1
(1
/
9)
i
=
k
1
/
9
1

1
/
9
=
k/
8
.
Thus,
k
= 8.
1