ECSE 2500: Engineering Probability
Exam 2 Solutions
April 13, 2010
1.
Solution:
(a)
X
is binomial random variable with
n
= 5 and
p
= 0
.
3
P
[
X
=
k
] =
5
k
!
(0
.
3)
k
(0
.
7)
5

k
, k
= 0
,
1
,...,
5
P
[
X
≥
2] = 1

P
[
X
= 0]

P
[
X
= 1] = 0
.
4718
(b)
E
[
X
] =
np
= 5(0
.
3) = 1
.
5
(c)
V ar
[
X
] =
np
(1

p
) = 5(0
.
3)(1

0
.
3) = 1
.
05
2.
Solution:
(a)
X
is Poisson random variable with
α
= 1(
call per minute
)
×
10(
minutes
) = 10. The
pmf of
X
is
P
[
X
=
k
] =
10
k
k
!
e

10
, k
= 0
,
1
,...
(b)
E
[
X
] =
α
= 10
(c)
V ar
[
X
] =
α
= 10
1
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F
X
(
x
) =
P
[
X
≤
x
] =
0
x <
0
b
x
c
X
k
=0
10
k
k
!
e

10
x
≥
0
where
b
x
c
is the largest integer less than or equal to
x
.
(e)
P
[
X
= 0] =
10
0
0!
e

10
=
1
1
e

10
=
e

10
= 4
.
54
×
10

5
3.
Solution:
(a) For uniform random variable
X
,
f
X
(
x
) =
(
1
2

1
< x <
1
0
otherwise
(b)
F
X
(
x
) =
0
x
≤ 
1
x
+1
2

1
< x <
1
1
x
≥
1
(c)
E
[
X
] =
1 + (

1)
2
= 0
(d)
V ar
[
X
] =
(1

(

1))
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 Spring '11
 Radke
 Poisson Distribution, Probability theory, Binomial distribution, Discrete probability distribution

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