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York University
MATH 2030 3.0 (Elementary Probability)
Assignment 7 – Solutions, Nov 2011
§
3.3 No. 3
(a)
E
[2
X
+ 3
Y
] = 2
E
[
X
] + 3
E
[
Y
] = 2
×
1 + 3
×
1 = 5
(b) Var[2
X
+ 3
Y
] = Var[2
X
] + Var[3
Y
] by independence. This then equals
2
2
Var[
X
] + 3
2
Var[
Y
] = 4
×
2 + 9
×
2 = 26.
(c) By independence
E
[
XY Z
] =
E
[
X
]
E
[
Y
]
E
[
Z
] = 1
×
1
×
1 = 1.
(d) Var[
XY Z
] =
E
[(
XY Z
)
2
]

E
[
XY Z
]
2
=
E
[
X
2
Y
2
Z
2
]

1
2
by part (c). By
independence this equals
E
[
X
2
]
E
[
Y
2
]
E
[
Z
2
]

1. But
E
[
X
2
] = Var[
X
] +
E
[
X
]
2
= 2 + 1
2
= 3 and the same is true for the other two variables.
Therefore Var[
XY Z
] = 3
×
3
×
3

1 = 26
.
§
3.3 No. 14
(a) Let
X
be the income of a family picked uniformly at random from the
given area. So
E
[
X
] = 10
,
000. We are asked to estimate the percentage
of families with incomes over 50,000, or in other words to estimate
P
(
X >
50
,
000). Since
X
≥
0 we can use Markov’s inequality, and get
P
(
X >
50
,
000)
≤
E
[
X
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This note was uploaded on 01/23/2012 for the course MATH 2030 taught by Professor Salisbury during the Spring '08 term at York University.
 Spring '08
 Salisbury
 Probability

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