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SIEO 3600 (IEOR Majors)
Solution to Assignment #9
Introduction to Probability and Statistics
April 11, 2010
Solution to Assignment #9
(
Uniform, Exponential and Normal random variables
)
1. If
X
is a normal random variable with parameters
μ
= 10,
σ
2
= 26, compute
(a)
P
(
X >
5);
(b)
P
(4
< X <
16);
(c)
P
(
X <
8);
(d)
P
(
X <
20);
(e)
P
(
X >
16).
Hint:
express these probabilities in terms of the standard normal CDF function Φ and use
the tables at the back of the textbook.
Solution:
Let
Z
be a standard normal random variable, i.e.,
Z
∼
N
(0
,
1).
(a)
P
(
X >
5) =
P
±
X

μ
σ
>
5

μ
σ
²
=
P
±
Z >
5

10
√
26
²
= 1

Φ
±
5

10
√
26
²
= 0
.
8366.
(b)
P
(4
< X <
16) =
P
±
4

μ
σ
<
X

μ
σ
<
16

μ
σ
²
=
P
±
4

μ
σ
< Z <
16

μ
σ
²
= Φ
±
16

10
√
26
²

Φ
±
4

10
√
26
²
= 0
.
7607.
(c)
P
(
X <
8) =
P
±
X

μ
σ
<
8

μ
σ
²
=
P
±
Z <
8

10
√
26
²
= Φ
±
8

10
√
26
²
= 0
.
3474.
(d)
P
(
X <
20) =
P
±
X

μ
σ
<
20

μ
σ
²
=
P
±
Z <
20

10
√
26
²
= Φ
±
20

10
√
26
²
= 0
.
9751.
(e)
P
(
X >
16) =
P
±
X

μ
σ
>
16

μ
σ
²
=
P
±
Z >
16

10
√
26
²
= 1

Φ
±
16

10
√
26
²
= 0
.
1197.
2. The annual rainfall (in inches) in a certain region is normally distributed with
μ
= 40,
σ
= 4.
What is the probability that in 2 of the next 4 years the rainfall will exceed 50 inches? Assume
that the rainfalls in diﬀerent years are independent.
Solution:
Let
Z
be a standard normal, i.e.,
Z
∼
N
(0
,
1). We have that the probability of
rainfall exceeding 50 in one year is
P
(
X >
50) =
P
³
X

40
4
>
50

40
4
´
=
P
(
Z >
2
.
5) = 1

Φ(2
.
5) = 1

0
.
9938 = 0
.
0062
≡
p.
Let
N
be the number of years in which the rainfall exceed 50 inches in
n
= 4 years. Then it
is clear that
N
is a
binomial
random variable with parameter
n
= 4 and
p
= 0
.
0062. Hence,
the desired probability is
P
(
N
= 2) =
³
4
2
´
p
2
(1

p
)
2
.
3. A random variable
X
is said to have a
lognormal
distribution if
Z
≡
log
X
is normally
distributed. If
X
is lognormal with
E
[
Z
] =
μ
and var(
Z
) =
σ
2
, then
(a) determine the probability distribution function (PDF) of
X
.
(b) What is
P
(
X
≤
x
)?
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SIEO 3600, Solution to Assignment #9
Solution:
Since
Z
= log
X
, we must have
X
=
e
Z
, where
Z
∼
N
(
μ,σ
2
). We ﬁrst solve (b).
(b) For
x >
0, we compute the CDF of
X
as
F
X
(
x
) =
P
(
X
≤
x
) =
P
(
e
Z
≤
x
) =
P
(
Z
≤
log
x
) =
Z
log
x
∞
1
√
2
πσ
2
e

(
z

μ
)
2
2
σ
2
dz.
(a) To compute the PDF of
X
, we diﬀerentiate its CDF obtained in (b), i.e., by the chain
rule of calculus, we have
f
X
(
x
) =
d
dx
F
X
(
x
) =
1
√
2
πσ
2
e

(log
x

μ
)
2
2
σ
2
·
(log
x
)
0
=
1
√
2
πσ
2
e

(log
x

μ
)
2
2
σ
2
·
1
x
.
4. In the midterm of a probability course, the scores that students get follow a
uniform
distri
bution from 20 to 100. Due to popular grieving, the professor decides to rescale the results.
The formula he uses is that the curved score
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This note was uploaded on 10/03/2011 for the course SIEO W3600 taught by Professor Yunanliu during the Spring '10 term at Columbia.
 Spring '10
 YUNANLIU

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