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Solution Manual for Probability and Statistical Inference, 10th Edition

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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 1 preview imageSOLUTIONSMANUALPROBABILITYANDSTATISTICALINFERENCETENTHEDITIONRobert V. HoggElliot A. TanisDale L. Zimmerman
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 2 preview image
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 3 preview imageiiiContentsPrefacev1Probability11.1Properties of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11.2Methods of Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21.3Conditional Probability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31.4Independent Events. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41.5Bayes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52Discrete Distributions72.1Random Variables of the Discrete Type. . . . . . . . . . . . . . . . . . . . . . . . . .72.2Mathematical Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .92.3Special Mathematical Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .112.4The Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .142.5The Hypergeometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .162.6The Negative Binomial Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . .172.7The Poisson Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .183Continuous Distributions193.1Random Variables of the Continuous Type . . . . . . . . . . . . . . . . . . . . . . . . .193.2The Exponential, Gamma, and Chi-Square Distributions . . . . . . . . . . . . . . . . .263.3The Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .283.4Additional Models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .304Bivariate Distributions334.1Bivariate Distributions of the Discrete Type . . . . . . . . . . . . . . . . . . . . . . . .334.2The Correlation Coefficient. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .344.3Conditional Distributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .364.4Bivariate Distributions of the Continuous Type . . . . . . . . . . . . . . . . . . . . . .374.5The Bivariate Normal Distribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . .415Distributions of Functions of Random Variables455.1Functions of One Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . .455.2Transformations of Two Random Variables. . . . . . . . . . . . . . . . . . . . . . . .475.3Several Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .525.4The Moment-Generating Function Technique. . . . . . . . . . . . . . . . . . . . . . .535.5Random Functions Associated with Normal Distributions. . . . . . . . . . . . . . . .555.6The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .585.7Approximations for Discrete Distributions . . . . . . . . . . . . . . . . . . . . . . . . .595.8Chebyshev’s Inequality and Convergence in Probability. . . . . . . . . . . . . . . . .615.9Limiting Moment-Generating Functions. . . . . . . . . . . . . . . . . . . . . . . . . .62
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 4 preview imageivContents6Point Estimation636.1Descriptive Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .636.2Exploratory Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .656.3Order Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .706.4Maximum Likelihood Estimation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .736.5A Simple Regression Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .766.6Asymptotic Distributions of MaximumLikelihood Estimators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .816.7Sufficient Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .816.8Bayesian Estimation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .847Interval Estimation877.1Confidence Intervals for Means. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .877.2Confidence Intervals for the Difference of Two Means . . . . . . . . . . . . . . . . . . .887.3Confidence Intervals For Proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . .907.4Sample Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .917.5Distribution-Free Confidence Intervals for Percentiles . . . . . . . . . . . . . . . . . . .927.6More Regression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .937.7Resampling Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .998Tests of Statistical Hypotheses1078.1Tests About One Mean. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1078.2Tests of the Equality of Two Means. . . . . . . . . . . . . . . . . . . . . . . . . . . .1098.3Tests for Variances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1118.4Tests about Proportions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1138.5Some Distribution-Free Tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1148.6Power of a Statistical Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1188.7Best Critical Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1218.8Likelihood Ratio Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1249More Tests1279.1Chi-Square Goodness-of-Fit Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1279.2Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1309.3One-Factor Analysis of Variance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1319.4Two-Way Analysis of Variance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1349.5General Factorial and 2kFactorial Designs . . . . . . . . . . . . . . . . . . . . . . . . .1359.6Tests Concerning Regression and Correlation. . . . . . . . . . . . . . . . . . . . . . .1369.7Statistical Quality Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .137
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 5 preview imageChapter 1 Probability1Chapter 1Probability1.1Properties of Probability1.1-2Sketch a figure and fill in the probabilities of each of the disjoint sets.LetA={insure more than one car},P(A) = 0.85.LetB={insure a sports car},P(B) = 0.23.LetC={insure exactly one car},P(C) = 0.15.It is also given thatP(A∩B) = 0.17. SinceA∩C=φ,P(A∩C) = 0. It follows thatP(A∩B∩C′) = 0.17. ThusP(A′∩B∩C′) = 0.06 andP(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′) + 0.4P(A∩B′)=0.1;(c)P(A′∪B′) =P[(A∩B)′] = 1−P(A∩B) = 1−0.4 = 0.6.1.1-8LetA={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=0.15.1.1-10A∪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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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 6 preview image2Section 1.2 Methods of Enumeration1.1-14P(A) = 2[r−r(√3/2)]2r= 1−√32.1.1-16Note that the respective probabilities arep0, p1=p0/4, p2=p0/42,· · ·.∞∑k=0p04k=1p01−1/4=1p0=341−p0−p1= 1−1516 =116 .1.2Methods of Enumeration1.2-2 (a)(4)(5)(2) = 40;(b)(2)(2)(2) = 8.1.2-4(a)4(63)= 80;(b)4(26) = 256;(c)(4−1 + 3)!(4−1)!3!= 20.1.2-6S={DDD,DDFD,DFDD,FDDD,DDFFD,DFDFD,FDDFD,DFFDD,FDFDD,FFDDD,FFF,FFDF,FDFF,DFFFFFDDF,FDFDF,DFFDF,FDDFF,DFDFF,DDFFF}so there are 20 possibilities.Note that thewinning player (2 choices) must win the last set and two of the previous sets, so thenumber of outcomes is2[(22)+(32)+(42)]= 20.1.2-83·3·212= 36,864.1.2-10(n−1r)+(n−1r−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)! =(nr).1.2-120=(1−1)n=n∑r=0(nr)(−1)r(1)n−r=n∑r=0(−1)r(nr).2n=(1 + 1)n=n∑r=0(nr)(1)r(1)n−r=n∑r=0(nr).1.2-14(5−1 + 2929)=33!29!4! = 40,920.1.2-16(a)(193)(52−196)(529)= 102,486351,325 = 0.2917;
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 7 preview imageChapter 1 Probability3(b)(193)(102)(71)(30)(51)(20)(62)(529)=7,6951,236,664 = 0.00622.1.2-18(a)P(A) =∑5n=1(1/2)n= 1−(1/2)5;(b)P(B) =∑10n=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)10.1.3Conditional Probability1.3-2(a)10411456 ;(b)392633 ;(c)649823 .(d)The proportion of women who favor a gun law is greater than the proportion of menwho favor a gun law.1.3-4(a)P(HH) = 1352·1251 =117 ;(b)P(HC) = 1352·1351 =13204 ;(c)P(Non-Ace Heart, Ace) +P(Ace of Hearts, Non-Heart Ace)= 1252·451 + 152·351 =5152·51 =152 .1.3-6LetH={died from heart disease};P={at least one parent had heart disease}.P(H|P′) =N(H∩P′)N(P′)= 110648.1.3-8(a)320·219·118 =11140 ;(b)(32)(171)(203)·117 =1380 ;(c)9∑k=1(32)(172k−2)(202k)·120−2k= 3576 = 0.4605;
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 8 preview image4Section 1.4 Independent Events(d)Draw second. The probability of winning is 1−0.4605 = 0.5395.1.3-10(a)P(A) = 5252·5152·5052·4952·4852·4752 =8,808,97511,881,376 = 0.74141;(b)P(A′) = 1−P(A) = 0.25859.1.3-12(a)It doesn’t matter becauseP(B1) =118, P(B5) =118, P(B18) =118 ;(b)P(B) =218 = 19on each draw.1.3-14(a)5·4·3 = 60;(b)5·5·5 = 125.1.3-1635·58 + 25·48 = 2340.1.4Independent Events1.4-2(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)=00.6 = 0.1.4-4Proof 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′).1.4-6P[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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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 9 preview imageChapter 1 Probability5P[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′).1.4-816·26·36 + 16·46·36 + 56·26·36 = 29 .1.4-10(a)34·34 =916 ;(b)14·34 + 34·24 =916 ;(c)24·14 + 24·44 = 1016 .1.4-12(a)(12)3(12)2;(b)(12)3(12)2;(c)(12)3(12)2;(d)5!3! 2!(12)3(12)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=015(45)2k= 59 ;(b)15 + 45·34·13 + 45·34·23·12·11 = 35.1.4-18 (a)7;(b)(1/2)7;(c)63;(d)No! (1/2)63= 1/9,223,372,036,854,775,808.1.4-20No. The equations that must hold are(1−p1)(1−p2) =p1(1−p2) +p2(1−p1) =p1p2.There are no real solutions.1.5Bayes’ Theorem1.5-2(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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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 10 preview image6Section 1.5 Bayes’ Theorem1.5-4Let eventBdenote an accident and letA1be the event that age of the driver is 16–25.ThenP(A1|B)=(0.1)(0.05)(0.1)(0.05) + (0.55)(0.02) + (0.20)(0.03) + (0.15)(0.04)=5050 + 110 + 60 + 60 =50280 = 0.179.1.5-6LetBbe the event that the policyholder dies.LetA1, A2, A3be the events that thedeceased is standard, preferred and ultra-preferred, respectively. ThenP(A1|B)=(0.60)(0.01)(0.60)(0.01) + (0.30)(0.008) + (0.10)(0.007)=6060 + 24 + 7 = 6091 = 0.659;P(A2|B)=2491 = 0.264;P(A3|B)=791 = 0.077.1.5-8LetAbe the event that the tablet is under warranty.P(B1|A)=(0.40)(0.10)(0.40)(0.10) + (0.30)(0.05) + (0.20)(0.03) + (0.10)(0.02)=4040 + 15 + 6 + 2 = 4063 = 0.635;P(B2|A)=1563 = 0.238;P(B3|A)=663 = 0.095;P(B4|A)=263 = 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+) = 0.04900.0674 = 0.727;P(A+|D+) = 0.01840.0674 = 0.273;(c)P(A−|D−) =(0.98)(0.95)(0.02)(0.08) + (0.98)(0.95) =931016 + 9310 = 0.998;P(A+|D−) = 0.002;(d)Yes, particularly those in part (b).1.5-12LetD={defective roll}. ThenP(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)=0.0180.018 + 0.004 = 0.0180.022 = 0.818.
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 11 preview imageChapter 2 Discrete Distributions7Chapter 2Discrete Distributions2.1Random Variables of the Discrete Type2.1-2(a)f(x) =0.6,x= 1,0.3,x= 5,0.1,x= 10;(b)f(x)x0.10.20.30.40.50.612345678910Figure 2.1–2: A probability histogram2.1-4(a)9∑x=1log10(x+ 1x)=9∑x=1[log10(x+ 1)−log10x]=log102−log101 + log103−log102 +· · ·+ log1010−log109=log1010 = 1;(b)F(x) =0,x <1,log10n,n−1≤x < n,n= 2,3, . . . ,9,1,9≤x.
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 12 preview image8Section 2.1 Random Variables of the Discrete Type2.1-6(a)f(x) =110,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)xf(x), h(x)0.020.040.060.080.100.120.14123456789Figure 2.1–6: Michigan daily lottery digits2.1-8(a)f(x) = 6− |7−x|36, x= 2,3,4,5,6,7,8,9,10,11,12;(b)xf(x)0.020.040.060.080.100.120.140.16123456789101112Figure 2.1–8: Probability histogram for the sum of a pair of dice
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 13 preview imageChapter 2 Discrete Distributions92.1-10(a)The space ofWisS={0,1,2,3,4,5,6,7}.P(W= 0) =P(X= 0, Y= 0) = 12·14 = 18 , assuming independence.P(W= 1) =P(X= 0, Y= 1) = 12·14 = 18 ,P(W= 2) =P(X= 2, Y= 0) = 12·14 = 18 ,P(W= 3) =P(X= 2, Y= 1) = 12·14 = 18 ,P(W= 4) =P(X= 0, Y= 4) = 12·14 = 18 ,P(W= 5) =P(X= 0, Y= 5) = 12·14 = 18 ,P(W= 6) =P(X= 2, Y= 4) = 12·14 = 18 ,P(W= 7) =P(X= 2, Y= 5) = 12·14 = 18 .That is,f(w) =P(W=w) = 18,w∈S.(b)w( )f w0.020.040.060.080.100.121234567Figure 2.1–10: Probability histogram of sum of two special dice2.1-12Letxequal the number of orange balls and 144−xthe number of blue balls. Thenx144·x−1143+ 144−x144·143−x143=x144·144−x143+ 144−x144·x143x2−x+ 144·143−144x−143x+x2=2·144x−2·x2x2−144x+ 5,148=0(x−78)(x−66)=0Thus there are 78 orange balls and 66 blue balls.2.2Mathematical Expectation2.2-2E(X) = (−1)(49)+ (0)(19)+ (1)(49)= 0;E(X2) = (−1)2(49)+ (0)2(19)+ (1)2(49)= 89 ;
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 14 preview image10Section 2.3 Special Mathematical ExpectationsE(3X2−2X+ 4) = 3(89)−2(0) + 4 = 203 .2.2-41=6∑x=0f(x) =910 +c(11 + 12 + 13 + 14 + 15 + 16)c=249 ;E(Payment)=249(1·12 + 2·13 + 3·14 + 4·15 + 5·16)=71490 units.2.2-6Note that∞∑x=16π2x2=6π2∞∑x=11x2=6π2π26= 1, so this is a pdf.E(X) =∞∑x=1x6π2x2=6π2∞∑x=11x= +∞and it is well known that the sum of this harmonic series is not finite.2.2-8E(|X−c|) = 17∑x∈S|x−c|, whereS={1,2,3,5,15,25,50}.Whenc= 5,E(|X−5|) = 17 [(5−1) + (5−2) + (5−3) + (5−5) + (15−5) + (25−5) + (50−5)].Ifcis either increased or decreased by 1, this expectation is increased by 1/7. Thusc= 5,the median, minimizes this expectation whileb=E(X) = 101/7 minimizesE[(X−b)2].2.2-10(1)·1536 + (−1)·2136 =−636 =−16 ;(1)·1536 + (−1)·2136 =−636 =−16 ;(4)·636 + (−1)·3036 =−636 =−16 .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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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 15 preview imageChapter 2 Discrete Distributions112.3Special Mathematical Expectations2.3-2(a)μ=E(X)=3∑x=1x3!x! (3−x)!(14)x(34)3−x=3(14)2∑k=02!k! (2−k)!(14)k(34)2−k=3(14)(14 + 34)2= 34 ;E[X(X−1)]=3∑x=2x(x−1)3!x! (3−x)!(14)x(34)3−x=2(3)(14)234 + 6(14)3=6(14)2= 2(14)(34);σ2=E[X(X−1)] +E(X)−μ2=(2)(34)(14)+(34)−(34)2=(2)(34)(14)+(34)(14)= 3(14)(34);(b)μ=E(X)=4∑x=1x4!x! (4−x)!(12)x(12)4−x=4(12)3∑k=03!k! (3−k)!(12)k(12)3−k=4(12)(12 + 12)3= 2;E[X(X−1)]=4∑x=2x(x−1)4!x! (4−x)!(12)x(12)4−x=2(6)(12)4+ (6)(4)(12)4+ (12)(12)4=48(12)4= 12(12)2;σ2=(12)(12)2+ 42−(42)2= 1.2.3-4E[(X−μ)/σ] = (1/σ)[E(X)−μ] = (1/σ)(μ−μ) = 0;E{[(X−μ)/σ]2}= (1/σ2)E[(X−μ)2] = (1/σ2)(σ2) = 1.2.3-6f(1) = 38, f(2) = 28, f(3) = 38,
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Solution Manual for Probability and Statistical Inference, 10th Edition - Page 16 preview image12Section 2.3 Special Mathematical Expectationsμ= 1·38 + 2·28 + 3·38 = 2,σ2= 12·38 + 22·28 + 32·38−22= 34 .2.3-8E(X)=4∑x=1x·2x−116=5016=3.125;E(X2)=4∑x=1x2·2x−116=858 ;Var(X)=858−(258)2=5564=0.8594;σ=√558=0.9270.2.3-10μ=E(X) = (−5)(1/16) + (−1)(5/8) + (3)(5/16) = 0 soE[(X−μ)3] =E(X3) = (−5)3(1/16)+(−1)3(5/8)+(33)(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) =(364365)x−1(1365),x= 1,2,3, . . .;(b)μ=11365= 365,σ2=364365(1365)2= 132,860,σ=364.500;(c)P(X >400)=(364365)400= 0.3337,P(X <300)=1−(364365)299= 0.5597.2.3-14P(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[etx]=∞∑x=2etx(1/2)x−1=2∞∑x=2(et/2)x=2(et/2)21−et/2=e2t2−et,t <ln 2;
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