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Introduction to Mathematical Statistics and Its Applications 6th Edition Larsen Solutions ManualFull Download: -edition-larseINSTRUCTOR’SSOLUTIONS MANUALA N I NTRODUCTION TOM ATHEMATICAL S TATISTICSAND I TS A PPLICATIONSSIXTH EDITIONRichard LarsenVanderbilt UniversityMorris MarxUniversity of West FloridaThis sample only, Download all chapters at: AlibabaDownload.com
The author and publisher of this book have used their best efforts in preparing this book. These efforts include thedevelopment, research, and testing of the theories and programs to determine their effectiveness. The author and publishermake no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in thisbook. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with,or arising out of, the furnishing, performance, or use of these programs.Reproduced by Pearson from electronic files supplied by the author.Copyright 2018, 2012, 2006 Pearson Education, Inc.Publishing as Pearson, 330 Hudson Street, NY NY 10013All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any formor by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of thepublisher. Printed in the United States of America.ISBN-13: 978-0-13-411427-9ISBN-10: 0-13-411427-2
ContentsChapter 2: Probability .12.22.32.42.52.62.7Samples Spaces and the Algebra of Sets . 1The Probability Function . 6Conditional Probability . 7Independence . 13Combinatorics . 17Combinatorial Probability . 24Chapter 3: Random Variables .273.23.33.43.53.63.73.83.93.103.113.12Binomial and Hypergeometric Probabilities . 27Discrete Random Variables . 36Continuous Random Variables . 41Expected Values . 44The Variance . 52Joint Densities . 57Transforming and Combining Random Variables. 69Further Properties of the Mean and Variance. 73Order Statistics . 79Conditional Densities . 82Moment-Generating Functions. 88Chapter 4: Special Distributions .934.24.34.44.54.6The Poisson Distribution . 93The Normal Distribution . 98The Geometric Distribution. 105The Negative Binomial Distribution . 107The Gamma Distribution . 109Chapter 5: Estimation.1115.25.35.45.55.65.75.8Estimating Parameters: The Method of Maximum Likelihoodand Method of Moments . 111Interval Estimation . 118Properties of Estimators . 123Minimum-Variance Estimators: The Cramér-Rao Lower Bound . 128Sufficient Estimators . 130Consistency . 133Bayesian Estimation . 135Copyright 2018 Pearson Education, Inc.iii
Chapter 6: Hypothesis Testing .1376.26.36.46.5The Decision Rule . 137Testing Binomial Data. 138Type I and Type II Errors . 139A Notion of Optimality: The Generalized Likelihood Ratio . 144Chapter 7: Inferences Based on the Normal Distribution .147Y S/ n. 1477.3Deriving the Distribution of7.4Drawing Inferences About . 1507.5Drawing Inferences About 2 . 156Chapter 8: Types of Data: A Brief Overview .1618.2Classifying Data . 161Chapter 9: Two-Sample Inference .1639.2Testing H 0 : X Y . 1639.39.49.5Testing H 0 : X2 Y2 —The F Test . 166Binomial Data: Testing H 0 : p X pY . 168Confidence Intervals for the Two-Sample Problem . 170Chapter 10: Goodness-of-Fit Tests .17310.210.310.410.5The Multinomial Distribution . 173Goodness-of-Fit Tests: All Parameters Known . 175Goodness-of-Fit Tests: Parameters Unknown . 178Contingency Tables . 185Chapter 11: Regression .18911.211.311.411.5The Method of Least Squares . 189The Linear Model . 199Covariance and Correlation . 204The Bivariate Normal Distribution . 208Chapter 12: The Analysis of Variance .21112.212.3The F test . 211Multiple Comparisons: Tukey’s Method. 214Copyright 2018 Pearson Education, Inc.iv
12.4Testing Subhypotheses with Constrasts . 21612.5Data Transformations . 218/( k 1)Appendix 12.A.2 The Distribution of SSTRSSE /( n k ) When H1 Is True . 218Chapter 13: Randomized Block Designs .22113.213.3The F Test for a Randomized Block Design . 221The Paired t Test. 224Chapter 14: Nonparametric Statistics.22914.214.314.414.514.6The Sign Test . 229Wilcoxon Tests . 232The Kruskal-Wallis Test . 236The Friedman Test. 239Testing for Randomness . 241Copyright 2018 Pearson Education, Inc.v
Chapter 2: ProbabilitySection 2.2: Sample Spaces and the Algebra of Sets2.2.1S ( s, s, s ), ( s, s, f ), ( s, f , s ), ( f , s, s ), ( s, f , f ), ( f , s, f ), ( f , f , s ), ( f , f , f ) A ( s, f , s), ( f , s, s) ; B ( f , f , f ) 2.2.2Let (x, y, z) denote a red x, a blue y, and a green z.Then A (2, 2,1), (2,1, 2), (1, 2, 2), (1,1,3), (1,3,1), (3,1,1) 2.2.3(1,3,4), (1,3,5), (1,3,6), (2,3,4), (2,3,5), (2,3,6)2.2.4There are 16 ways to get an ace and a 7, 16 ways to get a 2 and a 6, 16 ways to get a 3 and a5, and 6 ways to get two 4’s, giving 54 total.2.2.5The outcome sought is (4, 4). It is “harder” to obtain than the set {(5, 3), (3, 5), (6, 2), (2, 6)}of other outcomes making a total of 8.2.2.6The set N of five card hands in hearts that are not flushes are called straight flushes. These arefive cards whose denominations are consecutive. Each one is characterized by the lowestvalue in the hand. The choices for the lowest value are A, 2, 3, , 10. (Notice that an ace canbe high or low). Thus, N has 10 elements.2.2.7P {right triangles with sides (5, a, b): a2 b2 25}2.2.8A {SSBBBB, SBSBBB, SBBSBB, SBBBSB, BSSBBB, BSBSBB, BSBBSB, BBSSBB, BBSBSB,BBBSSB}2.2.9(a) S {(0, 0, 0, 0) (0, 0, 0, 1), (0, 0, 1, 0), (0, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1), (0, 1, 1, 0),(0, 1, 1, 1), (1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1, ), (1, 1, 0, 0), (1, 1, 0, 1),(1, 1, 1, 0), (1, 1, 1, 1, )}(b) A {(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0, )}(c) 1 k2.2.10(a) S {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}(b) {2, 3, 4, 5, 6, 8}2.2.11Let p1 and p2 denote the two perpetrators and i1, i2, and i3, the three in the lineup who areinnocent.Then S ( p1 , i1 ), ( p1 , i2 ), ( p1 , i3 ), ( p2 , i1 ), ( p2 , i2 ), ( p2 , i3 ), ( p1 , p2 ), (i1 , i2 ), (i1 , i3 ), (i2 , i3 ) .The event A contains every outcome in S except (p1, p2).2.2.12The quadratic equation will have complex roots—that is, the event A will occur—ifb2 4ac 0.2.2.13In order for the shooter to win with a point of 9, one of the following (countably infinite)sequences of sums must be rolled: (9,9), (9, no 7 or no 9,9), (9, no 7 or no 9, no 7 or no 9,9), Copyright 2018 Pearson Education, Inc.1
2Chapter 2: Probability2.2.14Let (x, y) denote the strategy of putting x white chips and y red chips in the first urn (whichresults in 10 x white chips and 10 y red chips being in the second urn). ThenS ( x, y ) : x 0,1,.,10, y 0,1,.,10, and 1 x y 19 . Intuitively, the optimal strategiesare (1, 0) and (9, 10).2.2.15Let Ak be the set of chips put in the urn at 1/2k minute until midnight. For example,A1 {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. Then the set of chips in the urn at midnight is (Ak {k 1}) .k 12.2.16move arrow on first figure raise B by 12.2.17If x2 2x 8, then (x 4)(x 2) 0 and A {x: 4 x 2}. Similarly, if x2 x 6, then(x 3)(x 2) 0 and B {x: 3 x 2). Therefore, A B {x: 3 x 2} andA B {x: 4 x 2}.2.2.18A B C {x: x 2, 3, 4}2.2.19The system fails if either the first pair fails or the second pair fails (or both pairs fail). Foreither pair to fail, though, both of its components must fail. Therefore,A (A11 A21) (A12 A22).2.2.20(a)(b)(c) empty set(d)2.2.21402.2.22(a) {E1, E2}2.2.23(a) If s is a member of A (B C) then s belongs to A or to B C. If it is a member of A orof B C, then it belongs to A B and to A C.Thus, it is a member of (A B) (A C).Conversely, choose s in (A B) (A C). If it belongs to A, then it belongs toA (B C). If it does not belong to A, then it must be a member of B C.In that case it also is a member of A (B C).(b) {S1, S2, T1, T2}(c) {A, I}Copyright 2018 Pearson Education, Inc.
Section 2.2: Sample Spaces and the Algebra of Sets3(b) If s is a member of A (B C) then s belongs to A and to B C. If it is a member of B,then it belongs to A B and, hence, (A B) (A C). Similarly, if it belongs to C, it isa member of (A B) (A C). Conversely, choose s in (A B) (A C). Then itbelongs to A. If it is a member of A B then it belongs to A (B C). Similarly, if itbelongs to A C, then it must be a member of A (B C).2.2.24Let B A1 A2 Ak. Then A1C A2C . AkC (A1 A2 Ak)C BC. Then theexpression is simply B BC S.2.2.25(a) Let s be a member of A (B C). Then s belongs to either A or B C (or both). If sbelongs to A, it necessarily belongs to (A B) C. If s belongs to B C, it belongs toB or C or both, so it must belong to (A B) C. Now, suppose s belongs to(A B) C. Then it belongs to either A B or C or both. If it belongs to C, it mustbelong to A (B C). If it belongs to A B, it must belong to either A or B or both, soit must belong to A (B C).(b) Suppose s belongs to A (B C), so it is a member of A and also B C. Then it isamember of A and of B and C. That makes it a member of (A B) C. Conversely, if sis a member of (A B) C, a similar argument shows it belongs to A (B C).2.2.26(a)(b)(c)(d)(e)2.2.27A is a subset of B.2.2.28(a)(b)(c)(d)(e)(f)2.2.29(a) B and C(b) B is a subset of A.2.2.30(a) A1 A2 A3(b) A1 A2 A3The second protocol would be better if speed of approval matters. For very important issues,the first protocol is superior.2.2.31Let A and B denote the students who saw the movie the first time and the second time,respectively. Then N(A) 850, N(B) 690, and N [( A B )C ] 4700(implying that N(A B) 1300). Therefore, N(A B) number who saw movie twice 850 690 1300 240.AC BC CCA B CA BC CC(A BC CC) (AC B CC) (AC BC C)(A B CC) (A BC C) (AC B C){0} {x: 5 x 10}{x: 3 x 5}{x: 0 x 7}{x: 0 x 3}{x: 3 x 10}{x: 7 x 10}Copyright 2018 Pearson Education, Inc.
4Chapter 2: Probability2.2.32(a)(b)2.2.33(a)(b)2.2.34(a)A (B C)(A B) C(b)A (B C)2.2.35(A B) CA and B are subsets of A B.Copyright 2018 Pearson Education, Inc.
Section 2.2: Sample Spaces and the Algebra of Sets2.2.365(a)( A B C ) C AC B(b)B ( A B )C AC B(c)A ( A B )C A B C2.2.37Let A be the set of those with MCAT scores 27 and B be the set of those with GPAs 3.5.We are given that N(A) 1000, N(B) 400, and N(A B) 300.Then N ( AC B C ) N [( A B )C ] 1200 N(A B) 1200 [(N(A) N(B) N(A B)] 1200 [(1000 400 300] 100. The requested proportion is 100/1200.2.2.38N(A B C) N(A) N(B) N(C) N(A B) N(A C) N(B C) N(A B C)2.2.39Let A be the set of those saying “yes” to the first question and B be the set of those saying“yes” to the second question. We are given that N(A) 600, N(B) 400, andN(AC B) 300. Then N(A B) N(B) N ( AC B ) 400 300 100. N ( A B C ) N(A) N(A B) 600 100 500.2.2.40N [( A B )C ] 120 N(A B) 120 [N( AC B) N(A B C ) N(A B)] 120 [50 15 2] 53Copyright 2018 Pearson Education, Inc.
6Chapter 2: ProbabilitySection 2.3: The Probability Function2.3.1Let L and V denote the sets of programs with offensive language and too much violence,respectively. Then P(L) 0.42, P(V) 0.27, and P(L V) 0.10.Therefore, P(program complies) P((L V)C) 1 [P(L) P(V) P(L V)] 0.41.2.3.2P(A or B but not both) P(A B) P(A B) P(A) P(B) P (A B) P(A B) 0.4 0.5 0.1 0.1 0.72.3.3(a) 1 P(A B)(b) P(B) P(A B)2.3.4P(A B) P(A) P(B) P(A B) 0.3; P(A) P(A B) 0.1. Therefore, P(B) 0.2.32.3.5 5 1No. P(A1 A2 A3) P(at least one “6” appears) 1 P(no 6’s appear) 1 . 6 2The Ai’s are not mutually exclusive, so P(A1 A2 A3) P(A1) P(A2) P(A3).2.3.6P(A or B but not both) 0.5 – 0.2 0.32.3.7By inspection, B (B A1) (B A2) (B An).2.3.8(a)(b)(b)Copyright 2018 Pearson Education, Inc.
Section 2.4: Conditional Probability72 3 8 42.3.9P(odd man out) 1 P(no odd man out) 1 P(HHH or TTT) 1 2.3.10A {2, 4, 6, , 24}; B {3, 6, 9, , 24); A B {6, 12, 18, 24}.12 84 16 .Therefore, P(A B) P(A) P(B) P(A B) 24 24 24 242.3.11Let A: State wins Saturday and B: State wins next Saturday. Then P(A) 0.10, P(B) 0.30,and P(lose both) 0.65 1 P(A B), which implies that P(A B) 0.35. Therefore,P(A B) 0.10 0.30 0.35 0.05, so P(State wins exactly once) P(A B) P(A B) 0.35 0.05 0.30.2.3.12Since A1 and A2 are mutually exclusive and cover the entire sample space, p1 p2 1.15But 3p1 p2 , so p2 .282.3.13Let F: female is hired and T: minority is hired. Then P(F) 0.60, P(T) 0.30, andP(FC TC) 0.25 1 P(F T). Since P(F T) 0.75, P(F T) 0.60 0.30 0.75 0.15.2.3.14The smallest value of P[(A B C)C] occurs when P(A B C) is as large as possible.This, in turn, occurs when A, B, and C are mutually disjoint. The largest value forP(A B C) is P(A) P(B) P(C) 0.2 0.1 0.3 0.6. Thus, the smallest value forP[(A B C)C] is 0.4.2.3.15(a) XC Y {(H, T, T, H), (T, H, H, T)}, so P(XC Y) 2/16(b) X YC {(H, T, T, T), (T, T, T, H), (T, H, H, H), (H, H, H, T)} so P(X YC) 4/162.3.16A {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}A BC {(1, 5), (3, 3), (5, 1)}, so P(A BC) 3/36 1/12.2.3.17A B, (A B) (A C), A, A B, S2.3.18Let A be the event of getting arrested for the first scam; B, for the second. We are givenP(A) 1/10, P(B) 1/30, and P(A B) 0.0025. Her chances of not getting arrested areP[(A B)C] 1 P(A B) 1 [P(A) P(B) P(A B)] 1 [1/10 1/30 0.0025] 0.869Section 2.4: Conditional Probability2.4.1P(sum 10 and sum exceeds 8)P (sum exceeds 8)P (sum 10)3 / 363 . P (sum 9, 10, 11, or 12) 4 / 36 3 / 36 2 / 36 1 / 36 10P(sum 10 sum exceeds 8) Copyright 2018 Pearson Education, Inc.
8Chapter 2: Probability2.4.2P(A B) P(B A) 0.75 P ( A B) P ( A B ) 10 P ( A B ) 5 P ( A B) , which implies4P( B)P ( A)that P(A B) 0.1.P( A B) P ( A) , then P(A B) P(A) P(B).P( B)P ( A B) P ( A) P( B) It follows that P(B A) P(B).P ( A)P ( A)2.4.3If P(A B) 2.4.4P(E A B) 2.4.5The answer would remain the same. Distinguishing only three family types does not makethem equally likely; (girl, boy) families will occur twice as often as either (boy, boy) or (girl,girl) families.2.4.6P(A B) 0.8 and P(A B) P(A B) 0.6, so P(A B) 0.2.P( A B)0.2 11 2Also, P(A B) 0.6 , so P(B) and P(A) 0.8 0.2 .0.6 33 3P( B)2.4.7Let Ri be the event that a red chip is selected on the ith draw, i 1, 2.3 1 3Then P(both are red) P(R1 R2) P(R2 R1)P(R1) .4 2 82.4.8P(A B) 2.4.9Let Wi be the event that a white chip is selected on the ith draw, i 1,2 .P(W1 W2 )Then P(W2 W1) . If both chips in the urn are white, P(W1) 1;P (W1 )1if one is white and one is black, P(W1) .21 1 1 3Since each chip distribution is equally likely, P(W1) 1 .2 2 2 41 1 1 55/8 5 .Similarly, P(W1 W2) 1 , so P(W2 W1) 2 4 2 83/ 4 62.4.10P[(A B) (A B)C] 2.4.11(a) P(AC BC) 1 P(A B) 1 [P(A) P(B) P(A B)] 1 [0.65 0.55 0.25] 0.05P ( E ( A B ))P( E )P ( A B ) P( A B) 0.4 0.1 3 .0.44P( A B)P( A B)P( A B)P ( A B) P ( A) P ( B ) P( A B) a b P ( A B ). P( B)P( B)ba b 1.But P(A B) 1, so P(A B) bP[( A B ) ( A B )C ]P ( ) 0CP[( A B ) ]P[( A B )C ]Copyright 2018 Pearson Education, Inc.
Introduction to Mathematical Statistics and Its Applications 6th Edition Larsen Solutions ManualFull Download: -edition-larseSection 2.4: Conditional Probability9(b) P[(AC B) (A BC)] P(AC B) P(A BC) [P(A) P(A B)] [P(B) P(A B)] [0.65 0.25] [0.55 0.25] 0.70(c) P(A B) 0.95(d) P[(A B)C] 1 P(A B) 1 0.25 0.75P[( AC B) ( A B C )](e) P{[(AC B) (A BC)] A B} 0.70/0.95 70/95P( A B)(f) P(A B) A B) P(A B)/P(A B) 0.25/0.95 25/95(g) P(B AC) P(AC B)/P(AC) ] [P(B) P(A B)]/[1 P(A)] [0.55 0.25]/[1 0.65] 30/352.4.12P(No. of heads 2 No. of heads 2) P(No. of heads 2 and No. of heads 2)/P(No. of heads 2) P(No. of heads 2)/P(No. of heads 2) (3/8)/(7/8) 3/72.4.13P(first die 4 sum 8) P(first die 4 and sum 8)/P(sum 8) P({(4, 4), (5, 3), (6, 2)}/P({(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}) 3/52.4.14There are 4 ways to choose three aces (count which one is left out). There are 48 ways tochoose the card that is not an ace, so there are 4 48 192 sets of cards where exactly threeare aces. That gives 193 sets where there are at least three aces. The conditional probability is(1/270,725)/(193/270,725) 1/193.2.4.15First note that P(A B) 1 P[(A B)C] 1 0.2 0.8.Then P(B) P(A B) P(A BC) P(A B) 0.8 0.3 0.1 0.5.Finally P(A B) P(A B)/P(B) 0.1/0.5 1/52.4.16P(A B) 0.5 implies P(A B) 0.5P(B). P(B A) 0.4 implies P(A B) (0.4)P(A).Thus, 0.5P(B) 0.4P(A) or P(B) 0.8P(A).Then, 0.9 P(A) P(B) P(A) 0.8P(A) or P(A) 0.9/1.8 0.5.2.4.17P[(A B)C] P[(A B)C] P(A BC) P(AC B) 0.2 0.1 0.3 0.6P(A B (A B)C) P[(A BC) (AC B)]/P((A B)C) [0.1 0.3]/0.6 2/32.4.18P(sum 8 at least one die shows 5) P(sum 8 and at least one die shows 5)/P(at least one die shows 5) P({(5, 3), (5, 4), (5, 6), (3, 5), (4, 5), (6, 5), (5, 5)})/(11/36) 7/112.4.19P(Outandout wins Australian Doll and Dusty Stake don’t win) P(Outandout wins and Australian Doll and Dusty Stake don’t win)/P(Australian Doll andDusty Stake don’t win) 0.20/0.55 20/552.4.20Suppose the guard will randomly choose to name Bob or Charley if they are the two to gofree. Then the probability the guard will name Bob, for example, isP(Andy, Bob) (1/2)P(Bob, Charley) 1/3 (1/2)(1/3) 1/2.The probability Andy will go free given the guard names Bob is P(Andy, Bob)/P(Guardnames Bob) (1/3)/(1/2) 2/3. A similar argument holds for the guard naming Charley.Andy’s concern is not justified.Copyright 2018 Pearson Education, Inc.This sample only, Download all chapters at: AlibabaDownload.com
INSTRUCTOR’S SOLUTIONS MANUAL AN INTRODUCTION TO MATHEMATICAL STATISTICS AND ITS APPLICATIONS SIXTH EDITION Richard Larsen Vanderbilt University Morris Marx University of West Florida Introduction to Mathematical Statistics and Its Applications 6th Edition Larsen Solutions Manual
*Marx. “On the Jewish Question.” Pp. 26-52. Avineri. “Marx and Jewish Emancipation.” Pp. 445-450. 2/2 “Marxism’s” genesis *Marx. “Economic and Philosophic Manuscripts of 1844.” Pp. 66-105. zank. “On the Origin of Species- eing: Marx Redefined.” Pp. 316-323. 2/9 The materialist conception of history *Marx. “Theses on .
years ago, David McLellan, a prolific student of Marx and Marxism, published a very good introduction to Marx’s life and thought. 2 He justified his book in noting it was the first since Mehring’s biography in 1918 and in the mean-time the Marx–Engels correspondence as well as several of Marx’s unpub-lished writings had become available.
– “Pulsed Power Formulary,” Richard J. Adler, North Star Power Engineering, 2001 edition, available . time interval between the beginning of one pulse and the beginning of the next . Marx – Basic Marx – Solid state Marx – Inversion generator – Stacked Blumlein – PFN Marx
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standard editions, of twelve essays by William Morris. The Introduction and Biographical Note by Norman Kelvin were prepared for the Dover edition. Library of Congress Cataloging-in-Publication Data Morris, William, 1834-1896. William Morris on art and socialism / William Morris ; edited and with an introduction by Norman Kelvin, p. cm.
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