Math 3C Homework 6 Solutions
Ilhwan Jo and Akemi Kashiwada
[email protected], [email protected]
Assignment:
Section 12.4 Problems 18, 29, 31, 32, 33, 34, 36, 38, 40
18.
Suppose that the probability mass functions of a discrete random variable
X
is given by the following
table. Find the mean, the variance, and the standard deviation of
X
.
x
P
(
X
=
x
)

1
0
.
1

0
.
5
0
.
2
0
.
1
0
.
1
0
.
5
0
.
25
1
0
.
35
Solution
EX
=
X
x
xP
(
X
=
x
)
= (

1)(0
.
1) + (

0
.
5)(0
.
2) + (0
.
1)(0
.
1) + (0
.
5)(0
.
25) + (1)(0
.
35)
= 0
.
2850
,
EX
2
=
X
x
x
2
P
(
X
=
x
)
= (

1)
2
(0
.
1) + (

0
.
5)
2
(0
.
2) + (0
.
1)
2
(0
.
1) + (0
.
5)
2
(0
.
25) + (1)
2
(0
.
35)
= 0
.
5635
,
var(
X
) =
EX
2

(
EX
)
2
= 0
.
5635

0
.
2850
2
≈
0
.
4823
,
σ
=
p
var(
X
) =
√
0
.
4823
≈
0
.
6945
.
29.
Toss a fair coin ten times. Let
X
be the number of heads.
(a)
Find
P
(
X
= 5).
(b)
Find
P
(
X
≥
8).
(c)
Find
P
(
X
≤
9).
Solution
(a)
X
is binomially distributed with probability
p
=
1
2
of having a head. The number of trials is
n
= 10.
P
(
X
= 5) =
10
5
¶
1
2
¶
5
1
2
¶
5
=
10
5
¶
1
2
¶
10
.
1
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(b)
P
(
X
≥
8) =
P
(
X
= 8) +
P
(
X
= 9) +
P
(
X
= 10)
=
10
8
¶
1
2
¶
10
+
10
9
¶
1
2
¶
10
+
10
10
¶
1
2
¶
10
=
10
8
¶
+
10
9
¶
+
10
10
¶¶
1
2
¶
10
=
10
·
9
2
+ 10 + 1
¶
1
2
¶
10
=
56
2
10
.
(c)
P
(
X
≤
9) = 1

P
(
X
= 10)
= 1

10
10
¶
1
2
¶
10
= 1

1
2
10
.
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 Spring '07
 SCHONMANN
 Probability, Probability theory

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