University of Illinois
Spring 2010
ECE 313:
Problem Set 9: Problems and Solutions
Probability Density, Poisson Processes
Due:
Wednesday March 31 at 4 p.m.
Reading:
Ross, Chapter 5; Powerpoint Lecture Slides, Sets 2225
Noncredit Exercises:
Chapter 5:
Problems 13, 5, 6, 1519, 2325, 3234;
Theoretical Exercises 1, 8; SelfTest Problems 14
1.
[Probability Density]
Random variable
X
has the following PDF:
f
X
(
u
) =
1
5
0
< u
≤
1
1
2
1
< u
≤
2
1
10
2
< u
≤
5
0
otherwise
Answer each of the following questions by shading and labeling the area of a region on a graph.
Do
not
perform any integrals.
(a) What is
P
(2
< X
≤
3)?
Solution:
1
/
10.
The plot should show shading between
u
= 2 and
u
= 3.
(b) What is
P
(

1
≤
X
≤
1
.
5)?
Solution:
9
/
20
(c) What is
P
(2
≤
X
≤
6)?
Solution:
3
/
10
(d) What is
P
(1
.
173
≤
X
≤
1
.
174)?
Solution:
0
.
0005. The graph should show a shaded rectangle of
width 0.001 and height 0.5.
(e) What is
F
X
(
v
)?
Solution:
F
X
(
v
) =
0
v
≤
0
v
5
0
≤
v
≤
1
1
5
+
v

1
2
1
≤
v
≤
2
7
10
+
v

2
10
2
≤
v
≤
5
1
5
≤
v
2.
[Black Swans]
A recent census conducted in the country of Metasylvania found that the body weight of adult Metasyl
vanians, measured in pounds, is a random variable
X
with the following probability density function:
f
X
(
u
) =
Ae

0
.
02(
u

80)
80
≤
u
0
else
The same survey found that the personal wealth of adult Metasylvanians, measured in Metasylvanian
dollars, is a random variable
Y
with the following probability density function:
f
Y
(
u
) =
B
(
u
100
)

1
.
5
100
≤
u
0
else
(a) Find the constants
A
and
B
.
Solution:
The PDFs must integrate to 1, therefore
A
= 0
.
02,
B
= 0
.
5.
(b) Notice that all Metasylvanians weigh at least 80 pounds.
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i. What is the probability that an individual Metasylvanian weighs at least 160 pounds?
So
lution:
Z
∞
160
0
.
02
e

0
.
02(
u

80)
du
=
e

80
/
50
≈
0
.
2
ii. What is the probability that an individual Metasylvanian weighs at least 800 pounds?
So
lution:
Z
∞
800
0
.
02
e

0
.
02(
u

80)
du
=
e

720
/
50
≈
5
.
6
×
10

7
iii. What is the expected weight of a Metasylvanian?
Solution:
Z
=
X

80 is an exponential
random variable with expectation of 50, therefore
E
[
X
] = 80 + 50 = 130 pounds.
(c) Notice that every Metasylvanian has a personal wealth of at least $100.
i. What is the probability that a Metasylvanian has a personal wealth of greater than $200?
Solution:
Z
∞
200
0
.
5
u
100

1
.
5
du
=
200
100

0
.
5
≈
0
.
7
ii. What is the probability that a Metasylvanian has a personal wealth of greater than $1000?
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 Spring '10
 S
 Probability theory, probability density function

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