Applied Statistics and Probability for Engineers Solution.pdf

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Applied Statistics
and Probability
for Engineers
Third Edition
Douglas C. Montgomery
Arizona State University
George C. Runger
Arizona State University
John Wiley & Sons, Inc.
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Library of Congress Cataloging-in-Publication Data
Montgomery, Douglas C.
Applied statistics and probability for engineers / Douglas C. Montgomery, George C.
Runger.—3rd ed.
p. cm.
Includes bibliographical references and index.
ISBN 0-471-20454-4 (acid-free paper)
1. Statistics. 2. Probabilities. I. Runger, George C. II. Title.
QA276.12.M645 2002
519.5—dc21
2002016765
Printed in the United States of America.
10 9 8 7 6 5 4 3 2 1
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Preface
The purpose of this Student Solutions Manual is to provide you with additional help in under-
standing the problem-solving processes presented in Applied Statistics and Probability for
Engineers. The Applied Statistics text includes a section entitled “Answers to Selected
Exercises,” which contains the final answers to most odd-numbered exercises in the book.
Within the text, problems with an answer available are indicated by the exercise number
enclosed in a box.
This Student Solutions Manual provides complete worked-out solutions to a subset of the
problems included in the “Answers to Selected Exercises.” If you are having difficulty reach-
ing the final answer provided in the text, the complete solution will help you determine the
correct way to solve the problem.
Those problems with a complete solution available are indicated in the “Answers to
Selected Exercises,” again by a box around the exercise number. The complete solutions to
this subset of problems may also be accessed by going directly to this Student Solutions
Manual.
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Chapter 2 Selected Problem Solutions
Section 2-2
2-43. 3 digits between 0 and 9, so the probability of any three numbers is 1/(10*10*10);
3 letters A to Z, so the probability of any three numbers is 1/(26*26*26); The probability your license plate
is chosen is then (1/10 3 )*(1/26 3 ) = 5.7 x 10 -8
Section 2-3
2-49.
a) P(A') = 1- P(A) = 0.7
b) P ( AB
∪
) = P(A) + P(B) - P( AB
∩
) = 0.3+0.2 - 0.1 = 0.4
c) P(
AB ) + P( AB
′ ∩
∩
) = P(B). Therefore, P(
AB ) = 0.2 - 0.1 = 0.1
′ ∩
d) P(A) = P( AB
∩
) + P( AB
∩ ′
). Therefore, P( AB
∩ ′
) = 0.3 - 0.1 = 0.2
e) P(( AB
∪
)') = 1 - P( AB
∪
) = 1 - 0.4 = 0.6
f) P(
AB ) = P(A') + P(B) - P(
′ ∪
AB ) = 0.7 + 0.2 - 0.1 = 0.8
′ ∩
Section 2-4
2-61.
Need data from example
a) P(A) = 0.05 + 0.10 = 0.15
b) P(A|B) =
P
(
A
B
B
)
=
0
.
04
+
0
07
=
0
.
153
P
(
)
0
.
72
c) P(B) = 0.72
d) P(B|A) =
P
(
A
B
B
)
=
0
.
04
+
0
07
=
0
.
733
P
(
)
0
15
e) P(A
∩
B) = 0.04 +0.07 = 0.11
f) P(A
∪
B) = 0.15 + 0.72 – 0.11 = 0.76
2-67. a) P(gas leak) = (55 + 32)/107 = 0.813
b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632
c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764
Section 2-5
2-73.
Let F denote the event that a roll contains a flaw.
Let C denote the event that a roll is cotton.
P
(
F
)
=
P
(
F
C
)
P
(
C
)
+
P
(
F
C
′
)
P
(
C
)
=
(
0
02
)(
0
70
)
+
(
0
03
)(
0
30
)
=
0
023
2-79.
Let A denote a event that the first part selected has excessive shrinkage.
Let B denote the event that the second part selected has excessive shrinkage.
a) P(B)= P( BA )P(A) + P( BA ')P(A')
= (4/24)(5/25) + (5/24)(20/25) = 0.20
b) Let C denote the event that the second part selected has excessive shrinkage.
PC PCA BPA B PCA B PA B
( )
= ∩∩+ ∩∩
(
) (
) (
') (
')
+ ∩∩+ ∩∩
PCABPAB PCABPAB
( ' ) ( ' ) ( ' ') ( ' ')
=
3
23
2
24
5
25
+
4
23
20
24
5
25
+
4
23
5
24
20
25
+
5
23
19
24
20
25
=
020
.
′
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Section 2-6
2-87.
It is useful to work one of these exercises with care to illustrate the laws of probability. Let H i denote the
event that the ith sample contains high levels of contamination.
a) PH H H H H
(
1
∩∩∩∩=
2
3
4
5
) ()()()()()
PH PH PH PH PH
1
2
3
4
5
by independence. Also, PH() .
'
=
09 . Therefore, the answer is 09 059
.
5
=
.
b) A HHHHH
1
=∩∩∩∩
(
2345
'
'
'
)
1
A HHHHH
2
=∩∩∩∩
(
1
345
'
'
)
2
A HHHHH
3
=∩∩∩∩
(
1
'
2
45
'
)
3
A HHHHH
4
=∩∩∩∩
(
1
'
2
3
5
)
4
A HHHHH
5
=∩∩∩∩
(
1
'
2
3
4
)
5
The requested probability is the probability of the union AAAAA
12 3 4 5
∪∪∪∪
and these events
are mutually exclusive. Also, by independence PA() . (.) .
=
09 01 00656
4
=
. Therefore, the answer is
5(0.0656) = 0.328.
c) Let B denote the event that no sample contains high levels of contamination. The requested
probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41.
2-89.
Let A denote the event that a sample is produced in cavity one of the mold.
a) By independence, PA A A A A
(
1
∩∩∩∩= =
2
3
4
5
) ( )
1
8
5
0 00003
.
b) Let B i be the event that all five samples are produced in cavity i. Because the B's are mutually
exclusive, PB B
(
... ) ( ) ( ) ... ( )
12 8 1 2
∪∪∪= + ++
B
PB PB
PB
8
From part a., PB() ()
=
1
8
5 . Therefore, the answer is 8
5
() .
1
8
=
0 00024
c) By independence, PAAAAA
(
1
∩∩∩∩=
2
3
4
5
) ()()
1
8
4
7
8
. The number of sequences in
which four out of five samples are from cavity one is 5. Therefore, the answer is 5
4
()() .
1
8
7
8
=
0 00107
.
Section 2-7
2-97.
Let G denote a product that received a good review. Let H, M, and P denote products that were high,
moderate, and poor performers, respectively.
a)
PG PGHPH PGMPM PGPPP
( ) ( )( ) ( )( ) ( )()
. (. ) . (. ) . (. )
.
=
+
+
=
0 95 0 40
+
0 60 0 35
+
0 10 0 25
0 615
b) Using the result from part a.,
PHG
=
(
)
=
PGHPH
PG
(
) ( )
()
=
095040
0 615
. (. )
.
=
0618
.
c) PHG
(
' )
=
PG HPH
PG
(' )()
(' )
=
005040
10 5
. (. )
.
−
=
0052
.
Supplemental
2-105. a) No, P(E 1
∩
E 2
∩
E 3 )
≠
0
b) No, E 1 ′
∩
E 2 ′
is not
∅
c) P(E 1 ′
∪
E 2 ′
∪
E 3 ′
) = P(E 1 ′
) + P(E 2 ′
) + P(E 3 ′
) – P(E 1 ′∩
E 2 ′
) - P(E 1 ′∩
E 3 ′
) - P(E 2 ′∩
E 3 ′
)
+ P(E 1 ′
∩
E 2 ′
∩
E 3 ′
)
= 40/240
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