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ISyE 3770 – Fall 2007
Homework #5 (Covers Modules 16–18) — Solutions
1. (Hines et al., 4–1). A refrigerator manufacturer subjects his ﬁnished products to
a ﬁnal inspection. Of interest are two categories of defects: scratches or ﬂaws in
the porcelain ﬁnish, and mechanical defects. The number of each type of defects
is a random variable. The results of inspecting 50 refrigerators are shown in the
following joint p.m.f. table, where
X
represents the occurrence of ﬁnish defects and
Y
represents the occurrence of mechanical defects.
Y
\
X
0
1
2
3
4
5
0
11/50 4/50 2/50 1/50 1/50 1/50
1
8/50
3/50 2/50 1/50 1/50
2
4/50
3/50 2/50 1/50
3
3/50
1/50
4
1/50
(a) Find the marginal distributions of
X
and
Y
.
Solution:
Let’s rewrite the table, this time including the marginals.
Y
\
X
0
1
2
3
4
5
f
Y
(
y
)
0
11/50
4/50
2/50 1/50 1/50 1/50
20/50
1
8/50
3/50
2/50 1/50 1/50
15/50
2
4/50
3/50
2/50 1/50
10/50
3
3/50
1/50
4/50
4
1/50
1/50
f
X
(
x
)
27/50 11/50 6/50 3/50 2/50 1/50
♦
(b) Find the marginal distribution of mechanical defects, given that there are no
ﬁnish defects.
Solution:
f
(
y

X
= 0) =
f
(0
,y
)
f
X
(0)
=
f
(0
,y
)
27
/
50
=
11
/
27 if
y
= 0
8
/
27
if
y
= 1
4
/
27
if
y
= 2
3
/
27
if
y
= 3
1
/
27
if
y
= 4
0
otherwise
♦
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(c) Find the marginal distribution of ﬁnish defects, given that there are no me
chanical defects.
Solution:
f
(
x

Y
= 0) =
f
(
x,
0)
f
Y
(0)
=
f
(
x,
0)
20
/
50
=
11
/
20 if
x
= 0
4
/
20
if
x
= 1
2
/
20
if
x
= 2
1
/
20
if
x
= 3
,
4
,
5
0
otherwise
♦
2. (Hines et al., 4–4). Consider a situation in which the surface tension and acidity
of a chemical product are measured. These variables are coded such that surface
tension is measured on a scale 0
≤
X
≤
2, and acidity is measured on a scale
2
≤
Y
≤
4. The probability density function of (
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This note was uploaded on 04/14/2008 for the course ISYE 3770 taught by Professor Goldsman during the Fall '07 term at Georgia Institute of Technology.
 Fall '07
 goldsman

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