Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions
: Yates and Goodman,8.1.1 8.1.3 8.2.2 8.3.2 8.3.3 and 9.1.2
Problem 8.1.1
Recall that
X
1
X
2
X
n
are independent exponential random variables with mean value
μ
X
5
so that for
x
0,
F
X
x
1
e
x
5
.
(a) Using Theorem 8.1,
σ
2
M
n
x
σ
2
X
n
. Realizing that
σ
2
X
25, we obtain
Var
M
9
X
σ
2
X
9
25
9
(b)
P X
1
7
1
P X
1
7
1
F
X
7
1
1
e
7 5
e
7 5
0 247
(c) First we express
P M
9
X
7 in terms of
X
1
X
9
.
P M
9
X
7
1
P M
9
X
7
1
P
X
1
X
9
63
Now the probability that
M
9
X
7 can be approximated using the Central Limit Theorem
(CLT).
P M
9
X
7
1
P
X
1
X
9
63
1
Φ
63
9
μ
X
9
σ
X
1
Φ
6 5
Consulting with Table 4.1 yields
P M
9
X
7
0 1151.
Problem 8.1.3
X
1
X
2
X
n
are independent uniform random variables with mean value
μ
X
7 and
σ
2
X
3
(a) Since
X
1
is a uniform random variable, it must have a uniform PDF over an interval
a b
. From
Appendix A, we can look up that
μ
X
a
b
2 and that Var
X
b
a
2
12. Hence, given
the mean and variance, we obtain the following equations for
a
and
b
.
b
a
2
12
3
a
b
2
7
Solving these equations yields
a
4 and
b
10 from which we can state the distribution of
X
.
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 Spring '08
 Ferry
 Central Limit Theorem, Standard Deviation, Variance, Probability theory

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