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CEE 3770  Summer 2009
Homework 2 (due on 06/02)
1. (10 points) Suppose that X is continuous with p.d.f.
f
(
x
) =
cx
2
,
0
≤
x
≤
1.
(a) Find the value of
c
that will make the p.d.f. integrate to 1.
Solution:
1 =
Z
1
0
cx
2
dx
=
c/
3
This implies that
c
= 3.
(b) Find the c.d.f.
F
(
x
) for all
x
.
Solution:
F
(
x
) =
Pr
(
X
≤
x
) =
0
if
x <
0
x
3
if 0
≤
x
≤
1
1
if
x
≥
1
(c) Calculate
E
[
X
].
Solution:
E
[
X
] =
Z
1
0
xf
(
x
)
dx
=
Z
1
0
x
3
x
3
dx
= 3
/
4
(d) Calculate
V ar
(
X
).
Solution:
E
[
X
2
] =
Z
1
0
x
2
f
(
x
)
dx
=
Z
1
0
3
x
4
dx
= 3
/
5
This implies that
V ar
(
X
) =
E
[
X
2
]

(
E
[
X
]
)
2
= 3
/
80.
(e) Calculate
Pr
(0
≤
X
≤
1
/
2).
Solution:
Pr
(0
≤
X
≤
1
/
2) =
F
(1
/
2)

F
(0) = 1
/
8
.
2. (14 points) Jake is chosen as a semiﬁnalist in GT Sweepstakes. This entitled him to enter the
grand prize drawing for $1 million dollars. Being skeptical, Jake sat down to ﬁgure out how
much money he could expect to win if he played. The prize breakdown is following:
Number of winners
Prize
Amount
1
Grand prize
$1 million
5
1st prize
$1000
25
2nd prize
$100
50
3rd prize
$10
1
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View Full Document(a) If 250,000 people have been selected as semiﬁnalists, how much money could Jake ap
proximately expect to win if he entered.
Solution:
Let
X
denote the the amount of money you may win.
X
=
$1
million
w.p. 1
/
250
,
000
$1
,
000
w.p. 5
/
250
,
000
$100
w.p. 25
/
250
,
000
$10
w.p. 50
/
250
,
000
$0
w.p. 249
,
919
/
250
,
000
Thus,
E
[
X
] = $1
million
×
1
/
250
,
000 + $1
,
000
×
5
/
250
,
000 + $100
×
25
/
250
,
000 + $10
×
50
/
250
,
000
= 4
.
032
(b) What is the variance of his possible gain?
Solution:
E
[
X
2
] = $1
million
2
×
1
/
250
,
000 + $1
,
000
2
×
5
/
250
,
000
+$100
2
×
25
/
250
,
000 + $10
2
×
50
/
250
,
000
= 4000021
.
02
V ar
(
X
) =
E
[
X
2
]

(
E
[
X
])
2
= 4000021
.
02

4
.
032
2
= 4000004
.
763
3. (15 points) Car arriving at an intersection may either turn or go straight. The probability of
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 Spring '08
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