Estadística
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
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Preface
The purpose of this Student Solutions Manual is to provide you withadditional help in understanding 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. ThisStudent 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 reaching 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.
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 probabilityyour license plate -8 is chosen is then (1/103)*(1/263) = 5.7 x 10
Section 2-3 2-49. a) P(A') = 1- P(A) = 0.7 b) P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B ) = 0.3+0.2 - 0.1 = 0.4 c) P( A ′ ∩ B ) + P( A ∩ B ) = P(B). Therefore, P( A ′ ∩ B ) = 0.2 - 0.1 = 0.1 d) P(A) = P( A ∩ B ) + P( A ∩ B ′ ). Therefore, P( A ∩ B ′ ) = 0.3 - 0.1 = 0.2 e) P(( A ∪ B )') = 1 - P( A ∪ B ) = 1 - 0.4 = 0.6 f) P( A ′ ∪ B )= P(A') + P(B) - P( A ′ ∩ B ) = 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 P( A ∩ B) 0.04 + 0.07 b) P(A|B) = = = 0.153 P( B ) 0.72 c) P(B) = 0.72 P( A ∩ B) 0.04 + 0.07 d) P(B|A) = = = 0.733 P( B ) 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...
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