Sample Solution manual of Stochastic Modeling and Mathematical Statistics : A Text for Statisticians and Quantitative Scientists PDF

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Authors: Francisco J. Samaniego
Published: Chapman & Hall/CRC 2014
Edition: 1st
Pages: 119
Type: pdf
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SOLUTIONS MANUAL FOR STOCHASTIC MODELING AND MATHEMATICAL STATISTICS A Text for Statisticians and Quantitative Scientists by

Francisco J. Samaniego

A CHAPMAN & HALL BOOK

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SOLUTIONS MANUAL FOR STOCHASTIC MODELING AND MATHEMATICAL STATISTICS A Text for Statisticians and Quantitative Scientists by

Francisco J. Samaniego

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

A CHAPMAN & HALL BOOK

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20140114 International Standard Book Number-13: 978-1-4665-6052-9 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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Contents Chapter 1 problems

2

Chapter 2 problems

15

Chapter 3 problems

23

Chapter 4 problems

35

Chapter 5 problems

46

Chapter 6 problems

50

Chapter 7 problems

60

Chapter 8 problems

75

Chapter 9 problems

80

Chapter 10 problems

87

Chapter 11 problems

102

Chapter 12 problems

109

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Chapter 1 problems 1.8.2 n [

Ai

i=1

!∁

n \

=

Ai∁

i=1

Let Ai∁ = Bi . Then, Bi∁ = (A∁i )∁ = Ai . Then, it follows that n [

B ∁i

i=1

!∁

n \

=

Bi .

i=1

By taking the complement on both sides,  

n [

B i∁

i=1

!∁ ∁

 =

n [

Bi∁

i=1

=

n \

Bi

i=1

!∁

.

Since Bi is arbitrary, we have shown that n \

i=1

Ai

!∁

=

n [

Ai∁

i=1

by replacing Bi with Ai above.

1.8.4 (a) Let Wi denote a win in the i-th outcome. Similarly, let Li denote a loss in the i-th outcome. Then, P (at least one win) = P (W1 ) + P (L1 )P (W2 ) =

75 37 1 1 = 2 = 0.0519. + 38 38 38 38

(b) P (win) =

75 76 2 = 2 > 2 = P (at least one win). 38 38 38

One way to look at the reason why we have a larger probability in (b) is as follows. P (losing in (b)) =

36 = 0.9474 < 0.9481 = 38



37 38

2

= P (losing in (a)).

1.8.6 (a) P (two aces) =

1752 = 0.0132. 132600

(b) Let A = Ace, F = Face card, and N = Neither. Possible outcomes for at least one A and one F . AAF NAF AFA NFA FAA ANF AFF AFN FAF FAN @solutionmanual1 FFA FNA

https://gioumeh.com/product/stochastic-modeling-and-mathematical-statistics-solution P (At least one A and one F ) 144 + 528 + 10368 11040 = = ≈ 0.0833. 132600 132600

= = =

P (A before F |At least one A and F selected) P (A before F ∩ At least one A and F selected) P (At least one A and F selected)   4·3·2  12·11·10  ·36 2 52·51·50 + 52·51·50 + 3 524·12 ·51·50 132600 6000 ≈ 0.04525. 132600

1.8.8 P

n \

i=1

A∁i

!

=

" #∁  n \ 1−P  Ai∁  i=1

=

n [

1−P

Ai

i=1

≥

1−

n X

!

P (Ai ) by countable subadditivity property.

i=1

1.8.10 P (largest is 4) =

1.8.12

4 2  33 3 3 0 − 3 0 = = 0.15. 6 6 20 3 3

Let C be cough, and S be sneeze. Then, we have the following probability table. EVENT C C∁ TOTAL

S 0.18 0.12 0.30

S∁ 0.22 0.48 0.70

TOTAL 0.40 0.60 1.00

Therefore, P (sneeze or cough, but not both) =

P (S ∩ C ∁ ) + P (S ∁ ∩ C) = 0.12 + 0.22 = 0.34.

Or, equivalently, P (sneeze or cough, but not both) =

P (S ∪ C) + P (S ∩ C) = 0.52 − 0.18 = 0.34.

1.8.14 P (At least one six in four rolls) =

1 − P (No six in four rolls)  4 5 ≈ 0.5177. = 1− 6

Also, P (At least one double six in twenty-four rolls) =

1 − P (No double six in twenty-four rolls)  24 35 ≈ 0.4914. = 1− 36

So, at least one six in four rolls is more likely to happen.

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https://gioumeh.com/product/stochastic-modeling-and-mathematical-statistics-solution 1.8.16 4 3 2 2

=

1

1

+

43 2 4 32 1 2 1 + 1 1 2 9 4

4 36 + 24 + 12 = ≈ 0.5714. 9 · 8 · 7 · 6/24 7

1.8.18 The basic approach is to consider 24 permutations of 4 hats. Person

Hat

1 2 2 2 3 3 3 4 4 4

2 1 3 4 1 4 4 3 3 1

3 4 4 1 4 2 1 1 2 2

4 3 1 3 2 1 2 2 1 3

Therefore, P (No match) =

9 = 0.375. 24

1.8.20 P (A|A ∪ B) =

P (A) P (A ∪ B)

and for a mutually exclusive event, P (A ∪ B) = P (A) + P (B). Thus, P (A|A ∪ B) =

P (A) . P (A) + P (B)

1.8.22       3 1 1 2 1 + = = 0.5. P (B) = 6 2 3 2 3                 1 1 1 2 1 1 2 2 1 1 1 1 + + + P (C) = 2 3 4 2 3 2 2 3 3 2 3 6 25 3 + 12 + 8 + 2 = ≈ 0.3472. =   72    72 1 1 1 1 P (A ∩ B ∩ C) = ≈ 0.0417. = 24 2 3 4 P (A|B ∩ C) = P (A ∩ B|C) = P (A|C) =

1  1 24    1 + 2 23 32 24 1 3 P (A ∩ B ∩ C) 24 = 0.12. = 25 = 25 P (C) 72 1 + 4 P (A ∩ C) 3 = 24 25 24 = = 0.6. P (C) 5 72

P (A ∩ B ∩ C) = P (B ∩ C)

=

1 24

1 24

@solutionmanual1

+

2 9

=

3 ≈ 0.1579. 19

https://gioumeh.com/product/stochastic-modeling-and-mathematical-statistics-solution 1.8.24 P (B) = 0.6,

P (L) = 0.3,

P (T ) = 0.1,

P (D|B) = 0.2,

P (D|L) = 0.3,

P (D|T ) = 0.9.

Therefore, (a) P (D) = P (D|B)P (B) + P (D|L)P (L) + P (D|T )P (T ) = 0.2 × 0.6 + 0.3 × 0.3 + 0.9 × 0.1 = 0.3. (b) P (B|D) =

P (D|B)P (B) 0.12 P (B ∩ D) = = = 0.4. P (D) P (D) 0.3

(c) Let Di be the event that i of them are detained. Then, P (D4 ∪ D5 ) =

P (D4 ) + P (D5 )  5 = [P (D|T )]4 [1 − P (D|T )] + [P (D|T )]5 4 = 5(0.9)4 (0.1) + (0.9)5 = 0.91854.

1.8.26 Let K be the event that he knows the answer, and R be the event that he gets the answer right. Then, P (K ) = 0.7,

P (K ∁ ) = 0.3,

P (R|K ) = 1,

P (R|K ∁ ) = 0.2.

(a) P (R) =

P (R|K )P (K ) + P (R|K ∁ )P (K ∁ ) = 1 × 0.7 + 0.2 × 0.3 = 0.76.

(b) P (K |R) =

P (K ∩ R) P (R|K )P (K ) 35 0.7 ≈ 0.9211. = = = 0.76 38 P (R) P (R)

(c) Let R2 be the event that he gets exactly two of the first three questions right. Then,   3 [P (R)]2 [1 − P (R)] P (R2 ) = 2   3 (0.76)2 (0.24) = 0.415872. = 2

1.8.28 Let D be an event such that the widget is defective. Then, P (D|A) = 0.2,

P (D|B) = 0.3,

P (D|C) = 0.5,

P (A) = 0.5,

P (B) = 0.3,

P (C) = 0.2.

(a) P (D) = =

P (D|A)P (A) + P (D|B)P (B) + P (D|C)P (C) 0.2 × 0.5 + 0.3 × 0.3 + 0.5 × 0.2 = 0.29.

(b) P (A|D) = P (B|D) = P (C|D) =

P (A ∩ D) 10 0.2 × 0.5 P (D|A)P (A) = = = ≈ 0.3448. 0.29 P (D) P (D) 29 9 0.3 × 0.3 P (B ∩ D) P (D|B)P (B) = = = ≈ 0.3103. 0.29 P (D) P (D) 29 P (C ∩ D) P (D|C )P (C) 10 0.5 × 0.2 = = = ≈ 0.3448. 0.29 P (D) P@solutionmanual1 (D) 29

https://gioumeh.com/product/stochastic-modeling-and-mathematical-statistics-solution 1.8.30 Let J = {have Jacuzzis} and N J = {have no Jacuzzis}. Then, (a) P (J ) = = =

P (J ∩ LBI) + P (J ∩ LW BI) P (LBI )P (J |LBI) + P (LW BI )P (J |LW BI)       2 1 2 1 1 + = = 0.4. 5 3 5 3 2

(b) P (LBI |N J ) =

P (N J ∩ LBI) = P (N J )

4 15 1 3

+

4 15

=

4 ≈ 0.444. 9

(c)  3   3 2 54 P (J in the 4-th visit) = [P (N J )] P (J ) = = 0.0864. = 5 625 5 3

1.8.32 Let T D be the event that Tom beats Dick, T H be the event that Tom beats Harry, and DH be the event that Dick beats Harry. Then, we have the following two possible outcomes (E1 and E2 ) so that each player wins (W) one and loses (L) one. E1 Tom Dick Harry

Tom L W

Dick W L

Harry L W -

E2 Tom Dick Harry

Tom W L

Dick L W

Harry W L -

Thus, P ({Each player wins one and loses one}) = = = =

P (E1 ) + P (E2 ) P (T D)P (T H ∁ )P (DH ) + P (T D∁ )P (T H )P (DH ∁ )         1 3 1 1 1 2 + 3 2 4 3 2 4 7 ≈ 0.2917. 24

1.8.34 Let P i = {Jim stays in prison}, P o = {Joe stays in prison}, and P e = {Jed stays in prison}. Assume P i , P o , and P e are equally likely. Suppose the guard reveals released prisoner if there is only one, and reveals each of two with probability 21 otherwise. Let “Joe” be the event that Jim is informed that Joe will be released and let “Jed” be the event that Jim is informed that Jed will be released.

P (P i |J oe) = P (P i |J ed) =

P (P i ∩ J oe) 1/6 1 = . = 1/2 P (J oe) 3 1 P (P i ∩ J ed) 1/6 = . = 1/2 3 P (J ed)

Thus, guard’s revelation does not alter Jim’s@solutionmanual1 chance of release.

https://gioumeh.com/product/stochastic-modeling-and-mathematical-statistics-solution 1.8.36 P (Two people have birthdays in the same month) = = =

1 − P (Nobody shares the same month)      11 12 − n + 1 12 · 1− 12 12 12 12! 11! . 1− =1− (12 − n)!12n−1 (12 − n)!12n

1.8.38 Think of the complementary event. That is, the first red ball is drawn before any white ball is drawn. Now, let Ri be the event that the first red ball is drawn at the i-th draw before any white ball is drawn. Then, the only possible balls drawn before the first red ball are the blue balls. Thus, P (The first red ball is drawn before any white ball is drawn) = = + + =

P (R1 ) + P (R2 ) + · · · + P (R7 )]             6 4 6 5 4 6 5 4 4 4 + + + 12 12 11 12 11 10 12 11 10 9              6 5 4 3 4 6 5 4 3 2 4 + 12 11 10 9 8 12 11 10 9 8 7         6 5 4 3 2 1 4 12 11 10 9 8 7 6 2 . 3

Therefore, P (A white ball is drawn before the first red ball is drawn) = 1 −

2 1 = . 3 3

1.8.40 Let C be “chew tobacco”, and B be “bad breath”. Then, we have the following probability table. EVENT B B∁ TOTAL

C 0.25 0.05 0.30

C∁ 0.15 0.55 0.70

TOTAL 0.40 0.60 1.00

(a) From the table, it follows that P (C ∩ B ∁ ) = 0.05, or P (C) = P (C ∩ B) + P (C ∩ B ∁ ), so P (C ∩ B ∁ ) = 0.3 − 0.25 = 0.05. (b) P (B|C ∁ ) =

0.15 P (B ∩ C ∁ ) ≈ 0.2142. = 0.17 P (C ∁ )

@solutionmanual1...