Probability and statistics for engineers and scientists solutions 9th edition

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INSTRUCTORS
SOLUTION MANUAL
KEYING YE AND SHARON MYERS
for
PROBABILITY & STATISTICS
FOR ENGINEERS & SCIENTISTS
EIGHTH EDITION
WALPOLE, MYERS, MYERS, YE
Contents
1 Introduction to Statistics and Data Analysis 1
2 Probability 11
3 Random Variables and Probability Distributions 29
4 Mathematical Expectation 45
5 Some Discrete Probability Distributions 59
6 Some Continuous Probability Distributions 71
7 Functions of Random Variables 85
8 Fundamental Sampling Distributions and Data Descriptions 91
9 One- and Two-Sample Estimation Problems 103
10 One- and Two-Sample Tests of Hypotheses 121
11 Simple Linear Regression and Correlation 149
12 Multiple Linear Regression and Certain Nonlinear Regression Models 171
13 One-Factor Experiments: General 185
14 Factorial Experiments (Two or More Factors) 213
15 2k Factorial Experiments and Fractions 237
16 Nonparametric Statistics 257
iii
iv CONTENTS
17 Statistical Quality Control 273
18 Bayesian Statistics 277
Chapter 1
Introduction to Statistics and Data
Analysis
1.1 (a) 15.
(b) x = 115(3.4 + 2.5 + 4.8 + + 4.8) = 3.787.
(c) Sample median is the 8th value, after the data is sorted from smallest to largest:3.6.
(d) A dot plot is shown below.
2.5 3.0 3.5 4.0 4.5 5.0 5.5
(e) After trimming total 40% of the data (20% highest and 20% lowest), the databecomes:
2.9 3.0 3.3 3.4 3.63.7 4.0 4.4 4.8
So. the trimmed mean is
xtr20 =1
9(2.9 + 3.0 + + 4.8) = 3.678.
1.2 (a) Mean=20.768 and Median=20.610.
(b) xtr10 = 20.743.
(c) A dot plot is shown below.
18 19 20 21 22 23
1
2 Chapter 1 Introduction to Statistics and Data Analysis
1.3 (a) A dot plot is shown below.
200 205 210 215 220 225 230
In the figure, represents the No aging group and represents the Aginggroup.
(b) Yes; tensile strength is greatly reduced due to the aging process.
(c) MeanAging = 209.90, and MeanNo aging = 222.10.
(d) MedianAging = 210.00, and MedianNo aging = 221.50. The means and medians foreach group are similar to each other.
1.4 (a) XA = 7.950 and XA = 8.250;XB = 10.260 and XB = 10.150.
(b) A dot plot is shown below.
6.5 7.5 8.5 9.5 10.5 11.5
In the figure, represents company A and represents company B. Thesteel rods made by company B show more flexibility.
1.5 (a) A dot plot is shown below.
10 0 10 20 30 40
In the figure, represents the control group and represents the treatmentgroup.
(b) XControl = 5.60, XControl = 5.00, and Xtr(10);Control = 5.13;
XTreatment = 7.60, XTreatment = 4.50, and Xtr(10);Treatment = 5.63.
(c) The difference of the means is 2.0 and the differences of the medians and thetrimmed means are 0.5, which are much smaller. The possible cause of this mightbe due to the extreme values (outliers) in the samples, especially the value of 37.
1.6 (a) A dot plot is shown below.
1.95 2.05 2.15 2.25 2.35 2.45 2.55
In the figure, represents the 20C group and represents the 45C group.(b) X20C = 2.1075, and X45C = 2.2350.
(c) Based on the plot, it seems that high temperature yields more high values oftensile strength, along with a few low values of tensile strength. Overall, thetemperature does have an influence on the tensile strength.
Solutions for Exercises in Chapter 1 3
(d) It also seems that the variation of the tensile strength gets larger when the curetemperature is increased.
1.7 s2 = 1151 [(3.43.787)2+(2.53.787)2+(4.83.787)2+ +(4.83.787)2] = 0.94284;
s =s2 =
0.9428 = 0.971.
1.8 s2 = 1201 [(18.71 20.768)2 + (21.41 20.768)2 + + (21.12 20.768)2] = 2.5345;
s =2.5345 = 1.592.
1.9 s2No Aging =1
101 [(227 222.10)2 + (222 222.10)2 + + (221 222.10)2] = 42.12;sNo Aging =
42.12 = 6.49.
s2Aging =1
101 [(219 209.90)2 + (214 209.90)2 + + (205 209.90)2] = 23.62;sAging =
23.62 = 4.86.
1.10 For company A: s2A = 1.2078 and sA =1.2078 = 1.099.
For company B: s2B = 0.3249 and sB =0.3249 = 0.570.
1.11 For the control group: s2Control = 69.39 and sControl = 8.33.For the treatment group: s2Treatment = 128.14 and sTreatment = 11.32.
1.12 For the cure temperature at 20C: s220C = 0.005 and s20C = 0.071.For the cure temperature at 45C: s245C = 0.0413 and s45C = 0.2032.The variation of the tensile strength is influenced by the increase of cure temperature.
1.13 (a) Mean = X = 124.3 and median = X = 120;
(b) 175 is an extreme observation.
1.14 (a) Mean = X = 570.5 and median = X = 571;
(b) Variance = s2 = 10; standard deviation= s = 3.162; range=10;
(c) Variation of the diameters seems too big.
1.15 Yes. The value 0.03125 is actually a P -value and a small value of this quantity meansthat the outcome (i.e., HHHHH) is very unlikely to happen with a fair coin.
1.16 The term on the left side can be manipulated to
ni=1
xi nx =ni=1
xi ni=1
xi = 0,
which is the term on the right side.
1.17 (a) Xsmokers = 43.70 and Xnonsmokers = 30.32;
(b) ssmokers = 16.93 and snonsmokers = 7.13;
4 Chapter 1 Introduction to Statistics and Data Analysis
(c) A dot plot is shown below.
10 20 30 40 50 60 70
In the figure, represents the nonsmoker group and represents the smokergroup.
(d) Smokers appear to take longer time to fall asleep and the time to fall asleep forsmoker group is more variable.
1.18 (a) A stem-and-leaf plot is shown below.
Stem Leaf Frequency1 057 32 35 23 246 34 1138 45 22457 56 00123445779 117 01244456678899 148 00011223445589 149 0258 4
(b) The following is the relative frequency distribution table.
Relative Frequency Distribution of GradesClass Interval Class Midpoint Frequency, f Relative Frequency
10 1920 2930 3940 4950 5960 6970 7980 8990 99
14.524.534.544.554.564.574.584.594.5
323451114144
0.050.030.050.070.080.180.230.230.07
(c) A histogram plot is given below.
14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5Final Exam Grades
Rel
ativ
e Fr
eque
ncy
Solutions for Exercises in Chapter 1 5
The distribution skews to the left.
(d) X = 65.48, X = 71.50 and s = 21.13.
1.19 (a) A stem-and-leaf plot is shown below.
Stem Leaf Frequency0 22233457 81 023558 62 035 33 03 24 057 35 0569 46 0005 4
(b) The following is the relative frequency distribution table.
Relative Frequency Distribution of YearsClass Interval Class Midpoint Frequency, f Relative Frequency
0.0 0.91.0 1.92.0 2.93.0 3.94.0 4.95.0 5.96.0 6.9
0.451.452.453.454.455.456.45
8632344

0.2670.2000.1000.0670.1000.1330.133
(c) X = 2.797, s = 2.227 and Sample range is 6.5 0.2 = 6.3.
1.20 (a) A stem-and-leaf plot is shown next.
Stem Leaf Frequency0* 34 20 56667777777889999 171* 0000001223333344 161 5566788899 102* 034 32 7 13* 2 1
(b) The relative frequency distribution table is shown next.
6 Chapter 1 Introduction to Statistics and Data Analysis
Relative Frequency Distribution of Fruit Fly LivesClass Interval Class Midpoint Frequency, f Relative Frequency
0 45 910 1415 1920 2425 2930 34
271217222732
2171610311
0.040.340.320.200.060.020.02
(c) A histogram plot is shown next.
2 7 12 17 22 27 32Fruit fly lives (seconds)
Rel
ativ
e Fr
eque
ncy
(d) X = 10.50.
1.21 (a) X = 1.7743 and X = 1.7700;
(b) s = 0.3905.
1.22 (a) X = 6.7261 and X = 0.0536.
(b) A histogram plot is shown next.
6.62 6.66 6.7 6.74 6.78 6.82Relative Frequency Histogram for Diameter
(c) The data appear to be skewed to the left.
1.23 (a) A dot plot is shown next.
0 100 200 300 400 500 600 700 800 900 1000
395.10160.15
(b) X1980 = 395.1 and X1990 = 160.2.
Solutions for Exercises in Chapter 1 7
(c) The sample mean for 1980 is over twice as large as that of 1990. The variabilityfor 1990 decreased also as seen by looking at the picture in (a). The gap representsan increase of over 400 ppm. It appears from the data that hydrocarbon emissionsdecreased considerably between 1980 and 1990 and that the extreme large emission(over 500 ppm) were no longer in evidence.
1.24 (a) X = 2.8973 and s = 0.5415.
(b) A histogram plot is shown next.
1.8 2.1 2.4 2.7 3 3.3 3.6 3.9Salaries
Rel
ativ
e Fr
eque
ncy
(c) Use the double-stem-and-leaf plot, we have the following.
Stem Leaf Frequency1 (84) 12* (05)(10)(14)(37)(44)(45) 62 (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) 113* (10)(13)(14)(22)(36)(37) 63 (51)(54)(57)(71)(79)(85) 6
1.25 (a) X = 33.31;
(b) X = 26.35;
(c) A histogram plot is shown next.
10 20 30 40 50 60 70 80 90Percentage of the families
Rel
ativ
e Fr
eque
ncy
8 Chapter 1 Introduction to Statistics and Data Analysis
(d) Xtr(10) = 30.97. This trimmed mean is in the middle of the mean and medianusing the full amount of data. Due to the skewness of the data to the right (seeplot in (c)), it is common to use trimmed data to have a more robust result.
1.26 If a model using the function of percent of families to predict staff salaries, it is likelythat the model would be wrong due to several extreme values of the data. Actually ifa scatter plot of these two data sets is made, it is easy to see that some outlier wouldinfluence the trend.
1.27 (a) The averages of the wear are plotted here.
700 800 900 1000 1100 1200 1300
250
300
350
load
wear
(b) When the load value increases, the wear value also increases. It does show certainrelationship.
(c) A plot of wears is shown next.
700 800 900 1000 1100 1200 1300
100
300
500
700
load
wear
(d) The relationship between load and wear in (c) is not as strong as the case in (a),especially for the load at 1300. One reason is that there is an extreme value (750)which influence the mean value at the load 1300.
1.28 (a) A dot plot is shown next.
71.45 71.65 71.85 72.05 72.25 72.45 72.65 72.85
LowHigh
In the figure, represents the low-injection-velocity group and representsthe high-injection-velocity group.
Solutions for Exercises in Chapter 1 9
(b) It appears that shrinkage values for the low-injection-velocity group is higher thanthose for the high-injection-velocity group. Also, the variation of the shrinkageis a little larger for the low injection velocity than that for the high injectionvelocity.
1.29 (a) A dot plot is shown next.
76 79 82 85 88 91 94
Low High
In the figure, represents the low-injection-velocity group and representsthe high-injection-velocity group.
(b) In this time, the shrinkage values are much higher for the high-injection-velocitygroup than those for the low-injection-velocity group. Also, the variation for theformer group is much higher as well.
(c) Since the shrinkage effects change in different direction between low mode tem-perature and high mold temperature, the apparent interactions between the moldtemperature and injection velocity are significant.
1.30 An interaction plot is shown next.
Low high injection velocitylow mold temp
high mold tempmean shrinkage value
It is quite obvious to find the interaction between the two variables. Since in this exper-imental data, those two variables can be controlled each at two levels, the interactioncan be investigated. However, if the data are from an observational studies, in whichthe variable values cannot be controlled, it would be difficult to study the interactionsamong these variables.
Chapter 2
Probability
2.1 (a) S = {8, 16, 24, 32, 40, 48}.(b) For x2 + 4x 5 = (x + 5)(x 1) = 0, the only solutions are x = 5 and x = 1.
S = {5, 1}.(c) S = {T,HT,HHT,HHH}.(d) S = {N. America, S. America,Europe,Asia,Africa,Australia,Antarctica}.(e) Solving 2x 4 0 gives x 2. Since we must also have x < 1, it follows that
S = .
2.2 S = {(x, y) | x2 + y2 < 9; x 0, y 0}.2.3 (a) A = {1, 3}.
(b) B = {1, 2, 3, 4, 5, 6}.(c) C = {x | x2 4x+ 3 = 0} = {x | (x 1)(x 3) = 0} = {1, 3}.(d) D = {0, 1, 2, 3, 4, 5, 6}. Clearly, A = C.
2.4 (a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
(b) S = {(x, y) | 1 x, y 6}.2.5 S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH,
5TT, 6H, 6T}.2.6 S = {A1A2, A1A3, A1A4, A2A3, A2A4, A3A4}.2.7 S1 = {MMMM,MMMF,MMFM,MFMM,FMMM,MMFF,MFMF,MFFM,
FMFM,FFMM,FMMF,MFFF, FMFF, FFMF, FFFM,FFFF}.S2 = {0, 1, 2, 3, 4}.
2.8 (a) A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.
11
12 Chapter 2 Probability
(b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4),(2, 5), (2, 6)}.
(c) C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.(d) A C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.(e) A B = .(f) B C = {(5, 2), (6, 2)}.(g) A Venn diagram is shown next.
A
A C
B
B C
C
S
2.9 (a) A = {1HH, 1HT, 1TH, 1TT, 2H, 2T}.(b) B = {1TT, 3TT, 5TT}.(c) A = {3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}.(d) A B = {3TT, 5TT}.(e) A B = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3TT, 5TT}.
2.10 (a) S = {FFF, FFN, FNF,NFF, FNN,NFN,NNF,NNN}.(b) E = {FFF, FFN, FNF,NFF}.(c) The second river was safe for fishing.
2.11 (a) S = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2, F1M1, F1M2, F1F2, F2M1, F2M2,F2F1}.
(b) A = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2}.(c) B = {M1F1,M1F2,M2F1,M2F2, F1M1, F1M2, F2M1, F2M2}.(d) C = {F1F2, F2F1}.(e) A B = {M1F1,M1F2,M2F1,M2F2}.(f) A C = {M1M2,M1F1,M1F2,M2M1,M2F1,M2F2, F1F2, F2F1}.
Solutions for Exercises in Chapter 2 13
(g)
A
A B
B
C
S
2.12 (a) S = {ZY F, ZNF,WY F,WNF, SY F, SNF, ZYM}.(b) A B = {ZY F, ZNF,WY F,WNF, SY F, SNF} = A.(c) A B = {WYF, SY F}.
2.13 A Venn diagram is shown next.
SP
F
S
2.14 (a) A C = {0, 2, 3, 4, 5, 6, 8}.(b) A B = .(c) C = {0, 1, 6, 7, 8, 9}.(d) C D = {1, 6, 7}, so (C D) B = {1, 3, 5, 6, 7, 9}.(e) (S C) = C = {0, 1, 6, 7, 8, 9}.(f) A C = {2, 4}, so A C D = {2, 4}.
2.15 (a) A = {nitrogen, potassium, uranium, oxygen}.(b) A C = {copper, sodium, zinc, oxygen}.(c) A B = {copper, zinc} and
C = {copper, sodium, nitrogen, potassium, uranium, zinc};so (A B) C = {copper, sodium, nitrogen, potassium, uranium, zinc}.
14 Chapter 2 Probability
(d) B C = {copper, uranium, zinc}.(e) A B C = .(f) A B = {copper, nitrogen, potassium, uranium, oxygen, zinc} and
A C = {oxygen}; so, (A B) (A C) = {oxygen}.2.16 (a) M N = {x | 0 < x < 9}.
(b) M N = {x | 1 < x < 5}.(c) M N = {x | 9 < x < 12}.
2.17 A Venn diagram is shown next.
A B
S
1 2 3 4
(a) From the above Venn diagram, (A B) contains the regions of 1, 2 and 4.(b) (A B) contains region 1.(c) A Venn diagram is shown next.
AB
C
S
1
2
3
4
5
6
7
8
(A C) B contains the regions of 3, 4, 5, 7 and 8.2.18 (a) Not mutually exclusive.
(b) Mutually exclusive.
(c) Not mutually exclusive.
(d) Mutually exclusive.
2.19 (a) The family will experience mechanical problems but will receive no ticket fortraffic violation and will not arrive at a campsite that has no vacancies.
(b) The family will receive a traffic ticket and arrive at a campsite that has no va-cancies but will not experience mechanical problems.
Solutions for Exercises in Chapter 2 15
(c) The family will experience mechanical problems and will arrive at a campsite thathas no vacancies.
(d) The family will receive a traffic ticket but will not arrive at a campsite that hasno vacancies.
(e) The family will not experience mechanical problems.
2.20 (a) 6;
(b) 2;
(c) 2, 5, 6;
(d) 4, 5, 6, 8.
2.21 With n1 = 6 sightseeing tours each available on n2 = 3 different days, the multiplicationrule gives n1n2 = (6)(3) = 18 ways for a person to arrange a tour.
2.22 With n1 = 8 blood types and n2 = 3 classifications of blood pressure, the multiplicationrule gives n1n2 = (8)(3) = 24 classifications.
2.23 Since the die can land in n1 = 6 ways and a letter can be selected in n2 = 26 ways, themultiplication rule gives n1n2 = (6)(26) = 156 points in S.
2.24 Since a student may be classified according to n1 = 4 class standing and n2 = 2 genderclassifications, the multiplication rule gives n1n2 = (4)(2) = 8 possible classificationsfor the students.
2.25 With n1 = 5 different shoe styles in n2 = 4 different colors, the multiplication rulegives n1n2 = (5)(4) = 20 different pairs of shoes.
2.26 Using Theorem 2.8, we obtain the followings.
(a) There are(75
)= 21 ways.
(b) There are(53
)= 10 ways.
2.27 Using the generalized multiplication rule, there are n1n2n3n4 = (4)(3)(2)(2) = 48different house plans available.
2.28 With n1 = 5 different manufacturers, n2 = 3 different preparations, and n3 = 2different strengths, the generalized multiplication rule yields n1n2n3 = (5)(3)(2) = 30different ways to prescribe a drug for asthma.
2.29 With n1 = 3 race cars, n2 = 5 brands of gasoline, n3 = 7 test sites, and n4 = 2 drivers,the generalized multiplication rule yields (3)(5)(7)(2) = 210 test runs.
2.30 With n1 = 2 choices for the first question, n2 = 2 choices for the second question,and so forth, the generalized multiplication rule yields n1n2 n9 = 29 = 512 ways toanswer the test.
16 Chapter 2 Probability
2.31 (a) With n1 = 4 possible answers for the first question, n2 = 4 possible answersfor the second question, and so forth, the generalized multiplication rule yields45 = 1024 ways to answer the test.
(b) With n1 = 3 wrong answers for the first question, n2 = 3 wrong answers for thesecond question, and so forth, the generalized multiplication rule yields
n1n2n3n4n5 = (3)(3)(3)(3)(3) = 35 = 243
ways to answer the test and get all questions wrong.
2.32 (a) By Theorem 2.3, 7! = 5040.
(b) Since the first letter must bem, the remaining 6 letters can be arranged in 6! = 720ways.
2.33 Since the first digit is a 5, there are n1 = 9 possibilities for the second digit and thenn2 = 8 possibilities for the third digit. Therefore, by the multiplication rule there aren1n2 = (9)(8) = 72 registrations to be checked.
2.34 (a) By Theorem 2.3, there are 6! = 720 ways.
(b) A certain 3 persons can follow each other in a line of 6 people in a specified order is4 ways or in (4)(3!) = 24 ways with regard to order. The other 3 persons can thenbe placed in line in 3! = 6 ways. By Theorem 2.1, there are total (24)(6) = 144ways to line up 6 people with a certain 3 following each other.
(c) Similar as in (b), the number of ways that a specified 2 persons can follow eachother in a line of 6 people is (5)(2!)(4!) = 240 ways. Therefore, there are 720 240 = 480 ways if a certain 2 persons refuse to follow each other.
2.35 The first house can be placed on any of the n1 = 9 lots, the second house on any of theremaining n2 = 8 lots, and so forth. Therefore, there are 9! = 362, 880 ways to placethe 9 homes on the 9 lots.
2.36 (a) Any of the 6 nonzero digits can be chosen for the hundreds position, and of theremaining 6 digits for the tens position, leaving 5 digits for the units position. So,there are (6)(5)(5) = 150 three digit numbers.
(b) The units position can be filled using any of the 3 odd digits. Any of the remaining5 nonzero digits can be chosen for the hundreds position, leaving a choice of 5digits for the tens position. By Theorem 2.2, there are (3)(5)(5) = 75 three digitodd numbers.
(c) If a 4, 5, or 6 is used in the hundreds position there remain 6 and 5 choices,respectively, for the tens and units positions. This gives (3)(6)(5) = 90 threedigit numbers beginning with a 4, 5, or 6. If a 3 is used in the hundreds position,then a 4, 5, or 6 must be used in the tens position leaving 5 choices for the unitsposition. In this case, there are (1)(3)(5) = 15 three digit number begin witha 3. So, the total number of three digit numbers that are greater than 330 is90 + 15 = 105.
Solutions for Exercises in Chapter 2 17
2.37 The first seat must be filled by any of 5 girls and the second seat by any of 4 boys.Continuing in this manner, the total number of ways to seat the 5 girls and 4 boys is(5)(4)(4)(3)(3)(2)(2)(1)(1) = 2880.
2.38 (a) 8! = 40320.
(b) There are 4! ways to seat 4 couples and then each member of a couple can beinterchanged resulting in 24(4!) = 384 ways.
(c) By Theorem 2.3, the members of each gender can be seated in 4! ways. Thenusing Theorem 2.1, both men and women can be seated in (4!)(4!) = 576 ways.
2.39 (a) Any of the n1 = 8 finalists may come in first, and of the n2 = 7 remaining finalistscan then come in second, and so forth. By Theorem 2.3, there 8! = 40320 possibleorders in which 8 finalists may finish the spelling bee.
(b) The possible orders for the first three positions are 8P3 =8!5!= 336.
2.40 By Theorem 2.4, 8P5 =8!3!= 6720.
2.41 By Theorem 2.4, 6P4 =6!2!= 360.
2.42 By Theorem 2.4, 40P3 =40!37!
= 59, 280.
2.43 By Theorem 2.5, there are 4! = 24 ways.
2.44 By Theorem 2.5, there are 7! = 5040 arrangements.
2.45 By Theorem 2.6, there are 8!3!2!
= 3360.
2.46 By Theorem 2.6, there are 9!3!4!2!
= 1260 ways.
2.47 By Theorem 2.7, there are(
127,3,2
)= 7920 ways.
2.48(
91,4,4
)+(
92,4,3
)+(
91,3,5
)+(
92,3,4
)+(
92,2,5
)= 4410.
2.49 By Theorem 2.8, there are(83
)= 56 ways.
2.50 Assume February 29th as March 1st for the leap year. There are total 365 days in ayear. The number of ways that all these 60 students will have different birth dates (i.e,arranging 60 from 365) is 365P60. This is a very large number.
2.51 (a) Sum of the probabilities exceeds 1.
(b) Sum of the probabilities is less than 1.
(c) A negative probability.
(d) Probability of both a heart and a black card is zero.
2.52 Assuming equal weights
18 Chapter 2 Probability
(a) P (A) = 518;
(b) P (C) = 13;
(c) P (A C) = 736.
2.53 S = {$10, $25, $100} with weights 275/500 = 11/20, 150/500 = 3/10, and 75/500 =3/20, respectively. The probability that the first envelope purchased contains less than$100 is equal to 11/20 + 3/10 = 17/20.
2.54 (a) P (S D) = 88/500 = 22/125.(b) P (E D S ) = 31/500.(c) P (S E ) = 171/500.
2.55 Consider the eventsS: industry will locate in Shanghai,B: industry will locate in Beijing.
(a) P (S B) = P (S) + P (B) P (S B) = 0.7 + 0.4 0.8 = 0.3.(b) P (S B) = 1 P (S B) = 1 0.8 = 0.2.
2.56 Consider the eventsB: customer invests in tax-free bonds,M : customer invests in mutual funds.
(a) P (B M) = P (B) + P (M) P (B M) = 0.6 + 0.3 0.15 = 0.75.(b) P (B M ) = 1 P (B M) = 1 0.75 = 0.25.
2.57 (a) Since 5 of the 26 letters are vowels, we get a probability of 5/26.
(b) Since 9 of the 26 letters precede j, we get a probability of 9/26.
(c) Since 19 of the 26 letters follow g, we get a probability of 19/26.
2.58 (a) Let A = Defect in brake system; B = Defect in fuel system; P (AB) = P (A) +P (B) P (A B) = 0.25 + 0.17 0.15 = 0.27.
(b) P (No defect) = 1 P (A B) = 1 0.27 = 0.73.2.59 By Theorem 2.2, there are N = (26)(25)(24)(9)(8)(7)(6) = 47, 174, 400 possible ways
to code the items of which n = (5)(25)(24)(8)(7)(6)(4) = 4, 032, 000 begin with a voweland end with an even digit. Therefore, n
N= 10
117.
2.60 (a) Of the (6)(6) = 36 elements in the sample space, only 5 elements (2,6), (3,5),(4,4), (5,3), and (6,2) add to 8. Hence the probability of obtaining a total of 8 isthen 5/36.
(b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a totalof at most is 10/36=5/18.
Solutions for Exercises in Chapter 2 19
2.61 Since there are 20 cards greater than 2 and less than 8, the probability of selecting twoof these in succession is (
20
52
)(19
51
)=
95
663.
2.62 (a)(11)(
82)
(93)= 1
3.
(b)(52)(
31)

(93)= 5
14.
2.63 (a)(43)(
482 )
(525 )= 94

54145.
(b)(134 )(
131 )
(525 )= 143
39984.
2.64 Any four of a kind, say four 2s and one 5 occur in(51
)= 5 ways each with probability
(1/6)(1/6)(1/6)(1/6)(1/6) = (1/6)5. Since there are 6P2 = 30 ways to choose variouspairs of numbers to constitute four of one kind and one of the other (we use permutationinstead of combination is because that four 2s and one 5, and four 5s and one 2 aretwo different ways), the probability is (5)(30)(1/6)5 = 25/1296.
2.65 (a) P (M H) = 88/100 = 22/25;(b) P (M H ) = 12/100 = 3/25;(c) P (H M ) = 34/100 = 17/50.
2.66 (a) 9;
(b) 1/9.
2.67 (a) 0.32;
(b) 0.68;
(c) office or den.
2.68 (a) 1 0.42 = 0.58;(b) 1 0.04 = 0.96.
2.69 P (A) = 0.2 and P (B) = 0.35
(a) P (A) = 1 0.2 = 0.8;(b) P (A B) = 1 P (A B) = 1 0.2 0.35 = 0.45;(c) P (A B) = 0.2 + 0.35 = 0.55.
2.70 (a) 0.02 + 0.30 = 0.32 = 32%;
(b) 0.32 + 0.25 + 0.30 = 0.87 = 87%;
20 Chapter 2 Probability
(c) 0.05 + 0.06 + 0.02 = 0.13 = 13%;
(d) 1 0.05 0.32 = 0.63 = 63%.2.71 (a) 0.12 + 0.19 = 0.31;
(b) 1 0.07 = 0.93;(c) 0.12 + 0.19 = 0.31.
2.72 (a) 1 0.40 = 0.60.(b) The probability that all six purchasing the electric oven or all six purchasing the
gas oven is 0.007 + 0.104 = 0.111. So the probability that at least one of eachtype is purchased is 1 0.111 = 0.889.
2.73 (a) P (C) = 1 P (A) P (B) = 1 0.990 0.001 = 0.009;(b) P (B) = 1 P (B) = 1 0.001 = 0.999;(c) P (B) + P (C) = 0.01.
2.74 (a) ($4.50 $4.00) 50, 000 = $25, 000;(b) Since the probability of underfilling is 0.001, we would expect 50, 0000.001 = 50
boxes to be underfilled. So, instead of having ($4.50 $4.00) 50 = $25 profitfor those 50 boxes, there are a loss of $4.00 50 = $200 due to the cost. So, theloss in profit expected due to underfilling is $25 + $200 = $250.
2.75 (a) 1 0.95 0.002 = 0.048;(b) ($25.00 $20.00) 10, 000 = $50, 000;(c) (0.05)(10, 000) $5.00 + (0.05)(10, 000) $20 = $12, 500.
2.76 P (AB) = 1P (AB) = 1(P (A)+P (B)P (AB) = 1+P (AB)P (A)P (B).2.77 (a) The probability that a convict who pushed dope, also committed armed robbery.
(b) The probability that a convict who committed armed robbery, did not push dope.
(c) The probability that a convict who did not push dope also did not commit armedrobbery.
2.78 P (S | A) = 10/18 = 5/9.2.79 Consider the events:

M : a person is a male;S: a person has a secondary education;C: a person has a college degree.
(a) P (M | S) = 28/78 = 14/39;(b) P (C | M ) = 95/112.
Solutions for Exercises in Chapter 2 21
2.80 Consider the events:A: a person is experiencing hypertension,B: a person is a heavy smoker,C: a person is a nonsmoker.
(a) P (A | B) = 30/49;(b) P (C | A) = 48/93 = 16/31.
2.81 (a) P (M P H) = 1068= 5
34;
(b) P (H M | P ) = P (HMP )P (P )
= 221010068 =
1232
= 38.
2.82 (a) (0.90)(0.08) = 0.072;
(b) (0.90)(0.92)(0.12) = 0.099.
2.83 (a) 0.018;
(b) 0.22 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614;
(c) 0.102/0.614 = 0.166;
(d) 0.102+0.0460.175+0.134
= 0.479.
2.84 Consider the events:C: an oil change is needed,F : an oil filter is needed.
(a) P (F | C) = P (FC)P (C)
= 0.140.25
= 0.56.
(b) P (C | F ) = P (CF )P (F )
= 0.140.40
= 0.35.

2.85 Consider the events:H : husband watches a certain show,W : wife watches the same show.
(a) P (W H) = P (W )P (H | W ) = (0.5)(0.7) = 0.35.(b) P (W | H) = P (WH)
P (H)= 0.35
0.4= 0.875.
(c) P (W H) = P (W ) + P (H) P (W H) = 0.5 + 0.4 0.35 = 0.55.2.86 Consider the events:
H : the husband will vote on the bond referendum,W : the wife will vote on the bond referendum.Then P (H) = 0.21, P (W ) = 0.28, and P (H W ) = 0.15.(a) P (H W ) = P (H) + P (W ) P (H W ) = 0.21 + 0.28 0.15 = 0.34.(b) P (W | H) = P (HW )
P (H)= 0.15
0.21= 5
7.
(c) P (H | W ) = P (HW )P (W )
= 0.060.72
= 112.
22 Chapter 2 Probability
2.87 Consider the events:A: the vehicle is a camper,B: the vehicle has Canadian license plates.
(a) P (B | A) = P (AB)P (A)
= 0.090.28
= 928.
(b) P (A | B) = P (AB)P (B)
= 0.090.12
= 34.
(c) P (B A) = 1 P (A B) = 1 0.09 = 0.91.2.88 Define
H : head of household is home,C: a change is made in long distance carriers.P (H C) = P (H)P (C | H) = (0.4)(0.3) = 0.12.
2.89 Consider the events:A: the doctor makes a correct diagnosis,B: the patient sues.P (A B) = P (A)P (B | A) = (0.3)(0.9) = 0.27.
2.90 (a) 0.43;
(b) (0.53)(0.22) = 0.12;
(c) 1 (0.47)(0.22) = 0.90.2.91 Consider the events:
A: the house is open,B: the correct key is selected.
P (A) = 0.4, P (A) = 0.6, and P (B) = (11)(
72)
(83)= 3
8= 0.375.
So, P [A (A B)] = P (A) + P (A)P (B) = 0.4 + (0.6)(0.375) = 0.625.2.92 Consider the events:
F : failed the test,P : passed the test.
(a) P (failed at least one tests) = 1 P (P1P2P3P4) = 1 (0.99)(0.97)(0.98)(0.99) =1 0.93 = 0.07,
(b) P (failed 2 or 3) = P (P1)P (P4)(1 P (P2P3)) = (0.99)(0.99)(1 (0.97)(0.98)) =0.0484.
(c) 100 0.07 = 7.(d) 0.25.
2.93 Let A and B represent the availability of each fire engine.
(a) P (A B) = P (A)P (B) = (0.04)(0.04) = 0.0016.(b) P (A B) = 1 P (A B) = 1 0.0016 = 0.9984.
Solutions for Exercises in Chapter 2 23
2.94 P (T N ) = P (T )P (N ) = (1 P (T ))(1 P (N)) = (0.3)(0.1) = 0.03.
2.95 Consider the events:A1: aspirin tablets are selected from the overnight case,A2: aspirin tablets are selected from the tote bag,L2: laxative tablets are selected from the tote bag,T1: thyroid tablets are selected from the overnight case,T2: thyroid tablets are selected from the tote bag.
(a) P (T1 T2) = P (T1)P (T2) = (3/5)(2/6) = 1/5.(b) P (T
1 T 2) = P (T 1)P (T 2) = (2/5)(4/6) = 4/15.(c) 1P (A1 A2)P (T1T2) = 1P (A1)P (A2)P (T1)P (T2) = 1 (2/5)(3/6)
(3/5)(2/6) = 3/5.
2.96 Consider the events:X: a person has an X-ray,C: a cavity is filled,T : a tooth is extracted.P (X C T ) = P (X)P (C | X)P (T | X C) = (0.6)(0.3)(0.1) = 0.018.
2.97 (a) P (Q1Q2Q3Q4) = P (Q1)P (Q2 | Q1)P (Q3 | Q1Q2)P (Q4 | Q1Q2Q3) =(15/20)(14/19)(13/18)(12/17) = 91/323.
(b) Let A be the event that 4 good quarts of milk are selected. Then
P (A) =
(154
)(204
) = 91323
.
2.98 P = (0.95)[1 (1 0.7)(1 0.8)](0.9) = 0.8037.
2.99 This is a parallel system of two series subsystems.
(a) P = 1 [1 (0.7)(0.7)][1 (0.8)(0.8)(0.8)] = 0.75112.(b) P = P (A
CDE)P system works
= (0.3)(0.8)(0.8)(0.8)0.75112
= 0.2045.
2.100 Define S: the system works.P (A | S ) = P (AS)
P (S)= P (A
)(1P (CDE))1P (S) =
(0.3)[1(0.8)(0.8)(0.8)]10.75112 = 0.588.
2.101 Consider the events:C: an adult selected has cancer,D: the adult is diagnosed as having cancer.P (C) = 0.05, P (D | C) = 0.78, P (C ) = 0.95 and P (D | C ) = 0.06. So, P (D) =P (C D) + P (C D) = (0.05)(0.78) + (0.95)(0.06) = 0.096.
24 Chapter 2 Probability
2.102 Let S1, S2, S3, and S4 represent the events that a person is speeding as he passes throughthe respective locations and let R represent the event that the radar traps is operatingresulting in a speeding ticket. Then the probability that he receives a speeding ticket:
P (R) =4i=1
P (R | Si)P (Si) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27.
2.103 P (C | D) = P (CD)P (D)
= 0.0390.096
= 0.40625.
2.104 P (S2 | R) = P (R S2)P (R) = 0.030.27 = 1/9.2.105 Consider the events:
A: no expiration date,B1: John is the inspector, P (B1) = 0.20 and P (A | B1) = 0.005,B2: Tom is the inspector, P (B2) = 0.60 and P (A | B2) = 0.010,B3: Jeff is the inspector, P (B3) = 0.15 and P (A | B3) = 0.011,B4: Pat is the inspector, P (B4) = 0.05 and P (A | B4) = 0.005,P (B1 | A) = (0.005)(0.20)(0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05) = 0.1124.
2.106 Consider the eventsE: a malfunction by other human errors,A: station A, B: station B, and C: station C.P (C |E) = P (E | C)P (C)
P (E | A)P (A)+P (E | B)P (B)+P (E | C)P (C) =(5/10)(10/43)
(7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43)=
0.11630.4419
= 0.2632.
2.107 (a) P (A B C) = P (C | A B)P (B | A)P (A) = (0.20)(0.75)(0.3) = 0.045.(b) P (B C) = P (A B C) + P (A B C) = P (C | A B)P (B | A)P (A) +
P (C | AB)P (B | A)P (A) = (0.80)(10.75)(0.3)+(0.90)(10.20)(10.3) =0.564.
(c) Use similar argument as in (a) and (b), P (C) = P (AB C)+P (AB C)+P (A B C) + P (A B C) = 0.045 + 0.060 + 0.021 + 0.504 = 0.630.
(d) P (A | B C) = P (A B C)/P (B C) = (0.06)(0.564) = 0.1064.2.108 Consider the events:
A: a customer purchases latex paint,A: a customer purchases semigloss paint,B: a customer purchases rollers.P (A | B) = P (B | A)P (A)
P (B | A)P (A)+P (B | A)P (A) =(0.60)(0.75)
(0.60)(0.75)+(0.25)(0.30)= 0.857.
2.109 Consider the events:G: guilty of committing a crime,I: innocent of the crime,i: judged innocent of the crime,g: judged guilty of the crime.P (I | g) = P (g | I)P (I)
P (g | G)P (G)+P (g | I)P (I) =(0.01)(0.95)
(0.05)(0.90)+(0.01)(0.95)= 0.1743.
Solutions for Exercises in Chapter 2 25
2.110 Let Ai be the event that the ith patient is allergic to some type of week.
(a) P (A1 A2 A3 A4) + P (A1 A2 A3 A4) + P (A1 A2 A3 A4) +P (A
1 A2 A3 A4) = P (A1)P (A2)P (A3)P (A4) + P (A1)P (A2)P (A3)P (A4) +P (A1)P (A
2)P (A3)P (A4) + P (A
1)P (A2)P (A3)P (A4) = (4)(1/2)4 = 1/4.
(b) P (A
1 A2 A3 A4) = P (A1)P (A2)P (A3)P (A4) = (1/2)4 = 1/16.2.111 No solution necessary.
2.112 (a) 0.28 + 0.10 + 0.17 = 0.55.
(b) 1 0.17 = 0.83.(c) 0.10 + 0.17 = 0.27.
2.113 P =(134 )(
136 )(
131 )(
132 )
(5213).
2.114 (a) P (M1 M2 M3 M4) = (0.1)4 = 0.0001, where Mi represents that ith personmake a mistake.
(b) P (J C R W ) = (0.1)(0.1)(0.9)(0.9) = 0.0081.2.115 Let R, S, and L represent the events that a client is assigned a room at the Ramada
Inn, Sheraton, and Lakeview Motor Lodge, respectively, and let F represents the eventthat the plumbing is faulty.
(a) P (F ) = P (F | R)P (R) + P (F | S)P (S) + P (F | L)P (L) = (0.05)(0.2) +(0.04)(0.4) + (0.08)(0.3) = 0.054.
(b) P (L | F ) = (0.08)(0.3)0.054
= 49.
2.116 (a) There are(93
)= 84 possible committees.
(b) There are(41
)(52
)= 40 possible committees.
(c) There are(31
)(11
)(51
)= 15 possible committees.
2.117 Denote by R the event that a patient survives. Then P (R) = 0.8.
(a) P (R1 R2 R3) + P (R1 R2 R3)P (R1 R2 R3) = P (R1)P (R2)P (R3) +P (R1)P (R
2)P (R3) + P (R
1)P (R2)P (R3) = (3)(0.8)(0.8)(0.2) = 0.384.
(b) P (R1 R2 R3) = P (R1)P (R2)P (R3) = (0.8)3 = 0.512.2.118 Consider events
M : an inmate is a male,N : an inmate is under 25 years of age.P (M N ) = P (M ) + P (N ) P (M N ) = 2/5 + 1/3 5/8 = 13/120.
2.119 There are(43
)(53
)(63
)= 800 possible selections.
26 Chapter 2 Probability
2.120 Consider the events:Bi: a black ball is drawn on the ith drawl,Gi: a green ball is drawn on the ith drawl.
(a) P (B1B2B3)+P (G1G2G3) = (6/10)(6/10)(6/10)+(4/10)(4/10)(4/10) =7/25.
(b) The probability that each color is represented is 1 7/25 = 18/25.2.121 The total number of ways to receive 2 or 3 defective sets among 5 that are purchased
is(32
)(93
)+(33
)(92
)= 288.
2.122 A Venn diagram is shown next.
AB
C
S
1
2
3
4
5
6
7
8
(a) (A B): 1, 2, 3, 6, 7, 8.(b) (A B): 1, 6.(c) (A C) B: 3, 4, 5, 7, 8.
2.123 Consider the events:O: overrun,A: consulting firm A,B: consulting firm B,C: consulting firm C.
(a) P (C |O) = P (O | C)P (C)P (O | A)P (A)+P (O | B)P (B)+P (O | C)P (C) =
(0.15)(0.25)(0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25)
=0.03750.0680
= 0.5515.
(b) P (A | O) = (0.05)(0.40)0.0680
= 0.2941.
2.124 (a) 36;
(b) 12;
(c) order is not important.
2.125 (a) 1(362 )
= 0.0016;

(b)(121 )(
241 )
(362 )= 288
630= 0.4571.
Solutions for Exercises in Chapter 2 27
2.126 Consider the events:C: a woman over 60 has the cancer,P : the test gives a positive result.So, P (C) = 0.07, P (P | C) = 0.1 and P (P | C ) = 0.05.P (C | P ) = P (P | C)P (C)
P (P | C)P (C)+P (P | C)P (C) =(0.1)(0.07)
(0.1)(0.07)+(10.05)(10.07) =0.0070.8905
= 0.00786.
2.127 Consider the events:A: two nondefective components are selected,N : a lot does not contain defective components, P (N) = 0.6, P (A | N) = 1,O: a lot contains one defective component, P (O) = 0.3, P (A | O) = (
192 )(202 )
= 910,
T : a lot contains two defective components,P (T ) = 0.1, P (A | T ) = (182 )(202 )
= 153190
.
(a) P (N | A) = P (A | N)P (N)P (A | N)P (N)+P (A | O)P (O)+P (A | T )P (T ) =
(1)(0.6)(1)(0.6)+(9/10)(0.3)+(153/190)(0.1)
= 0.60.9505
= 0.6312;
(b) P (O | A) = (9/10)(0.3)0.9505
= 0.2841;
(c) P (T | A) = 1 0.6312 0.2841 = 0.0847.
2.128 Consider events:D: a person has the rare disease, P (D) = 1/500,P : the test shows a positive result, P (P | D) = 0.95 and P (P | D) = 0.01.P (D | P ) = P (P | D)P (D)
P (P | D)P (D)+P (P | D)P (D) =(0.95)(1/500)
(0.95)(1/500)+(0.01)(11/500) = 0.1599.
2.129 Consider the events:1: engineer 1, P (1) = 0.7, and 2: engineer 2, P (2) = 0.3,E: an error has occurred in estimating cost, P (E | 1) = 0.02 and P (E | 2) = 0.04.P (1 | E) = P (E | 1)P (1)
P (E | 1)P (1)+P (E | 2)P (2) =(0.02)(0.7)
(0.02)(0.7)+(0.04)(0.3)= 0.5385, and
P (2 | E) = 1 0.5385 = 0.4615. So, more likely engineer 1 did the job.
2.130 Consider the events: D: an item is defective
(a) P (D1D2D3) = P (D1)P (D2)P (D3) = (0.2)3 = 0.008.
(b) P (three out of four are defectives) =(43
)(0.2)3(1 0.2) = 0.0256.
2.131 LetA be the event that an injured worker is admitted to the hospital andN be the eventthat an injured worker is back to work the next day. P (A) = 0.10, P (N) = 0.15 andP (AN) = 0.02. So, P (AN) = P (A)+P (N)P (AN) = 0.1+0.150.02 = 0.23.
2.132 Consider the events:T : an operator is trained, P (T ) = 0.5,M an operator meets quota, P (M | T ) = 0.9 and P (M | T ) = 0.65.P (T | M) = P (M | T )P (T )
P (M | T )P (T )+P (M | T )P (T ) =(0.9)(0.5)
(0.9)(0.5)+(0.65)(0.5)= 0.5807.
28 Chapter 2 Probability
2.133 Consider the events:A: purchased from vendor A,D: a customer is dissatisfied.Then P (A) = 0.2, P (A | D) = 0.5, and P (D) = 0.1.So, P (D | A) = P (A | D)P (D)
P (A)= (0.5)(0.1)
0.2= 0.25.
2.134 (a) P (Union member | New company (same field)) = 1313+10
= 1323= 0.5652.
(b) P (Unemployed | Union member) = 240+13+4+2
= 259= 0.034.
2.135 Consider the events:C: the queen is a carrier, P (C) = 0.5,D: a prince has the disease, P (D | C) = 0.5.P (C | D1D2D3) = P (D
1D
2D
3 | C)P (C)P (D
1D
2D
3 | C)P (C)+P (D
1D
2D
3 | C)P (C)= (0.5)
3(0.5)(0.5)3(0.5)+1(0.5)
= 19.
2.136 Using the solutions to Exercise 2.50, we know that there are total 365P60 ways that notwo students have the same birth date. Since the total number of ways of the birthdates that 60 students can have is 36560, the probability that at least two studentswill have the same birth date in a class of 60 is P = 1 365P60
36560. To compute this

number, regular calculator may not be able to handle it. Using approximation (suchas Stirlings approximation formula), we obtain P = 0.9941, which is quite high.
Chapter 3
Random Variables and Probability
Distributions
3.1 Discrete; continuous; continuous; discrete; discrete; continuous.
3.2 A table of sample space and assigned values of the random variable is shown next.
Sample Space xNNNNNBNBNBNNNBBBNBBBNBBB
01112223
3.3 A table of sample space and assigned values of the random variable is shown next.
Sample Space wHHHHHTHTHTHHHTTTHTTTHTTT
3111
1113
3.4 S = {HHH, THHH,HTHHH, TTHHH, TTTHHH,HTTHHH, THTHHH,HHTHHH, . . .}; The sample space is discrete containing as many elements as thereare positive integers.
29
30 Chapter 3 Random Variables and Probability Distributions
3.5 (a) c = 1/30 since 1 =3
x=0
c(x2 + 4) = 30c.
(b) c = 1/10 since
1 =
2x=0

c
(2
x
)(3
3 x)= c
[(2
0
)(3
3
)+
(2
1
)(3
2
)+
(2
2
)(3
1
)]= 10c.
3.6 (a) P (X > 200) =200
20000(x+100)3
dx = 10000(x+100)2
200
= 19.
(b) P (80 < X < 200) = 12080
20000(x+100)3
dx = 10000(x+100)2
12080
= 10009801
= 0.1020.
3.7 (a) P (X < 1.2) = 10x dx+
1.21
(2 x) dx = x22
10+(2x x2
2
)1.21
= 0.68.
(b) P (0.5 < X < 1) = 10.5x dx = x
2
2
10.5
= 0.375.
3.8 Referring to the sample space in Exercise 3.3 and making use of the fact that P (H) =2/3 and P (T ) = 1/3, we haveP (W = 3) = P (TTT ) = (1/3)3 = 1/27;P (W = 1) = P (HTT ) + P (THT ) + P (TTH) = 3(2/3)(1/3)2 = 2/9;P (W = 1) = P (HHT ) + P (HTH) + P (THH) = 3(2/3)2(1/3) = 2/9;P (W = 3) = P (HHH) = (2/3)3 = 8/27;The probability distribution for W is then
w 3 1 1 3P (W = w) 1/27 2/9 2/9 8/27
3.9 (a) P (0 < X < 1) = 10
2(x+2)5
dx = (x+2)2
5
10= 1.
(b) P (1/4 < X < 1/2) = 1/21/4
2(x+2)5
dx = (x+2)2
5
1/21/4
= 19/80.
3.10 The die can land in 6 different ways each with probability 1/6. Therefore, f(x) = 16,
for x = 1, 2, . . . , 6.
3.11 We can select x defective sets from 2, and 3 x good sets from 5 in (2x
)(5
3x)ways. A
random selection of 3 from 7 sets can be made in(73
)ways. Therefore,
f(x) =
(2x
)(5
3x)(
73
) , x = 0, 1, 2.In tabular form
x 0 1 2f(x) 2/7 4/7 1/7
Solutions for Exercises in Chapter 3 31
The following is a probability histogram:
1 2 3xx
f(x)
f(x)
1/7
2/7
3/7
4/7
3.12 (a) P (T = 5) = F (5) F (4) = 3/4 1/2 = 1/4.(b) P (T > 3) = 1 F (3) = 1 1/2 = 1/2.(c) P (1.4 < T < 6) = F (6) F (1.4) = 3/4 1/4 = 1/2.
3.13 The c.d.f. of X is
F (x) =
0, for x < 0,
0.41, for 0 x < 1,0.78, for 1 x < 2,0.94, for 2 x < 3,0.99, for 3 x < 4,1, for x 4.
3.14 (a) P (X < 0.2) = F (0.2) = 1 e1.6 = 0.7981;(b) f(x) = F (x) = 8e8x. Therefore, P (X < 0.2) = 8
0.20
e8x dx = e8x|0.20 =0.7981.
3.15 The c.d.f. of X is
F (x) =
0, for x < 0,
2/7, for 0 x < 1,6/7, for 1 x < 2,1, for x 2.
(a) P (X = 1) = P (X 1) P (X 0) = 6/7 2/7 = 4/7;(b) P (0 < X 2) = P (X 2) P (X 0) = 1 2/7 = 5/7.
32 Chapter 3 Random Variables and Probability Distributions
3.16 A graph of the c.d.f. is shown next.
F(x)
F(x)
xx
1/7
2/7
3/7
4/7
5/7
6/7
1
0 1 2
3.17 (a) Area = 31(1/2) dx = x
2
31= 1.
(b) P (2 < X < 2.5) 2.52
(1/2) dx = x2
2.52
= 14.
(c) P (X 1.6) = 1.61
(1/2) dx = x2
1.61

= 0.3.
3.18 (a) P (X < 4) = 42
2(1+x)27
dx = (1+x)2
27
42= 16/27.
(b) P (3 X < 4) = 43
2(1+x)27
dx = (1+x)2
27

43= 1/3.
3.19 F (x) = x1(1/2) dt = x1
2,
P (2 < X < 2.5) = F (2.5) F (2) = 1.52 1
2= 1
4.
3.20 F (x) = 227
x2(1 + t) dt = 2
27
(t+ t
2
2
)x2= (x+4)(x2)
27,
P (3 X < 4) = F (4) F (3) = (8)(2)27
(7)(1)27
= 13.
3.21 (a) 1 = k 10
x dx = 2k
3x3/2
10= 2k
3. Therefore, k = 3
2.
(b) F (x) = 32
x0
t dt = t3/2
x0= x3/2.
P (0.3 < X < 0.6) = F (0.6) F (0.3) = (0.6)3/2 (0.3)3/2 = 0.3004.
3.22 Denote by X the number of spades int he three draws. Let S and N stand for a spadeand not a spade, respectively. ThenP (X = 0) = P (NNN) = (39/52)(38/51)(37/50) = 703/1700,P (X = 1) = P (SNN) + P (NSN) + P (NNS) = 3(13/52)(39/51)(38/50) = 741/1700,P (X = 3) = P (SSS) = (13/52)(12/51)(11/50) = 11/850, andP (X = 2) = 1 703/1700 741/1700 11/850 = 117/850.The probability mass function for X is then
x 0 1 2 3f(x) 703/1700 741/1700 117/850 11/850
Solutions for Exercises in Chapter 3 33
3.23 The c.d.f. of X is
F (x) =
0, for w < 3,1/27, for 3 w < 1,7/27, for 1 w < 1,19/27, for 1 w < 3,1, for w 3,
(a) P (W > 0 = 1 P (W 0) = 1 7/27 = 20/27.(b) P (1 W < 3) = F (2) F (3) = 19/27 1/27 = 2/3.
3.24 There are(104
)ways of selecting any 4 CDs from 10. We can select x jazz CDs from 5
and 4 x from the remaining CDs in (5x
)(5
4x)ways. Hence
f(x) =
(5x
)(5
4x)(
104
) , x = 0, 1, 2, 3, 4.3.25 Let T be the total value of the three coins. Let D and N stand for a dime and nickel,
respectively. Since we are selecting without replacement, the sample space containingelements for which t = 20, 25, and 30 cents corresponding to the selecting of 2 nickels
and 1 dime, 1 nickel and 2 dimes, and 3 dimes. Therefore, P (T = 20) =(22)(
41)
(63)= 1
5,
P (T = 25) =(21)(
42)
(63)= 3
5,
P (T = 30) =(43)(63)
= 15,
and the probability distribution in tabular form is
t 20 25 30P (T = t) 1/5 3/5 1/5
As a probability histogram
20 25 30xx
f(x)
f(x)
1/5
2/5
3/5
34 Chapter 3 Random Variables and Probability Distributions
3.26 Denote by X the number of green balls in the three draws. Let G and B stand for thecolors of green and black, respectively.
Simple Event x P (X = x)BBBGBBBGBBBGBGGGBGGGBGGG
01112223
(2/3)3 = 8/27(1/3)(2/3)2 = 4/27(1/3)(2/3)2 = 4/27(1/3)(2/3)2 = 4/27(1/3)2(2/3) = 2/27(1/3)2(2/3) = 2/27(1/3)2(2/3) = 2/27(1/3)3 = 1/27
The probability mass function for X is then
x 0 1 2 3P (X = x) 8/27 4/9 2/9 1/27
3.27 (a) For x 0, F (x) = x0
12000

exp(t/2000) dt = exp(t/2000)|x0= 1 exp(x/2000). So
F (x) =
{0, x < 0,
1 exp(x/2000), x 0.
(b) P (X > 1000) = 1 F (1000) = 1 [1 exp(1000/2000)] = 0.6065.(c) P (X < 2000) = F (2000) = 1 exp(2000/2000) = 0.6321.
3.28 (a) f(x) 0 and 26.2523.75
25dx = 2
5t26.2523.75
= 2.52.5
= 1.
(b) P (X < 24) = 2423.75
25dx = 2
5(24 23.75) = 0.1.
(c) P (X > 26) = 26.2526
25dx = 2
5(26.25 26) = 0.1. It is not extremely rare.
3.29 (a) f(x) 0 and 1
3x4 dx = 3 x33
1= 1. So, this is a density function.
(b) For x 1, F (x) = x13t4 dt = 1 x3. So,
F (x) =
{0, x < 1,
1 x3, x 1.
(c) P (X > 4) = 1 F (4) = 43 = 0.0156.
3.30 (a) 1 = k 11(3 x2) dx = k
(3x x3
3
)11
= 163k. So, k = 3
16.
Solutions for Exercises in Chapter 3 35
(b) For 1 x < 1, F (x) = 316
x1(3 t2) dt =
(3t 1
3t3)x1 =

12+ 9
16x x3
16.
So, P(X < 1
2
)= 1
2 ( 9
16
) (12
) 116
(12
)3= 99
128.
(c) P (|X| < 0.8) = P (X < 0.8) + P (X > 0.8) = F (0.8) + 1 F (0.8)= 1 +
(12 9
160.8 + 1
160.83
) (12+ 9
160.8 1
160.83
)= 0.164.
3.31 (a) For y 0, F (y) = 14
y0et/4 dy = 1 ey/4. So, P (Y > 6) = e6/4 = 0.2231. This
probability certainly cannot be considered as unlikely.

(b) P (Y 1) = 1 e1/4 = 0.2212, which is not so small either.
3.32 (a) f(y) 0 and 105(1 y)4 dy = (1 y)5|10 = 1. So, this is a density function.
(b) P (Y < 0.1) = (1 y)5|0.10 = 1 (1 0.1)5 = 0.4095.(c) P (Y > 0.5) = (1 0.5)5 = 0.03125.
3.33 (a) Using integral by parts and setting 1 = k 10y4(1 y)3 dy, we obtain k = 280.
(b) For 0 y < 1, F (y) = 56y5(1 Y )3 + 28y6(1 y)2 + 8y7(1 y) + y8. So,P (Y 0.5) = 0.3633.
(c) Using the cdf in (b), P (Y > 0.8) = 0.0563.
3.34 (a) The event Y = y means that among 5 selected, exactly y tubes meet the spec-ification (M) and 5 y (M ) does not. The probability for one combination ofsuch a situation is (0.99)y(1 0.99)5y if we assume independence among thetubes. Since there are 5!
y!(5y)! permutations of getting y Ms and 5 y M s, theprobability of this event (Y = y) would be what it is specified in the problem.
(b) Three out of 5 is outside of specification means that Y = 2. P (Y = 2) = 9.8106which is extremely small. So, the conjecture is false.
3.35 (a) P (X > 8) = 1 P (X 8) =8
x=0
e6 6x
x!= e6
(60
0!+ 6
1
1!+ + 68
8!
)= 0.1528.
(b) P (X = 2) = e6 62
2!= 0.0446.
3.36 For 0 < x < 1, F (x) = 2 x0(1 t) dt = (1 t)2|x0 = 1 (1 x)2.
(a) P (X 1/3) = 1 (1 1/3)2 = 5/9.(b) P (X > 0.5) = (1 1/2)2 = 1/4.(c) P (X < 0.75 | X 0.5) = P (0.5X
36 Chapter 3 Random Variables and Probability Distributions
xf(x, y) 0 1 2 3
0 0 1/30 2/30 3/30y 1 1/30 2/30 3/30 4/30
2 2/30 3/30 4/30 5/30
(a) P (X 2, Y = 1) = f(0, 1) + f(1, 1) + f(2, 1) = 1/30 + 2/30 + 3/30 = 1/5.(b) P (X > 2, Y 1) = f(3, 0) + f(3, 1) = 3/30 + 4/30 = 7/30.(c) P (X > Y ) = f(1, 0) + f(2, 0) + f(3, 0) + f(2, 1) + f(3, 1) + f(3, 2)
= 1/30 + 2/30 + 3/30 + 3/30 + 4/30 + 5/30 = 3/5.
(d) P (X + Y = 4) = f(2, 2) + f(3, 1) = 4/30 + 4/30 = 4/15.
3.39 (a) We can select x oranges from 3, y apples from 2, and 4 x y bananas from 3in(3x
)(2y
)(3
4xy)ways. A random selection of 4 pieces of fruit can be made in
(84
)ways. Therefore,
f(x, y) =
(3x
)(2y
)(3
4xy)(
84
) , x = 0, 1, 2, 3; y = 0, 1, 2; 1 x+ y 4.(b) P [(X, Y ) A] = P (X + Y 2) = f(1, 0) + f(2, 0) + f(0, 1) + f(1, 1) + f(0, 2)
= 3/70 + 9/70 + 2/70 + 18/70 + 3/70 = 1/2.
3.40 (a) g(x) = 23
10(x+ 2y) dy = 2
3(x+ 1), for 0 x 1.
(b) h(y) = 23
10(x+ 2y) dy = 1
3(1 + 4y), for 0 y 1.
(c) P (X < 1/2) = 23
1/20
(x+ 1) dx = 512.
3.41 (a) P (X + Y 1/2) = 1/20
1/2y0
24xy dx dy = 12 1/20
(12 y)2 y dy = 1
16.
(b) g(x) = 1x0
24xy dy = 12x(1 x)2, for 0 x < 1.(c) f(y|x) = 24xy
12x(1x)2 =2y
(1x)2 , for 0 y 1 x.Therefore, P (Y < 1/8 | X = 3/4) = 32 1/8
0y dy = 1/4.
3.42 Since h(y) = ey0ex dx = ey, for y > 0, then f(x|y) = f(x, y)/h(y) = ex, for
x > 0. So, P (0 < X < 1 | Y = 2) = 10ex dx = 0.6321.
3.43 (a) P (0 X 1/2, 1/4 Y 1/2) = 1/20
1/21/4
4xy dy dx = 3/8 1/20
x dx = 3/64.
(b) P (X < Y ) = 10
y04xy dx dy = 2
10y3 dy = 1/2.
3.44 (a) 1 = k 5030
5030(x2 + y2) dx dy = k(50 30)
( 5030x2 dx+
5030y2 dy
)= 392k
3 104.
So, k = 3392
104.
Solutions for Exercises in Chapter 3 37
(b) P (30 X 40, 40 Y 50) = 3392
104 4030
5040(x2 + y2) dy dx
= 3392
103( 4030x2 dx+
5040y2 dy) = 3
392 103
(403303
3+ 50
34033
)= 49
196.
(c) P (30 X 40, 30 Y 40) = 3392
104 4030
4030(x2 + y2) dx dy
= 2 3392
104(40 30) 4030x2 dx = 3
196 103 403303
3= 37
196.
3.45 P (X + Y > 1/2) = 1 P (X + Y < 1/2) = 1 1/40
1/2xx
1ydy dx
= 1 1/40
[ln(12 x) ln x] dx = 1 + [(1
2 x) ln (1
2 x) x ln x]1/4
0
= 1 + 14ln(14
)= 0.6534.
3.46 (a) From the column totals of Exercise 3.38, we have
x 0 1 2 3g(x) 1/10 1/5 3/10 2/5
(b) From the row totals of Exercise 3.38, we have
y 0 1 2h(y) 1/5 1/3 7/15
3.47 (a) g(x) = 2 1xdy = 2(1 x) for 0 < x < 1;
h(y) = 2 y0dx = 2y, for 0 < y < 1.
Since f(x, y) 6= g(x)h(y), X and Y are not independent.(b) f(x|y) = f(x, y)/h(y) = 1/y, for 0 < x < y.

Therefore, P (1/4 < X < 1/2 | Y = 3/4) = 43
1/21/4
dx = 13.
3.48 (a) g(2) =2
y=0
f(2, y) = f(2, 0) + f(2, 1) + f(2, 2) = 9/70 + 18/70 + 3/70 = 3/7. So,
f(y|2) = f(2, y)/g(2) = (7/3)f(2, y).f(0|2) = (7/3)f(2, 0) = (7/3)(9/70) = 3/10, f(1|2) = 3/5 and f(2|2) = 1/10. Intabular form,
y 0 1 2f(y|2) 3/10 3/5 1/10
(b) P (Y = 0 | X = 2) = f(0|2) = 3/10.
3.49 (a)x 1 2 3
g(x) 0.10 0.35 0.55
(b)y 1 2 3
h(y) 0.20 0.50 0.30

(c) P (Y = 3 | X = 2) = 0.20.05+0.10+0.20
= 0.5714.
38 Chapter 3 Random Variables and Probability Distributions
3.50
xf(x, y) 2 4 h(y)
1 0.10 0.15 0.25y 3 0.20 0.30 0.50
5 0.10 0.15 0.25g(x) 0.40 0.60
(a)x 2 4
g(x) 0.40 0.60
(b)y 1 3 5
h(y) 0.25 0.50 0.25
3.51 (a) Let X be the number of 4s and Y be the number of 5s. The sample spaceconsists of 36 elements each with probability 1/36 of the form (m,n) wherem is the outcome of the first roll of the die and n is the value obtained onthe second roll. The joint probability distribution f(x, y) is defined for x =0, 1, 2 and y = 0, 1, 2 with 0 x + y 2. To find f(0, 1), for example,consider the event A of obtaining zero 4s and one 5 in the 2 rolls. ThenA = {(1, 5), (2, 5), (3, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 6)}, so f(0, 1) = 8/36 = 2/9.In a like manner we find f(0, 0) = 16/36 = 4/9, f(0, 2) = 1/36, f(1, 0) = 2/9,f(2, 0) = 1/36, and f(1, 1) = 1/18.
(b) P [(X, Y ) A] = P (2X + Y < 3) = f(0, 0) + f(0, 1) + f(0, 2) + f(1, 0) =4/9 + 1/9 + 1/36 + 2/9 = 11/12.
3.52 A tabular form of the experiment can be established as,
Sample Space x yHHHHHTHTHTHHHTTTHTTTHTTT
32221110
3111
1113
So, the joint probability distribution is,
xf(x, y) 0 1 2 3
3 1/8y 1 3/8
1 3/83 1/8
Solutions for Exercises in Chapter 3 39
3.53 (a) If (x, y) represents the selection of x kings and y jacks in 3 draws, we must havex = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0 x + y 3. Therefore, (1, 2) represents theselection of 1 king and 2 jacks which will occur with probability
f(1, 2) =
(41
)(42
)(123
) = 655.
Proceeding in a similar fashion for the other possibilities, we arrive at the follow-ing joint probability distribution:
xf(x, y) 0 1 2 3
0 1/55 6/55 6/55 1/55y 1 6/55 16/55 6/55
2 6/55 6/553 1/55
(b) P [(X, Y ) A] = P (X +Y 2) = 1P (X +Y < 2) = 1 1/55 6/55 6/55 =42/55.
3.54 (a) P (H) = 0.4, P (T ) = 0.6, and S = {HH,HT, TH, TT}. Let (W,Z) representa typical outcome of the experiment. The particular outcome (1, 0) indicating atotal of 1 head and no heads on the first toss corresponds to the event TH . There-fore, f(1, 0) = P (W = 1, Z = 0) = P (TH) = P (T )P (H) = (0.6)(0.4) = 0.24.Similar calculations for the outcomes (0, 0), (1, 1), and (2, 1) lead to the followingjoint probability distribution:
wf(w, z) 0 1 2

z 0 0.36 0.241 0.24 0.16
(b) Summing the columns, the marginal distribution of W is
w 0 1 2g(w) 0.36 0.48 0.16
(c) Summing the rows, the marginal distribution of Z is
z 0 1h(z) 0.60 0.40
(d) P (W 1) = f(1, 0) + f(1, 1) + f(2, 1) = 0.24 + 0.24 + 0.16 = 0.64.
40 Chapter 3 Random Variables and Probability Distributions
3.55 g(x) = 18
42(6 x y) dy = 3x
4, for 0 < x < 2.
So, f(y|x) = f(x,y)g(x)
= 6xy2(3x) , for 2 < y < 4,
and P (1 < Y < 3 | X = 1) = 14
32(5 y) dy = 5
8.
3.56 Since f(1, 1) 6= g(1)h(1), the variables are not independent.3.57 X and Y are independent since f(x, y) = g(x)h(y) for all (x, y).
3.58 (a) h(y) = 6 1y0
x dx = 3(1 y)2, for 0 < y < 1. Since f(x|y) = f(x,y)h(y)
= 2x(1y)2 , for
0 < x < 1 y, involves the variable y, X and Y are not independent.(b) P (X > 0.3 | Y = 0.5) = 8 0.5
0.3x dx = 0.64.
3.59 (a) 1 = k 10
10
20xy2z dx dy dz = 2k
10
10y2z dy dz = 2k
3
10z dz = k
3. So, k = 3.
(b) P(X < 1
4, Y > 1
2, 1 < Z < 2
)= 3
1/40
11/2
21xy2z dx dy dz = 9
2
1/40
11/2
y2z dy dz
= 2116
1/40
z dz = 21512
.
3.60 g(x) = 4 10xy dy = 2x, for 0 < x < 1; h(y) = 4
10xy dx = 2y, for 0 < y < 1. Since
f(x, y) = g(x)h(y) for all (x, y), X and Y are independent.
3.61 g(x) = k 5030(x2 + y2) dy = k
(x2y + y
3
3
)5030= k
(20x2 + 98,000
3
), and
h(y) = k(20y2 + 98,000
3
).
Since f(x, y) 6= g(x)h(y), X and Y are not independent.
3.62 (a) g(y, z) = 49
10xyz2 dx = 2
9yz2, for 0 < y < 1 and 0 < z < 3.
(b) h(y) = 29
30yz2 dz = 2y, for 0 < y < 1.
(c) P(14< X < 1
2, Y > 1
3, Z < 2
)= 4
9

21
11/3
1/21/4
xyz2 dx dy dz = 7162
.
(d) Since f(x|y, z) = f(x,y,z)g(y,z)
= 2x, for 0 < x < 1, P(0 < X < 1
2| Y = 1
4, Z = 2
)=
2 1/20
x dx = 14.
3.63 g(x) = 24 1x0
xy dy = 12x(1 x)2, for 0 < x < 1.
(a) P (X 0.5) = 12 10.5x(1 x)2 dx = 1
0.5(12x 24x2 + 12x3) dx = 5
16= 0.3125.
(b) h(y) = 24 1y0
xy dx = 12y(1 y)2, for 0 < y < 1.(c) f(x|y) = f(x,y)
h(y)= 24xy
12y(1y)2 =2x
(1y)2 , for 0 < x < 1 y.So, P
(X < 1
8| Y = 3
4
)= 1/80
2x1/16
dx = 32 1/80
= 0.25.
3.64 (a)x 1 3 5 7
f(x) 0.4 0.2 0.2 0.2
(b) P (4 < X 7) = P (X 7) P (X 4) = F (7) F (4) = 1 0.6 = 0.4.
Solutions for Exercises in Chapter 3 41
3.65 (a) g(x) =0yey(1+x) dy = 1
1+xyey(1+x)
0+ 1
1+x
0ey(1+x) dy
= 1(1+x)2
ey(1+x)0
= 1(1+x)2
, for x > 0.
h(y) = yey0eyx dx = ey eyx|0 = ey, for y > 0.
(b) P (X 2, Y 2) = 2
2yey(1+x) dx dy =
2ey eyx|2 dy =
2e3ydy
= 13e3y
2= 1
3e6.
3.66 (a) P(X 1
2, Y 1
2
)= 3

2
1/20
1/20
(x2 + y2) dxdy = 32
1/20
(x2y + y
3
3
)1/20
dx
= 34
1/20
(x2 + 1
12
)dx = 1
16.
(b) P(X 3
4
)= 3
2
13/4
(x2 + 1
3
)dx = 53
128.
3.67 (a)x 0 1 2 3 4 5 6
f(x) 0.1353 0.2707 0.2707 0.1804 0.0902 0.0361 0.0120
(b) A histogram is shown next.
1 2 3 4 5 6 7xx
f(x)
f(x)
0.0
0.1
0.2
0.3
(c)x 0 1 2 3 4 5 6
F (x) 0.1353 0.4060 0.6767 0.8571 0.9473 0.9834 0.9954
3.68 (a) g(x) = 10(x+ y) dy = x+ 1
2, for 0 < x < 1, and h(y) = y + 1
2for 0 < y < 1.
(b) P (X > 0.5, Y > 0.5) = 10.5
10.5(x+ y) dx dy =
10.5
(x2
2+ xy
)10.5
dy
= 10.5
[(12+ y
) (18+ y
2
)]dy = 3
8.
3.69 f(x) =(5x
)(0.1)x(1 0.1)5x, for x = 0, 1, 2, 3, 4, 5.
3.70 (a) g(x) = 21
(3xy9
)dy = 3xyy
2/29
21= x
3 1
6, for 1 < x < 3, and
h(y) = 31
(3xy9
)dx = 4
3 2
9y, for 1 < y < 2.
(b) No, since g(x)h(y) 6= f(x, y).(c) P (X > 2) =
32
(x3 1
6

)dx =
(x2
6 x
6
)32= 2
3.
42 Chapter 3 Random Variables and Probability Distributions
3.71 (a) f(x) = ddxF (x) = 1
50ex/50, for x > 0.
(b) P (X > 70) = 1 P (X 70) = 1 F (70) = 1 (1 e70/50) = 0.2466.3.72 (a) f(x) = 1
10, for x = 1, 2, . . . , 10.
(b) A c.d.f. plot is shown next.F(
x)F(
x)
xx
00.10.20.30.40.50.60.70.80.91.0
1 2 3 4 5 6 7 8 9 10
3.73 P (X 3) = 12
3ey/2 = e3/2 = 0.2231.
3.74 (a) f(x) 0 and 100
110dx = 1. This is a continuous uniform distribution.
(b) P (X 7) = 110
70dx = 0.7.
3.75 (a) f(y) 0 and 10f(y) dy = 10
10(1 y)9 dy = 10
10(1 y)101
0= 1.
(b) P (Y > 0.6) = 10.6f(y) dy = (1 y)10|10.6 = (1 0.6)10 = 0.0001.
3.76 (a) P (Z > 20) = 110
20ez/10 dz = ez/10
20= e20/10 = 0.1353.
(b) P (Z 10) = ez/10100= 1 e10/10 = 0.6321.
3.77 (a) g(x1) = 1x12 dx2 = 2(1 x1), for 0 < x1 < 1.
(b) h(x2) = x20
2 dx1 = 2x2, for 0 < x2 < 1.
(c) P (X1 < 0.2, X2 > 0, 5) = 10.5
0.20
2 dx1 dx2 = 2(1 0.5)(0.2 0) = 0.2.(d) fX1|X2(x1|x2) = f(x1,x2)h(x2) = 22x2 = 1x2 , for 0 < x1 < x2.
3.78 (a) fX1(x1) = x10
6x2 dx2 = 3x21, for 0 < x1 < 1. Apparently, fX1(x1) 0 and 1
0fX1(x1) dx1 =
103x21 dx1 = 1. So, fX1(x1) is a density function.
(b) fX2|X1(x2|x1) = f(x1,x2)fX1 (x1) =6x23x21
= 2x2x21, for 0 < x2 < x1.
So, P (X2 < 0.5 | X1 = 0.7) = 20.72 0.50
x2 dx2 =2549.
Solutions for Exercises in Chapter 3 43
3.79 (a) g(x) = 9(16)4y
x=0
14x
= 9(16)4y
111/4 =
34 14x, for x = 0, 1, 2, . . . ; similarly, h(y) = 3
4 14y,
for y = 0, 1, 2, . . . . Since f(x, y) = g(x)h(y), X and Y are independent.
(b) P (X + Y < 4) = f(0, 0)+ f(0, 1) + f(0, 2)+ f(0, 3)+ f(1, 0)+ f(1, 1)+ f(1, 2)+f(2, 0) + f(2, 1) + f(3, 0) = 9
16
(1 + 1
4+ 1
42+ 1
43+ 1
4+ 1
42+ 1
43+ 1
r2+ 1
43+ 1
43
)=
916
(1 + 2
4+ 3
42+ 4
43
)= 63
64.
3.80 P (the system works) = P (all components work) = (0.95)(0.99)(0.92) = 0.86526.
3.81 P (the system does not fail) = P (at least one of the components works)= 1P (all components fail) = 1 (10.95)(10.94)(10.90)(10.97) = 0.999991.
3.82 Denote by X the number of components (out of 5) work.Then, P (the system is operational) = P (X 3) = P (X = 3) + P (X = 4) + P (X =5) =
(53
)(0.92)3(1 0.92)2 + (5
4
)(0.92)4(1 0.92) + (5
5
)(0.92)5 = 0.9955.
Chapter 4
Mathematical Expectation
4.1 E(X) = 1pia2
aaa2y2a2y2
x dx dy = 1pia2
[(a2y2
2
)(a2y2
2
)]dy = 0.
4.2 E(X) =3
x=0
x f(x) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 3/4.
4.3 = E(X) = (20)(1/5) + (25)(3/5) + (30)(1/5) = 25 cents.
4.4 Assigning wrights of 3w and w for a head and tail, respectively. We obtain P (H) = 3/4and P (T ) = 1/4. The sample space for the experiment is S = {HH,HT, TH, TT}.Now if X represents the number of tails that occur in two tosses of the coin, we have
P (X = 0) = P (HH) = (3/4)(3/4) = 9/16,
P (X = 1) = P (HT ) + P (TH) = (2)(3/4)(1/4) = 3/8,
P (X = 2) = P (TT ) = (1/4)(1/4) = 1/16.
The probability distribution for X is then
x 0 1 2f(x) 9/16 3/8 1/16
from which we get = E(X) = (0)(9/16) + (1)(3/8) + (2)(1/16) = 1/2.
4.5 = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.
4.6 = E(X) = ($7)(1/12)+($9)(1/12)+($11)(1/4)+($13)(1/4)+($15)(1/6)+($17)(1/6)= $12.67.
4.7 Expected gain = E(X) = (4000)(0.3) + (1000)(0.7) = $500.4.8 Let X = profit. Then
= E(X) = (250)(0.22) + (150)(0.36) + (0)(0.28) + (150)(0.14) = $88.
45
46 Chapter 4 Mathematical Expectation
4.9 Let c = amount to play the game and Y = amount won.
y 5 c 3 c cf(y) 2/13 2/13 9/13
E(Y ) = (5 c)(2/13) + (3 c)(2/13) + (c)(9/13) = 0. So, 13c = 16 which impliesc = $1.23.
4.10 X =
xg(x) = (1)(0.17) + (2)(0.5) + (3)(0.33) = 2.16,Y =
yh(y) = (1)(0.23) + (2)(0.5) + (3)(0.27) = 2.04.
4.11 For the insurance of $200,000 pilot, the distribution of the claim the insurance companywould have is as follows:
Claim Amount $200,000 $100,000 $50,000 0f(x) 0.002 0.01 0.1 0.888
So, the expected claim would be
($200, 000)(0.002) + ($100, 000)(0.01) + ($50, 000)(0.1) + ($0)(0.888) = $6, 400.
Hence the insurance company should charge a premium of $6, 400 + $500 = $6, 900.
4.12 E(X) = 102x(1 x) dx = 1/3. So, (1/3)($5, 000) = $1, 667.67.
4.13 E(X) = 4pi
10
x1+x2
dx = ln 4pi.
4.14 E(X) = 10
2x(x+2)5
dx = 815.
4.15 E(X) = 10x2 dx +
21x(2 x) dx = 1. Therefore, the average number of hours per
year is (1)(100) = 100 hours.
4.16 P (X1 +X2 = 1) = P (X1 = 1, X2 = 0) + P (X1 = 0, X2 = 1)
=(9801 )(
201 )
(10002 )+(9801 )(
201 )
(10002 )= (2)(0.0392) = 0.0784.
4.17 The probability density function is,
x 3 6 9f(x) 1/6 1/2 1/3g(x) 25 169 361
g(X) = E[(2X + 1)2] = (25)(1/6) + (169)(1/2) + (361)(1/3) = 209.
4.18 E(X2) = (0)(27/64) + (1)(27/64) + (4)(9/64) + (9)(1/64) = 9/8.
4.19 Let Y = 1200X 50X2 be the amount spent.
Solutions for Exercises in Chapter 4 47
x 0 1 2 3f(x) 1/10 3/10 2/5 1/5
y = g(x) 0 1150 2200 3150
Y = E(1200X 50X2) = (0)(1/10) + (1150)(3/10) + (2200)(2/5) + (3150)(1/5)= $1, 855.
4.20 E[g(X)] = E(e2X/3) =0e2x/3ex dx =
0ex/3 dx = 3.
4.21 E(X2) = 102x2(1 x) dx = 1
6. Therefore, the average profit per new automobile is
(1/6)($5000.00) = $833.33.
4.22 E(Y ) = E(X + 4) =0
32(x+ 4) 1(x+4)3
dx = 8 days.
4.23 (a) E[g(X, Y )] = E(XY 2) =x
y
xy2f(x, y)
= (2)(1)2(0.10) + (2)(3)2(0.20) + (2)(5)2(0.10) + (4)(1)2(0.15) + (4)(3)2(0.30)+ (4)(5)2(0.15) = 35.2.
(b) X = E(X) = (2)(0.40) + (4)(0.60) = 3.20,Y = E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3.00.
4.24 (a) E(X2Y 2XY ) =3
x=0
2y=0
(x2y 2xy)f(x, y) = (1 2)(18/70) + (4 4)(18/70) + + (8 8)(3/70) = 3/7.
(b)x 0 1 2 3
g(x) 5/70 30/70 30/70 5/70y 0 1 2
h(y) 15/70 40/70 15/70X = E(X) = (0)(5/70) + (1)(30/70) + (2)(30/70) + (3)(5/70) = 3/2,Y = E(Y ) = (0)(15/70) + (1)(40/70) + (2)(15/70) = 1.
4.25 X+Y = E(X + Y ) =3
x=0
3y=0
(x+ y)f(x, y) = (0+ 0)(1/55)+ (1+ 0)(6/55)+ + (0+3)(1/55) = 2.
4.26 E(Z) = E(X2 + Y 2) =
10
104xy
x2 + y2 dx dy = 4
3
10[y(1 + y2)3/2 y4] dy
= 8(23/2 1)/15 = 0.9752.4.27 E(X) = 1
2000
0x exp(x/2000) dx = 2000
0y exp(y) dy = 2000.
4.28 (a) The density function is shown next.
f(x)
23.75 26.25
2/5
48 Chapter 4 Mathematical Expectation
(b) E(X) = 25
26.2523.75
x dx = 15(26.252 23.752) = 25.
(c) The mean is exactly in the middle of the interval. This should not be surpriseddue to the symmetry of the density at 25.
4.29 (a) The density function is shown nextf(x
)
1 1.5 2 2.5 3 3.5 40
1
2
3
(b) = E(X) =1
3x3 dx = 32.
4.30 E(Y ) = 14
0yey/4 dy = 4.
4.31 (a) = E(Y ) = 5 10y(1 y)4 dy = 1
0y d(1 y)5 =
0(1 y)5 dy = 1
6.
(b) P (Y > 1/6) = 11/6
5(1 y)4 dy = (1 y)5|11/6 = (1 1/6)5 = 0.4019.
4.32 (a) A histogram is shown next.
1 2 3 4 5xx
f(x)
f(x)
0
0.1
0.2
0.3
0.4
(b) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.
(c) E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.

(d) V ar(X) = 1.62 0.882 = 0.8456.4.33 = $500. So, 2 = E[(X )2] =
x
(x )2f(x) = (1500)2(0.7) + (3500)2(0.3) =$5, 250, 000.
Solutions for Exercises in Chapter 4 49
4.34 = (2)(0.3) + (3)(0.2) + (5)(0.5) = 2.5 andE(X2) = (2)2(0.3) + (3)2(0.2) + (5)2(0.5) = 15.5.So, 2 = E(X2) 2 = 9.25 and = 3.041.
4.35 = (2)(0.01) + (3)(0.25) + (4)(0.4) + (5)(0.3) + (6)(0.04) = 4.11,E(X2) = (2)2(0.01) + (3)2(0.25) + (4)2(0.4) + (5)2(0.3) + (6)2(0.04) = 17.63.So, 2 = 17.63 4.112 = 0.74.
4.36 = (0)(0.4) + (1)(0.3) + (2)(0.2) + (3)(0.1) = 1.0,and E(X2) = (0)2(0.4) + (1)2(0.3) + (2)2(0.2) + (3)2(0.1) = 2.0.So, 2 = 2.0 1.02 = 1.0.
4.37 It is know = 1/3.
So, E(X2) = 102x2(1 x) dx = 1/6 and 2 = 1/6 (1/3)2 = 1/18. So, in the actual
profit, the variance is 118(5000)2.
4.38 It is known = 8/15.
Since E(X2) = 10
25x2(x+ 2) dx = 11
30, then 2 = 11/30 (8/15)2 = 37/450.
4.39 It is known = 1.Since E(X2) =
10x2 dx+

21x2(2 x) dx = 7/6, then 2 = 7/6 (1)2 = 1/6.
4.40 g(X) = E[g(X)] = 10(3x2 + 4)
(2x+45
)dx = 1
5
10(6x3 + 12x2 + 8x+ 16) dx = 5.1.
So, 2 = E[g(X) ]2 = 10(3x2 + 4 5.1)2 (2x+4
5
)dx
= 10(9x4 6.6x2 + 1.21) (2x+4
5
)dx = 0.83.
4.41 It is known g(X) = E[(2X + 1)2] = 209. Hence
2g(X) =x
[(2X + 1)2 209]2g(x)= (25 209)2(1/6) + (169 209)2(1/2) + (361 209)2(1/3) = 14, 144.So, g(X) =
14, 144 = 118.9.
4.42 It is known g(X) = E(X2) = 1/6. Hence
2g(X) = 102(x2 1
6
)2(1 x) dx = 7/180.
4.43 Y = E(3X 2) = 140(3x 2)ex/4 dx = 10. So
2Y = E{[(3X 2) 10]2} = 940(x 4)2ex/4 dx = 144.
4.44 E(XY ) =x
y
xyf(x, y) = (1)(1)(18/70) + (2)(1)(18/70)
+ (3)(1)(2/70) + (1)(2)(9/70) + (2)(2)(3/70) = 9/7;X =
x
y
xf(x, y) = (0)f(0, 1) + (0)f(0, 2) + (1)f(1, 0) + + (3)f(3, 1) = 3/2,and Y = 1.So, XY = E(XY ) XY = 9/7 (3/2)(1) = 3/14.
50 Chapter 4 Mathematical Expectation
4.45 X =x
xg(x) = 2.45, Y =y
yh(y) = 2.10, and
E(XY ) =x
x
xyf(x, y) = (1)(0.05) + (2)(0.05) + (3)(0.10) + (2)(0.05)
+ (4)(0.10) + (6)(0.35) + (3)(0) + (6)(0.20) + (9)(0.10) = 5.15.So, XY = 5.15 (2.45)(2.10) = 0.005.
4.46 From previous exercise, k =(
3392
)104, and g(x) = k
(20x2 + 98000
3
), with
X = E(X) = 5030xg(x) dx = k
5030
(20x3 + 98000
3x)dx = 40.8163.

Similarly, Y = 40.8163. On the other hand,E(XY ) = k
5030
5030xy(x2 + y2) dy dx = 1665.3061.
Hence, XY = E(XY ) XY = 1665.3061 (40.8163)2 = 0.6642.4.47 g(x) = 2
3
10(x+ 2y) dy = 2
3(x+ 1, for 0 < x < 1, so X =
23
10x(x+ 1) dx = 5
9;
h(y) = 23
10(x+ 2y) dx = 2
3
(12+ 2y
), so Y =
23
10y(12+ 2y
)dy = 11
18; and
E(XY ) = 23
10
10xy(x+ 2y) dy dx = 1
3.
So, XY = E(XY ) XY = 13 (59
) (1118
)= 0.0062.
4.48 Since XY = Cov(a+ bX,X) = b2X and
2Y = b
22X , then
= XYX
Y =b2X2Xb
22X= b|b| = sign of b.
Hence = 1 if b > 0 and = 1 if b < 0.4.49 E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88
and E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.So, V ar(X) = 1.62 0.882 = 0.8456 and = 0.8456 = 0.9196.
4.50 E(X) = 2 10x(1 x) dx = 2
(x2
2 x3
3
)10= 1
3and
E(X2) = 2 10x2(1 x) dx = 2
(x3
3 x4
4
)10= 1
6. Hence,
V ar(X) = 16 (1
3
)2= 1
18, and =
1/18 = 0.2357.
4.51 Previously we found = 4.11 and 2 = 0.74, Therefore,g(X) = E(3X 2) = 3 2 = (3)(4.11) 2 = 10.33 and g(X) = 92 = 6.66.
4.52 Previously we found = 1 and 2 = 1. Therefore,g(X) = E(5X + 3) = 5+ 3 = (5)(1) + 3 = 8 and g(X) = 25
2 = 25.
4.53 Let X = number of cartons sold and Y = profit.We can write Y = 1.65X + (0.90)(5X) 6 = 0.75X 1.50. NowE(X) = (0)(1/15)+(1)(2/15)+(2)(2/15)+(3)(3/15)+(4)(4/15)+(5)(3/15) = 46/15,and E(Y ) = (0.75)E(X) 1.50 = (0.75)(46/15) 1.50 = $0.80.
4.54 X = E(X) =14
0xex/4 dx = 4.
Therefore, Y = E(3X 2) = 3E(X) 2 = (3)(4) 2 = 10.Since E(X2) = 1
4
0x2ex/4 dx = 32, therefore, 2X = E(X
2) 2X = 32 16 = 16.Hence 2Y = 9
2X = (9)(16) = 144.
Solutions for Exercises in Chapter 4 51
4.55 E(X) = (3)(1/6) + (6)(1/2) + (9)(1/3) = 11/2,E(X2) = (3)2(1/6) + (6)2(1/2) + (9)2(1/3) = 93/2. So,E[(2X + 1)2] = 4E(X2) + 4E(X) + 1 = (4)(93/2) + (4)(11/2) + 1 = 209.
4.56 Since E(X) = 10x2 dx+
21x(2 x) dx = 1, and
E(X2) = 10x32 dx+
21x2(2 x) dx = 7/6,then
E(Y ) = 60E(X2) + 39E(X) = (60)(7/6) + (39)(1) = 109 kilowatt hours.
4.57 The equations E[(X 1)2] = 10 and E[(X 2)2] = 6 may be written in the form:
E(X2) 2E(X) = 9, E(X2) 4E(X) = 2.

Solving these two equations simultaneously we obtain

E(X) = 7/2, and E(X2) = 16.
Hence = 7/2 and 2 = 16 (7/2)2 = 15/4.4.58 E(X) = (2)(0.40) + (4)(0.60) = 3.20, and
E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3. So,
(a) E(2X 3Y ) = 2E(X) 3E(Y ) = (2)(3.20) (3)(3.00) = 2.60.(b) E(XY ) = E(X)E(Y ) = (3.20)(3.00) = 9.60.
4.59 E(2XY 2 X2Y ) = 2E(XY 2) E(X2Y ). Now,E(XY 2) =
2x=0
2y=0
xy2f(x, y) = (1)(1)2(3/14) = 3.14, and
E(X2Y ) =2
x=0

2y=0
x2yf(x, y) = (1)2(1)(3/14) = 3.14.
Therefore, E(2XY 2 X2Y ) = (2)(3/14) (3/14) = 3/14.4.60 Using = 60 and = 6 and Chebyshevs theorem
P ( k < X < + k) 1 1k2,
since from + k = 84 we obtain k = 4.So, P (X < 84) P (36 < X < 84) 1 1
42= 0.9375. Therefore,
P (X 84) 1 0.9375 = 0.0625.
Since 1000(0.0625) = 62.5, we claim that at most 63 applicants would have a score as84 or higher. Since there will be 70 positions, the applicant will have the job.
4.61 = 900 hours and = 50 hours. Solving k = 700 we obtain k = 4.So, using Chebyshevs theorem with P ( 4 < X < + 4) 1 1/42 = 0.9375,we obtain P (700 < X < 1100) 0.9375. Therefore, P (X 700) 0.03125.
52 Chapter 4 Mathematical Expectation
4.62 = 52 and = 6.5. Solving + k = 71.5 we obtain k = 3. So,
P ( 3 < X < + 3) 1 132
= 0.8889,
which is
P (32.5 < X < 71.5) 0.8889.
we obtain P (X > 71.5) < 10.88892

= 0.0556 using the symmetry.
4.63 n = 500, = 4.5 and = 2.8733. Solving + k(/500) = 5 we obtain
k =5 4.5
2.87333/500
=0.5
0.1284= 3.8924.
So, P (4 X 5) 1 1k2
= 0.9340.
4.64 2Z = 22X+4Y 3 = 4
2X + 16
2Y = (4)(5) + (16)(3) = 68.
4.65 2Z = 22X+4Y 3 = 4
2X + 16
2Y 16XY = (4)(5) + (16)(3) (16)(1) = 52.
4.66 (a) P (6 < X < 18) = P [12 (2)(3) < X < 12 + (2)(3)] 1 122= 3
4.
(b) P (3 < X < 21) = P [12 (3)(3) < X < 12 + (3)(3)] 1 132= 8
9.
4.67 (a) P (|X 10| 3) = 1 P (|X 10| < 3)= 1 P [10 (3/2)(2) < X < 10 + (3/2)(2)] 1
[1 1
(3/2)2
]= 4
9.
(b) P (|X 10| < 3) = 1 P (|X 10| 3) 1 49= 5
9.
(c) P (5 < X < 15) = P [10 (5/2)(2) < X < 10 + (5/2)(2)] 1 1(5/2)2
= 2125.
(d) P (|X 10| c) 0.04 implies that P (|X 10| < c) 1 0.04 = 0.96.Solving 0.96 = 1 1
k2we obtain k = 5. So, c = k = (5)(2) = 10.
4.68 = E(X) = 6 10x2(1 x) dx = 0.5, E(X2) = 6 1
0x3(1 x) dx = 0.3, which imply
2 = 0.3 (0.5)2 = 0.05 and = 0.2236. Hence,
P ( 2 < X < + 2) = P (0.5 0.4472 < X < 0.5 + 0.4472)
= P (0.0528 < X < 0.9472) = 6
0.94720.0528
x(1 x) dx = 0.9839,
compared to a probability of at least 0.75 given by Chebyshevs theorem.
4.69 It is easy to see that the expectations of X and Y are both 3.5. So,
(a) E(X + Y ) = E(X) + E(Y ) = 3.5 + 3.5 = 7.0.
(b) E(X Y ) = E(X) E(Y ) = 0.
Solutions for Exercises in Chapter 4 53
(c) E(XY ) = E(X)E(Y ) = (3.5)(3.5) = 12.25.
4.70 E(Z) = E(XY ) = E(X)E(Y ) = 10
2
16xy(y/x3) dx dy = 8/3.
4.71 E[g(X, Y )] = E(X/Y 3 +X2Y ) = E(X/Y 3) + E(X2Y ).
E(X/Y 3) = 21
10
2x(x+2y)7y3
dx dy = 27
21
(13y3
+ 1y2
)dy = 15
84;
E(X2Y ) = 21
10
2x2y(x+2y)7
dx dy = 27
21y(14+ 2y
3
)dy = 139
252.
Hence, E[g(X, Y )] = 1584+ 139
252= 46
63.
4.72 X = Y = 3.5. 2X =
2Y = [(1)
2 + (2)2 + + (6)2](1/6) (3.5)2 = 3512.
(a) 2XY = 42X + 2Y =
17512;
(b) X+3Y 5 = 2X + 92Y =
1756.
4.73 (a) = 15
50x dx = 2.5, 2 = E(X2) 2 = 1
5x2 50x2 dx 2.52 = 2.08.
So, =2 = 1.44.
(b) By Chebyshevs theorem,
P [2.5 (2)(1.44) < X < 2.5 + (2)(1.44)] = P (0.38 < X < 5.38) 0.75.

Using integration, P (0.38 < X < 5.38) = 1 0.75;
P [2.5 (3)(1.44) < X < 2.5 + (3)(1.44)] = P (1.82 < X < 6.82) 0.89.
Using integration, P (1.82 < X < 6.82) = 1 0.89.
4.74 P = I2R with R = 50, I = E(I) = 15 and 2I = V ar(I) = 0.03.
E(P ) = E(I2R) = 50E(I2) = 50[V ar(I) + 2I ] = 50(0.03 + 152) = 11251.5. If we use
the approximation formula, with g(I) = I2, g(I) = 2I and g(I) = 2, we obtain,
E(P ) 50[g(I) + 2
2I2
]= 50(152 + 0.03) = 11251.5.
Since V ar[g(I)] [g(i)i
]2i=I
2I , we obtain
V ar(P ) = 502V ar(I2) = 502(2I)22I = 50
2(30)2(0.03) = 67500.
4.75 For 0 < a < 1, since g(a) =x=0
ax = 11a , g
(a) =x=1
xax1 = 1(1a)2 and
g(a) =x=2
x(x 1)ax2 = 2(1a)3 .
54 Chapter 4 Mathematical Expectation
(a) E(X) = (3/4)x=1
x(1/4)x = (3/4)(1/4)x=1
x(1/4)x1 = (3/16)[1/(1 1/4)2]= 1/3, and E(Y ) = E(X) = 1/3.
E(X2)E(X) = E[X(X 1)] = (3/4)x=2
x(x 1)(1/4)x
= (3/4)(1/4)2x=2
x(x 1)(1/4)x2 = (3/43)[2/(1 1/4)3] = 2/9.So, V ar(X) = E(X2) [E(X)]2 = [E(X2) E(X)] + E(X) [E(X)]22/9 + 1/3 (1/3)2 = 4/9, and V ar(Y ) = 4/9.
(b) E(Z) = E(X) + E(Y ) = (1/3) + (1/3) = 2/3, andV ar(Z) = V ar(X + Y ) = V ar(X) + V ar(Y ) = (4/9)+ (4/9) = 8/9, since X andY are independent (from Exercise 3.79).
4.76 (a) g(x) = 32
10(x2 + y2) dy = 1
2(3x2 + 1) for 0 < x < 1 and
h(y) = 12(3y2 + 1) for 0 < y < 1.
Since f(x, y) 6= g(x)h(y), X and Y are not independent.(b) E(X + Y ) = E(X) + E(Y ) = 2E(X) =
10x(3x2 + 1) dx = 3/4 + 1/2 = 5/4.
E(XY ) = 32
10
10xy(x2 + y2) dx dy = 3
2
10y(14+ y
2
2
)dy
= 32
[(14
) (12
)+(12
) (14
)]= 3
8.
(c) V ar(X) = E(X2) [E(X)]2 = 12
10x2(3x2 + 1) dx (5
8
)2= 7
15 25
64= 73
960, and
V ar(Y ) = 73960

. Also, Cov(X, Y ) = E(XY )E(X)E(Y ) = 38 (5
8
)2= 1
64.
(d) V ar(X + Y ) = V ar(X) + V ar(Y ) + 2Cov(X, Y ) = 2 73960
2 164= 29
240.
4.77 (a) E(Y ) =0yey/4 dy = 4.
(b) E(Y 2) =0y2ey/4 dy = 32 and V ar(Y ) = 32 42 = 16.
4.78 (a) The density function is shown next.
x
f(x)
7 80
1
(b) E(Y ) = 87y dy = 1
2[82 72] = 15
2= 7.5,
E(Y 2) = 87y2 dy = 1
3[83 73] = 169
3, and V ar(Y ) = 169

3 (15
2
)2= 1
12.
4.79 Using the exact formula, we have
E(eY ) =
87
ey dy = ey|87 = 1884.32.
Solutions for Exercises in Chapter 4 55
Using the approximation, since g(y) = ey, so g(y) = ey. Hence, using the approxima-tion formula,
E(eY ) eY + eY 2Y
2=
(1 +
1
24
)e7.5 = 1883.38.
The approximation is very close to the true value.
4.80 Using the exact formula, E(Z2) = 87e2y dy = 1
2e2y|87 = 3841753.12. Hence,
V ar(Z) = E(Z2) [E(Z)]2 = 291091.3.Using the approximation formula, we have
V ar(eY ) = (eY )2V ar(Y ) =e(2)(7.5)
12= 272418.11.
The approximation is not so close to each other. One reason is that the first orderapproximation may not always be good enough.
4.81 Define I1 = {xi| |xi | < k} and I2 = {xi| |xi | k}. Then
2 = E[(X )2] =x
(x )2f(x) =xiI1
(xi )2f(xi) +xiI2
(xi )2f(xi)
xiI2
(xi )2f(xi) k22xiI2

f(xi) = k22P (|X | k),
which implies
P (|X | k) 1k2.
Hence, P (|X | < k) 1 1k2.
4.82 E(XY ) = 10
10xy(x+y) dx dy = 1
3, E(X) =
10
10x(x+y) dx dy = 7
12and E(Y ) = 7
12.
Therefore, XY = E(XY ) XY = 13 (712
)2= 1
144.
4.83 E(Y X) = 10
y02(y x) dx dy = 1
0y2 dy = 1
3. Therefore, the average amount of
kerosene left in the tank at the end of each day is (1/3)(1000) = 333 liters.
4.84 (a) E(X) =0
x5ex/5 dx = 5.
(b) E(X2) =0
x2
5ex/5 dx = 50, so V ar(X) = 50 52 = 25, and = 5.
(c) E[(X + 5)2] = E{[(X 5) + 10]2} = E[(X 5)2] + 102 + 20E(X 5)= V ar(X) + 100 = 125.
4.85 E(XY ) = 24 10
1y0
x2y2 dx dy = 8 10y2(1 y)3 dy = 2
15,
X = 24 10
1y0
x2y dx dy = 25and Y = 24
10
1y0
xy2 dx dy = 25. Therefore,
XY = E(XY ) XY = 215 (25
)2= 2
75.
56 Chapter 4 Mathematical Expectation
4.86 E(X + Y ) = 10
1y0
24(x+ y)xy dx dy = 45.
4.87 (a) E(X) =0
x900
ex/900 dx = 900 hours.
(b) E(X2) =0
x2
900ex/900 dx = 1620000 hours2.
(c) V ar(X) = E(X2) [E(X)]2 = 810000 hours2 and = 900 hours.4.88 It is known g(x) = 2
3(x+ 1), for 0 < x < 1, and h(y) = 1
3(1 + 4y), for 0 < y < 1.
(a) X = 10
23x(x+ 1) dx = 5
9and Y =
10
13y(1 + 4y) dy = 11
18.
(b) E[(X + Y )/2] = 12[E(X) + E(Y )] = 7
12.
4.89 Cov(aX, bY ) = E[(aX aX)(bY bY )] = abE[(X X)(Y Y )] = abCov(X, Y ).4.90 It is known = 900 and = 900. For k = 2,
P ( 2 < X < + 2) = P (900 < X < 2700) 0.75
using Chebyshevs theorem. On the other hand,
P ( 2 < X < + 2) = P (900 < X < 2700) = 1 e3 = 0.9502.
For k = 3, Chebyshevs theorem yields
P ( 3 < X < + 3) = P (1800 < X < 3600) 0.8889,
while P (1800 < X < 3600) = 1 e4 = 0.9817.
4.91 g(x) = 10
16yx3
dy = 8x3, for x > 2, with X =
2
8x2
dx = 8x
2= 4,
h(y) =2
16yx3
dx = 8yx2
2= 2y, for 0 < y < 1, with Y =
102y2 = 2
3, and
E(XY ) =2
10
16y2
x2dy dx = 16
3
2
1x2
dx = 83. Hence,
XY = E(XY ) XY = 83 (4)(23
)= 0.
4.92 Since XY = 1, 2X = 5 and
2Y = 3, we have =
XYXY
= 1(5)(3)
= 0.2582.
4.93 (a) From Exercise 4.37, we have 2 = 1/18, so = 0.2357.
(b) Also, X = 1/3 from Exercise 4.12. So,
P ( 2 < X < + 2) = P [1/3 (2)(0.2357) < X < 1/3 + (2)(0.2357)]= P (0 < X < 0.8047) =
0.80470
2(1 x) dx = 0.9619.
Using Chebyshevs theorem, the probability of this event should be larger than0.75, which is true.
(c) P (profit > $500) = P (X > 0.1) = 10.1
2(1 x) = 0.81.
Solutions for Exercises in Chapter 4 57
4.94 Since g(0)h(0) = (0.17)(0.23) 6= 0.10 = f(0, 0), X and Y are not independent.4.95 E(X) = (5000)(0.2) + (10000)(0.5) + (30000)(0.3) = $13, 000.4.96 (a) f(x) =
(3x
)(0.15)x(0.85)3x, for x = 0, 1, 2, 3.
x 0 1 2 3f(x) 0.614125 0.325125 0.057375 0.003375
(b) E(X) = 0.45.
(c) E(X2) = 0.585, so V ar(X) = 0.585 0.452 = 0.3825.(d) P (X 2) = 1 P (X = 3) = 1 0.003375 = 0.996625.(e) 0.003375.
(f) Yes.
4.97 (a) E(X) = ($15k)(0.05)+($15k)(0.15)+($25k)(0.30)+($40k)(0.15)+($50k)(0.10)+($100k)(0.05) + ($150k)(0.03) + ($200k)(0.02) = $33.5k.
(b) E(X2) = 2, 697, 500, 000 dollars2. So, =E(X2) [E(X)]2 = $39.689k.
4.98 (a) E(X) = 34503
5050 x(50
2 x2) dx = 0.(b) E(X2) = 3
4503 5050 x
2(502 x2) dx = 500.(c) =
E(X2) [E(X)]2 = 500 0 = 22.36.
4.99 (a) The marginal density of X is
x1 0 1 2 3 4fX1(x1) 0.13 0.21 0.31 0.23 0.12
(b) The marginal density of Y is
x2 0 1 2 3 4fX2(x2) 0.10 0.30 0.39 0.15 0.06
(c) Given X2 = 3, the conditional density function of X1 is f(x1, 3)/0.15. So
x1 0 1 2 3 4fX2(x2)
715
15
115
15
115
(d) E(X1) = (0)(0.13) + (1)(0.21) + (2)(0.31) + (3)(0.23) + (4)(0.12) = 2.
(e) E(X2) = (0)(0.10) + (1)(0.30) + (2)(0.39) + (3)(0.15) + (4)(0.06) = 1.77.
(f) E(X1|X2 = 3) = (0)(715
)+ (1)
(15
)+ (2)
(115
)+ (3)
(15
)+ (4)
(115
)= 18
15= 6
5= 1.2.
(g) E(X21 ) = (0)2(0.13) + (1)2(0.21) + (2)2(0.31) + (3)2(0.23) + (4)2(0.12) = 5.44.
So, X1 =E(X21 ) [E(X1)]2 =
5.44 22 = 1.44 = 1.2.
4.100 (a) The marginal densities of X and Y are, respectively,
58 Chapter 4 Mathematical Expectation
x 0 1 2g(x) 0.2 0.32 0.48
y 0 1 2h(y) 0.26 0.35 0.39
The conditional density of X given Y = 2 is
x 0 1 2fX|Y=2(x|2) 439 539 3039
(b) E(X) = (0)(0.2) + (1)(0.32) + (2)(0.48) = 1.28,E(X2) = (0)2(0.2) + (1)2(0.32) + (2)2(0.48) = 2.24, andV ar(X) = 2.24 1.282 = 0.6016.
(c) E(X|Y = 2) = (1) 539+ (2)30
39= 65
39and E(X2|Y = 2) = (1)2 5
39+ (2)2 30
39= 125
39. So,
V ar(X) = 12539 (65
39
)2= 650
1521= 50
117.
4.101 The profit is 8X + 3Y 10 for each trip. So, we need to calculate the average of thisquantity. The marginal densities of X and Y are, respectively,
x 0 1 2g(x) 0.34 0.32 0.34

y 0 1 2 3 4 5h(y) 0.05 0.18 0.15 0.27 0.19 0.16
So, E(8X+3Y 10) = (8)[(1)(0.32)+(2)(0.34)]+(3)[(1)(0.18)+(2)(0.15)+(3)(0.27)+(4)(0.19) + (5)(0.16)] 10 = $6.55.
4.102 Using the approximation formula, V ar(Y ) ki=1 [h(x1,x2,...,xk)xi ]2xi=i, 1ik
2i , we
have
V ar(Y ) 2i=0
(eb0+b1k1+b2k2
bi
)2bi=i, 0i2
2bi = e2(0+k11+k22)(20 + k
21
21 + k
22
22).
4.103 (a) E(Y ) = 10 10y(1 y)9 dy = y(1 y)10|10 +
10(1 y)10 dy = 1
11.
(b) E(1 Y ) = 1 E(Y ) = 1011.
(c) V ar(Z) = V ar(1 Y ) = V ar(Y ) = E(Y 2) [E(Y )]2 = 1011212 = 0.006887.
Chapter 5
Some Discrete Probability
Distributions
5.1 This is a uniform distribution: f(x) = 110, for x = 1, 2, . . . , 10.
Therefore P (X < 4) =3
x=1
f(x) = 310.
5.2 Binomial distribution with n = 12 and p = 0.5. HenceP (X = 3) = P (X 3) P (X 2) = 0.0730 0.0193 = 0.0537.
5.3 =10x=1
x10= 5.5, and 2 =
10x=1
(x5.5)210
= 8.25.
5.4 For n = 5 and p = 3/4, we have
(a) P (X = 2) =(52
)(3/4)2(1/4)3 = 0.0879,
(b) P (X 3) =3
x=0
b(x; 5, 3/4) = 1 P (X = 4) P (X = 5)= 1 (5
4
)(3/4)4(1/4)1 (5
5
)(3/4)5(1/4)0 = 0.3672.
5.5 We are considering a b(x; 20, 0.3).
(a) P (X 10) = 1 P (X 9) = 1 0.9520 = 0.0480.(b) P (X 4) = 0.2375.(c) P (X = 5) = 0.1789. This probability is not very small so this is not a rare event.
Therefore, P = 0.30 is reasonable.
5.6 For n = 6 and p = 1/2.
(a) P (2 X 5) = P (X 5) P (X 1) = 0.9844

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