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Solution Manual for Probability and Statistical Inference, 10th Edition

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    S OLUTIONS M ANUAL P ROBABILITY AND S TATISTICAL I NFERENCE T ENTH E DITION Robert V. Hogg Elliot A. Tanis Dale L. Zimmerman

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    iii Contents Preface v 1 Probability 1 1.1 Properties of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Methods of Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.4 Independent Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.5 Bayes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2 Discrete Distributions 7 2.1 Random Variables of the Discrete Type . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Mathematical Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.3 Special Mathematical Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.4 The Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 The Hypergeometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.6 The Negative Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.7 The Poisson Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3 Continuous Distributions 19 3.1 Random Variables of the Continuous Type . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.2 The Exponential, Gamma, and Chi-Square Distributions . . . . . . . . . . . . . . . . . 26 3.3 The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.4 Additional Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4 Bivariate Distributions 33 4.1 Bivariate Distributions of the Discrete Type . . . . . . . . . . . . . . . . . . . . . . . . 33 4.2 The Correlation Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.3 Conditional Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.4 Bivariate Distributions of the Continuous Type . . . . . . . . . . . . . . . . . . . . . . 37 4.5 The Bivariate Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 5 Distributions of Functions of Random Variables 45 5.1 Functions of One Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 5.2 Transformations of Two Random Variables . . . . . . . . . . . . . . . . . . . . . . . . 47 5.3 Several Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 5.4 The Moment-Generating Function Technique . . . . . . . . . . . . . . . . . . . . . . . 53 5.5 Random Functions Associated with Normal Distributions . . . . . . . . . . . . . . . . 55 5.6 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 5.7 Approximations for Discrete Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 59 5.8 Chebyshev’s Inequality and Convergence in Probability . . . . . . . . . . . . . . . . . 61 5.9 Limiting Moment-Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 62

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    iv Contents 6 Point Estimation 63 6.1 Descriptive Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 6.2 Exploratory Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 6.3 Order Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 6.4 Maximum Likelihood Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 6.5 A Simple Regression Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 6.6 Asymptotic Distributions of Maximum Likelihood Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 6.7 Sufficient Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 6.8 Bayesian Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 7 Interval Estimation 87 7.1 Confidence Intervals for Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 7.2 Confidence Intervals for the Difference of Two Means . . . . . . . . . . . . . . . . . . . 88 7.3 Confidence Intervals For Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 7.4 Sample Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 7.5 Distribution-Free Confidence Intervals for Percentiles . . . . . . . . . . . . . . . . . . . 92 7.6 More Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 7.7 Resampling Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 8 Tests of Statistical Hypotheses 107 8.1 Tests About One Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 8.2 Tests of the Equality of Two Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 8.3 Tests for Variances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 8.4 Tests about Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 8.5 Some Distribution-Free Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 8.6 Power of a Statistical Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 8.7 Best Critical Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 8.8 Likelihood Ratio Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 9 More Tests 127 9.1 Chi-Square Goodness-of-Fit Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 9.2 Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 9.3 One-Factor Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 9.4 Two-Way Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 9.5 General Factorial and 2 k Factorial Designs . . . . . . . . . . . . . . . . . . . . . . . . . 135 9.6 Tests Concerning Regression and Correlation . . . . . . . . . . . . . . . . . . . . . . . 136 9.7 Statistical Quality Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

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    Chapter 1 Probability 1 Chapter 1 Probability 1.1 Properties of Probability 1.1-2 Sketch a figure and fill in the probabilities of each of the disjoint sets. Let A = { insure more than one car } , P ( A ) = 0 . 85. Let B = { insure a sports car } , P ( B ) = 0 . 23. Let C = { insure exactly one car } , P ( C ) = 0 . 15. It is also given that P ( A ∩ B ) = 0 . 17. Since A ∩ C = φ , P ( A ∩ C ) = 0. It follows that P ( A ∩ B ∩ C ′ ) = 0 . 17. Thus P ( A ′ ∩ B ∩ C ′ ) = 0 . 06 and P ( B ′ ∩ C ) = 0 . 09. 1.1-4 (a) S = { HHHH , HHHT , HHTH , HTHH , THHH , HHTT , HTTH , TTHH , HTHT , THTH , THHT , HTTT , THTT , TTHT , TTTH , TTTT } ; (b) (i) 5/16, (ii) 0, (iii) 11/16, (iv) 4/16, (v) 4/16, (vi) 9/16, (vii) 4/16. 1.1-6 (a) P ( A ∪ B ) = 0 . 5 + 0 . 6 − 0 . 4 = 0 . 7; (b) A = ( A ∩ B ′ ) ∪ ( A ∩ B ) P ( A ) = P ( A ∩ B ′ ) + P ( A ∩ B ) 0 . 5 = P ( A ∩ B ′ ) + 0 . 4 P ( A ∩ B ′ ) = 0 . 1; (c) P ( A ′ ∪ B ′ ) = P [( A ∩ B ) ′ ] = 1 − P ( A ∩ B ) = 1 − 0 . 4 = 0 . 6. 1.1-8 Let A = { lab work done } , B = { referral to a specialist } , P ( A ) = 0 . 41 , P ( B ) = 0 . 53 , P ([ A ∪ B ] ′ ) = 0 . 21. P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) 0 . 79 = 0 . 41 + 0 . 53 − P ( A ∩ B ) P ( A ∩ B ) = 0 . 41 + 0 . 53 − 0 . 79 = 0 . 15 . 1.1-10 A ∪ B ∪ C = A ∪ ( B ∪ C ) P ( A ∪ B ∪ C ) = P ( A ) + P ( B ∪ C ) − P [ A ∩ ( B ∪ C )] = P ( A ) + P ( B ) + P ( C ) − P ( B ∩ C ) − P [( A ∩ B ) ∪ ( A ∩ C )] = P ( A ) + P ( B ) + P ( C ) − P ( B ∩ C ) − P ( A ∩ B ) − P ( A ∩ C ) + P ( A ∩ B ∩ C ) . 1.1-12 (a) 1/3; (b) 2/3; (c) 0; (d) 1/2.

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    2 Section 1.2 Methods of Enumeration 1.1-14 P ( A ) = 2[ r − r ( √ 3 / 2)] 2 r = 1 − √ 3 2 . 1.1-16 Note that the respective probabilities are p 0 , p 1 = p 0 / 4 , p 2 = p 0 / 4 2 , · · · . ∞ ∑ k =0 p 0 4 k = 1 p 0 1 − 1 / 4 = 1 p 0 = 3 4 1 − p 0 − p 1 = 1 − 15 16 = 1 16 . 1.2 Methods of Enumeration 1.2-2 (a) (4)(5)(2) = 40; (b) (2)(2)(2) = 8. 1.2-4 (a) 4 ( 6 3 ) = 80; (b) 4(2 6 ) = 256; (c) (4 − 1 + 3)! (4 − 1)!3! = 20. 1.2-6 S = { DDD, DDFD, DFDD, FDDD, DDFFD, DFDFD, FDDFD, DFFDD, FDFDD, FFDDD, FFF, FFDF, FDFF, DFFF FFDDF, FDFDF, DFFDF, FDDFF, DFDFF, DDFFF } so there are 20 possibilities. Note that the winning player (2 choices) must win the last set and two of the previous sets, so the number of outcomes is 2 [( 2 2 ) + ( 3 2 ) + ( 4 2 )] = 20 . 1.2-8 3 · 3 · 2 12 = 36 , 864 . 1.2-10 ( n − 1 r ) + ( n − 1 r − 1 ) = ( n − 1)! r !( n − 1 − r )! + ( n − 1)! ( r − 1)!( n − r )! = ( n − r )( n − 1)! + r ( n − 1)! r !( n − r )! = n ! r !( n − r )! = ( n r ) . 1.2-12 0 = (1 − 1) n = n ∑ r =0 ( n r ) ( − 1) r (1) n − r = n ∑ r =0 ( − 1) r ( n r ) . 2 n = (1 + 1) n = n ∑ r =0 ( n r ) (1) r (1) n − r = n ∑ r =0 ( n r ) . 1.2-14 ( 5 − 1 + 29 29 ) = 33! 29!4! = 40 , 920. 1.2-16 (a) ( 19 3 )( 52 − 19 6 ) ( 52 9 ) = 102 , 486 351 , 325 = 0 . 2917;

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    Chapter 1 Probability 3 (b) ( 19 3 )( 10 2 )( 7 1 )( 3 0 )( 5 1 )( 2 0 )( 6 2 ) ( 52 9 ) = 7 , 695 1 , 236 , 664 = 0 . 00622 . 1.2-18 (a) P ( A ) = ∑ 5 n =1 (1 / 2) n = 1 − (1 / 2) 5 ; (b) P ( B ) = ∑ 10 n =1 (1 / 2) n = 1 − (1 / 2) 10 ; (c) P ( A ∪ B ) = P ( B ) = 1 − (1 / 2) 10 ; (d) P ( A ∩ B ) = P ( A ) = 1 − (1 / 2) 5 ; (e) P ( C ) = P ( B ) − P ( A ) = (1 / 2) 5 − (1 / 2) 10 ; (f ) P ( B ′ ) = 1 − P ( B ) = (1 / 2) 10 . 1.3 Conditional Probability 1.3-2 (a) 1041 1456 ; (b) 392 633 ; (c) 649 823 . (d) The proportion of women who favor a gun law is greater than the proportion of men who favor a gun law. 1.3-4 (a) P (HH) = 13 52 · 12 51 = 1 17 ; (b) P (HC) = 13 52 · 13 51 = 13 204 ; (c) P (Non-Ace Heart, Ace) + P (Ace of Hearts, Non-Heart Ace) = 12 52 · 4 51 + 1 52 · 3 51 = 51 52 · 51 = 1 52 . 1.3-6 Let H = { died from heart disease } ; P = { at least one parent had heart disease } . P ( H | P ′ ) = N ( H ∩ P ′ ) N ( P ′ ) = 110 648 . 1.3-8 (a) 3 20 · 2 19 · 1 18 = 1 1140 ; (b) ( 3 2 )( 17 1 ) ( 20 3 ) · 1 17 = 1 380 ; (c) 9 ∑ k =1 ( 3 2 )( 17 2 k − 2 ) ( 20 2 k ) · 1 20 − 2 k = 35 76 = 0 . 4605;

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    4 Section 1.4 Independent Events (d) Draw second. The probability of winning is 1 − 0 . 4605 = 0 . 5395. 1.3-10 (a) P ( A ) = 52 52 · 51 52 · 50 52 · 49 52 · 48 52 · 47 52 = 8 , 808 , 975 11 , 881 , 376 = 0 . 74141; (b) P ( A ′ ) = 1 − P ( A ) = 0 . 25859. 1.3-12 (a) It doesn’t matter because P ( B 1 ) = 1 18 , P ( B 5 ) = 1 18 , P ( B 18 ) = 1 18 ; (b) P ( B ) = 2 18 = 1 9 on each draw. 1.3-14 (a) 5 · 4 · 3 = 60; (b) 5 · 5 · 5 = 125. 1.3-16 3 5 · 5 8 + 2 5 · 4 8 = 23 40 . 1.4 Independent Events 1.4-2 (a) P ( A ∩ B ) = P ( A ) P ( B ) = (0 . 3)(0 . 6) = 0 . 18; P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) = 0 . 3 + 0 . 6 − 0 . 18 = 0 . 72; (b) P ( A | B ) = P ( A ∩ B ) P ( B ) = 0 0 . 6 = 0 . 1.4-4 Proof of (b) : P ( A ′ ∩ B ) = P ( B ) P ( A ′ | B ) = P ( B )[1 − P ( A | B )] = P ( B )[1 − P ( A )] = P ( B ) P ( A ′ ) . Proof of (c) : P ( A ′ ∩ B ′ ) = P [( A ∪ B ) ′ ] = 1 − P ( A ∪ B ) = 1 − P ( A ) − P ( B ) + P ( A ∩ B ) = 1 − P ( A ) − P ( B ) + P ( A ) P ( B ) = [1 − P ( A )][1 − P ( B )] = P ( A ′ ) P ( B ′ ) . 1.4-6 P [ A ∩ ( B ∩ C )] = P [ A ∩ B ∩ C ] = P ( A ) P ( B ) P ( C ) = P ( A ) P ( B ∩ C ) . P [ A ∩ ( B ∪ C )] = P [( A ∩ B ) ∪ ( A ∩ C )] = P ( A ∩ B ) + P ( A ∩ C ) − P ( A ∩ B ∩ C ) = P ( A ) P ( B ) + P ( A ) P ( C ) − P ( A ) P ( B ) P ( C ) = P ( A )[ P ( B ) + P ( C ) − P ( B ∩ C )] = P ( A ) P ( B ∪ C ) . P [ A ′ ∩ ( B ∩ C ′ )] = P ( A ′ ∩ C ′ ∩ B ) = P ( B )[ P ( A ′ ∩ C ′ ) | B ] = P ( B )[1 − P ( A ∪ C | B )] = P ( B )[1 − P ( A ∪ C )] = P ( B ) P [( A ∪ C ) ′ ] = P ( B ) P ( A ′ ∩ C ′ ) = P ( B ) P ( A ′ ) P ( C ′ ) = P ( A ′ ) P ( B ) P ( C ′ ) = P ( A ′ ) P ( B ∩ C ′ ) .

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    Chapter 1 Probability 5 P [ A ′ ∩ B ′ ∩ C ′ ] = P [( A ∪ B ∪ C ) ′ ] = 1 − P ( A ∪ B ∪ C ) = 1 − P ( A ) − P ( B ) − P ( C ) + P ( A ) P ( B ) + P ( A ) P ( C )+ P ( B ) P ( C ) − P A ) P ( B ) P ( C ) = [1 − P ( A )][1 − P ( B )][1 − P ( C )] = P ( A ′ ) P ( B ′ ) P ( C ′ ) . 1.4-8 1 6 · 2 6 · 3 6 + 1 6 · 4 6 · 3 6 + 5 6 · 2 6 · 3 6 = 2 9 . 1.4-10 (a) 3 4 · 3 4 = 9 16 ; (b) 1 4 · 3 4 + 3 4 · 2 4 = 9 16 ; (c) 2 4 · 1 4 + 2 4 · 4 4 = 10 16 . 1.4-12 (a) ( 1 2 ) 3 ( 1 2 ) 2 ; (b) ( 1 2 ) 3 ( 1 2 ) 2 ; (c) ( 1 2 ) 3 ( 1 2 ) 2 ; (d) 5! 3! 2! ( 1 2 ) 3 ( 1 2 ) 2 . 1.4-14 (a) 1 − (0 . 4) 3 = 1 − 0 . 064 = 0 . 936; (b) 1 − (0 . 4) 8 = 1 − 0 . 00065536 = 0 . 99934464. 1.4-16 (a) ∞ ∑ k =0 1 5 ( 4 5 ) 2 k = 5 9 ; (b) 1 5 + 4 5 · 3 4 · 1 3 + 4 5 · 3 4 · 2 3 · 1 2 · 1 1 = 3 5 . 1.4-18 (a) 7; (b) (1 / 2) 7 ; (c) 63; (d) No! (1 / 2) 63 = 1 / 9 , 223 , 372 , 036 , 854 , 775 , 808. 1.4-20 No. The equations that must hold are (1 − p 1 )(1 − p 2 ) = p 1 (1 − p 2 ) + p 2 (1 − p 1 ) = p 1 p 2 . There are no real solutions. 1.5 Bayes’ Theorem 1.5-2 (a) P ( G ) = P ( A ∩ G ) + P ( B ∩ G ) = P ( A ) P ( G | A ) + P ( B ) P ( G | B ) = (0 . 40)(0 . 85) + (0 . 60)(0 . 75) = 0 . 79; (b) P ( A | G ) = P ( A ∩ G ) P ( G ) = (0 . 40)(0 . 85) 0 . 79 = 0 . 43 .

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    6 Section 1.5 Bayes’ Theorem 1.5-4 Let event B denote an accident and let A 1 be the event that age of the driver is 16–25. Then P ( A 1 | B ) = (0 . 1)(0 . 05) (0 . 1)(0 . 05) + (0 . 55)(0 . 02) + (0 . 20)(0 . 03) + (0 . 15)(0 . 04) = 50 50 + 110 + 60 + 60 = 50 280 = 0 . 179 . 1.5-6 Let B be the event that the policyholder dies. Let A 1 , A 2 , A 3 be the events that the deceased is standard, preferred and ultra-preferred, respectively. Then P ( A 1 | B ) = (0 . 60)(0 . 01) (0 . 60)(0 . 01) + (0 . 30)(0 . 008) + (0 . 10)(0 . 007) = 60 60 + 24 + 7 = 60 91 = 0 . 659; P ( A 2 | B ) = 24 91 = 0 . 264; P ( A 3 | B ) = 7 91 = 0 . 077 . 1.5-8 Let A be the event that the tablet is under warranty. P ( B 1 | A ) = (0 . 40)(0 . 10) (0 . 40)(0 . 10) + (0 . 30)(0 . 05) + (0 . 20)(0 . 03) + (0 . 10)(0 . 02) = 40 40 + 15 + 6 + 2 = 40 63 = 0 . 635; P ( B 2 | A ) = 15 63 = 0 . 238; P ( B 3 | A ) = 6 63 = 0 . 095; P ( B 4 | A ) = 2 63 = 0 . 032 . 1.5-10 (a) P ( D + ) = (0 . 02)(0 . 92) + (0 . 98)(0 . 05) = 0 . 0184 + 0 . 0490 = 0 . 0674; (b) P ( A − | D + ) = 0 . 0490 0 . 0674 = 0 . 727; P ( A + | D + ) = 0 . 0184 0 . 0674 = 0 . 273; (c) P ( A − | D − ) = (0 . 98)(0 . 95) (0 . 02)(0 . 08) + (0 . 98)(0 . 95) = 9310 16 + 9310 = 0 . 998; P ( A + | D − ) = 0 . 002; (d) Yes, particularly those in part (b). 1.5-12 Let D = { defective roll } . Then P ( I | D ) = P ( I ∩ D ) P ( D ) = P ( I ) · P ( D | I ) P ( I ) · P ( D | I ) + P ( II ) · P ( D | II ) = (0 . 60)(0 . 03) (0 . 60)(0 . 03) + (0 . 40)(0 . 01) = 0 . 018 0 . 018 + 0 . 004 = 0 . 018 0 . 022 = 0 . 818 .

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    Chapter 2 Discrete Distributions 7 Chapter 2 Discrete Distributions 2.1 Random Variables of the Discrete Type 2.1-2 (a) f ( x ) =    0 . 6 , x = 1 , 0 . 3 , x = 5 , 0 . 1 , x = 10; (b) f(x) x 0.1 0.2 0.3 0.4 0.5 0.6 1 2 3 4 5 6 7 8 9 10 Figure 2.1–2: A probability histogram 2.1-4 (a) 9 ∑ x =1 log 10 ( x + 1 x ) = 9 ∑ x =1 [log 10 ( x + 1) − log 10 x ] = log 10 2 − log 10 1 + log 10 3 − log 10 2 + · · · + log 10 10 − log 10 9 = log 10 10 = 1; (b) F ( x ) =    0 , x < 1 , log 10 n, n − 1 ≤ x < n, n = 2 , 3 , . . . , 9 , 1 , 9 ≤ x.

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    8 Section 2.1 Random Variables of the Discrete Type 2.1-6 (a) f ( x ) = 1 10 , x = 0 , 1 , 2 , · · · , 9; (b) N ( { 0 } ) / 150 = 11 / 150 = 0 . 073; N ( { 5 } ) / 150 = 13 / 150 = 0 . 087; N ( { 1 } ) / 150 = 14 / 150 = 0 . 093; N ( { 6 } ) / 150 = 22 / 150 = 0 . 147; N ( { 2 } ) / 150 = 13 / 150 = 0 . 087; N ( { 7 } ) / 150 = 16 / 150 = 0 . 107; N ( { 3 } ) / 150 = 12 / 150 = 0 . 080; N ( { 8 } ) / 150 = 18 / 150 = 0 . 120; N ( { 4 } ) / 150 = 16 / 150 = 0 . 107; N ( { 9 } ) / 150 = 15 / 150 = 0 . 100. (c) x f(x), h(x) 0.02 0.04 0.06 0.08 0.10 0.12 0.14 1 2 3 4 5 6 7 8 9 Figure 2.1–6: Michigan daily lottery digits 2.1-8 (a) f ( x ) = 6 − | 7 − x | 36 , x = 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12; (b) x f(x) 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 1 2 3 4 5 6 7 8 9 10 11 12 Figure 2.1–8: Probability histogram for the sum of a pair of dice

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    Chapter 2 Discrete Distributions 9 2.1-10 (a) The space of W is S = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 } . P ( W = 0) = P ( X = 0 , Y = 0) = 1 2 · 1 4 = 1 8 , assuming independence. P ( W = 1) = P ( X = 0 , Y = 1) = 1 2 · 1 4 = 1 8 , P ( W = 2) = P ( X = 2 , Y = 0) = 1 2 · 1 4 = 1 8 , P ( W = 3) = P ( X = 2 , Y = 1) = 1 2 · 1 4 = 1 8 , P ( W = 4) = P ( X = 0 , Y = 4) = 1 2 · 1 4 = 1 8 , P ( W = 5) = P ( X = 0 , Y = 5) = 1 2 · 1 4 = 1 8 , P ( W = 6) = P ( X = 2 , Y = 4) = 1 2 · 1 4 = 1 8 , P ( W = 7) = P ( X = 2 , Y = 5) = 1 2 · 1 4 = 1 8 . That is, f ( w ) = P ( W = w ) = 1 8 , w ∈ S . (b) w ( ) f w 0.02 0.04 0.06 0.08 0.10 0.12 1 2 3 4 5 6 7 Figure 2.1–10: Probability histogram of sum of two special dice 2.1-12 Let x equal the number of orange balls and 144 − x the number of blue balls. Then x 144 · x − 1 143 + 144 − x 144 · 143 − x 143 = x 144 · 144 − x 143 + 144 − x 144 · x 143 x 2 − x + 144 · 143 − 144 x − 143 x + x 2 = 2 · 144 x − 2 · x 2 x 2 − 144 x + 5 , 148 = 0 ( x − 78)( x − 66) = 0 Thus there are 78 orange balls and 66 blue balls. 2.2 Mathematical Expectation 2.2-2 E ( X ) = ( − 1) ( 4 9 ) + (0) ( 1 9 ) + (1) ( 4 9 ) = 0; E ( X 2 ) = ( − 1) 2 ( 4 9 ) + (0) 2 ( 1 9 ) + (1) 2 ( 4 9 ) = 8 9 ;

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    10 Section 2.3 Special Mathematical Expectations E (3 X 2 − 2 X + 4) = 3 ( 8 9 ) − 2(0) + 4 = 20 3 . 2.2-4 1 = 6 ∑ x =0 f ( x ) = 9 10 + c ( 1 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 ) c = 2 49 ; E (Payment) = 2 49 ( 1 · 1 2 + 2 · 1 3 + 3 · 1 4 + 4 · 1 5 + 5 · 1 6 ) = 71 490 units . 2.2-6 Note that ∞ ∑ x =1 6 π 2 x 2 = 6 π 2 ∞ ∑ x =1 1 x 2 = 6 π 2 π 2 6 = 1, so this is a pdf. E ( X ) = ∞ ∑ x =1 x 6 π 2 x 2 = 6 π 2 ∞ ∑ x =1 1 x = + ∞ and it is well known that the sum of this harmonic series is not finite. 2.2-8 E ( | X − c | ) = 1 7 ∑ x ∈ S | x − c | , where S = { 1 , 2 , 3 , 5 , 15 , 25 , 50 } . When c = 5, E ( | X − 5 | ) = 1 7 [(5 − 1) + (5 − 2) + (5 − 3) + (5 − 5) + (15 − 5) + (25 − 5) + (50 − 5)] . If c is either increased or decreased by 1, this expectation is increased by 1 / 7. Thus c = 5, the median, minimizes this expectation while b = E ( X ) = 101 / 7 minimizes E [( X − b ) 2 ]. 2.2-10 (1) · 15 36 + ( − 1) · 21 36 = − 6 36 = − 1 6 ; (1) · 15 36 + ( − 1) · 21 36 = − 6 36 = − 1 6 ; (4) · 6 36 + ( − 1) · 30 36 = − 6 36 = − 1 6 . 2.2-12 (a) The average class size is (16)(25) + (3)(100) + (1)(300) 20 = 50; (b) f ( x ) =    0 . 4 , x = 25 , 0 . 3 , x = 100 , 0 . 3 , x = 300; (c) E ( X ) = 25(0 . 4) + 100(0 . 3) + 300(0 . 3) = 130 .

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    Chapter 2 Discrete Distributions 11 2.3 Special Mathematical Expectations 2.3-2 (a) μ = E ( X ) = 3 ∑ x =1 x 3! x ! (3 − x )! ( 1 4 ) x ( 3 4 ) 3 − x = 3 ( 1 4 ) 2 ∑ k =0 2! k ! (2 − k )! ( 1 4 ) k ( 3 4 ) 2 − k = 3 ( 1 4 )( 1 4 + 3 4 ) 2 = 3 4 ; E [ X ( X − 1)] = 3 ∑ x =2 x ( x − 1) 3! x ! (3 − x )! ( 1 4 ) x ( 3 4 ) 3 − x = 2(3) ( 1 4 ) 2 3 4 + 6 ( 1 4 ) 3 = 6 ( 1 4 ) 2 = 2 ( 1 4 )( 3 4 ) ; σ 2 = E [ X ( X − 1)] + E ( X ) − μ 2 = (2) ( 3 4 )( 1 4 ) + ( 3 4 ) − ( 3 4 ) 2 = (2) ( 3 4 )( 1 4 ) + ( 3 4 )( 1 4 ) = 3 ( 1 4 )( 3 4 ) ; (b) μ = E ( X ) = 4 ∑ x =1 x 4! x ! (4 − x )! ( 1 2 ) x ( 1 2 ) 4 − x = 4 ( 1 2 ) 3 ∑ k =0 3! k ! (3 − k )! ( 1 2 ) k ( 1 2 ) 3 − k = 4 ( 1 2 )( 1 2 + 1 2 ) 3 = 2; E [ X ( X − 1)] = 4 ∑ x =2 x ( x − 1) 4! x ! (4 − x )! ( 1 2 ) x ( 1 2 ) 4 − x = 2(6) ( 1 2 ) 4 + (6)(4) ( 1 2 ) 4 + (12) ( 1 2 ) 4 = 48 ( 1 2 ) 4 = 12 ( 1 2 ) 2 ; σ 2 = (12) ( 1 2 ) 2 + 4 2 − ( 4 2 ) 2 = 1 . 2.3-4 E [( X − μ ) /σ ] = (1 /σ )[ E ( X ) − μ ] = (1 /σ )( μ − μ ) = 0; E { [( X − μ ) /σ ] 2 } = (1 /σ 2 ) E [( X − μ ) 2 ] = (1 /σ 2 )( σ 2 ) = 1. 2.3-6 f (1) = 3 8 , f (2) = 2 8 , f (3) = 3 8 ,

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    12 Section 2.3 Special Mathematical Expectations μ = 1 · 3 8 + 2 · 2 8 + 3 · 3 8 = 2, σ 2 = 1 2 · 3 8 + 2 2 · 2 8 + 3 2 · 3 8 − 2 2 = 3 4 . 2.3-8 E ( X ) = 4 ∑ x =1 x · 2 x − 1 16 = 50 16 = 3 . 125; E ( X 2 ) = 4 ∑ x =1 x 2 · 2 x − 1 16 = 85 8 ; Var( X ) = 85 8 − ( 25 8 ) 2 = 55 64 = 0 . 8594; σ = √ 55 8 = 0 . 9270 . 2.3-10 μ = E ( X ) = ( − 5)(1 / 16) + ( − 1)(5 / 8) + (3)(5 / 16) = 0 so E [( X − μ ) 3 ] = E ( X 3 ) = ( − 5) 3 (1 / 16)+( − 1) 3 (5 / 8)+(3 3 )(5 / 16) = ( − 125 − 10+135) / 16 = 0. Thus the index of skewness likewise equals 0. The distribution is not symmetric, however. 2.3-12 (a) f ( x ) = ( 364 365 ) x − 1 ( 1 365 ) , x = 1 , 2 , 3 , . . . ; (b) μ = 1 1 365 = 365 , σ 2 = 364 365 ( 1 365 ) 2 = 132 , 860 , σ = 364 . 500; (c) P ( X > 400) = ( 364 365 ) 400 = 0 . 3337 , P ( X < 300) = 1 − ( 364 365 ) 299 = 0 . 5597 . 2.3-14 P ( X ≥ 100) = P ( X > 99) = (0 . 99) 99 = 0 . 3697 . 2.3-16 (a) f ( x ) = (1 / 2) x − 1 , x = 2 , 3 , 4 , . . . ; (b) M ( t ) = E [ e tx ] = ∞ ∑ x =2 e tx (1 / 2) x − 1 = 2 ∞ ∑ x =2 ( e t / 2) x = 2( e t / 2) 2 1 − e t / 2 = e 2 t 2 − e t , t < ln 2;

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