Probability and statistics for engineers and scientists answers

Solutions manual for probability and statistics for engineers and scientists 9th edition by walpole

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Class Notes in Discrete Mathematics, Operations Research, Statistics and Probability (Fourth Edition, v2)

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Contents 1 Introduction to Statistics and Data Analysis

2 Probability

3 Random Variables and Probability Distributions

4 Mathematical Expectation

5 Some Discrete Probability Distributions

6 Some Continuous Probability Distributions

7 Functions of Random Variables

8 Fundamental Sampling Distributions and Data Descriptions

9 One- and Two-Sample Estimation Problems

10 One- and Two-Sample Tests of Hypotheses

113

11 Simple Linear Regression and Correlation

139

12 Multiple Linear Regression and Certain Nonlinear Regression Models

161

13 One-Factor Experiments: General

175

14 Factorial Experiments (Two or More Factors)

197

15 2k Factorial Experiments and Fractions

219

16 Nonparametric Statistics

233

17 Statistical Quality Control

247

18 Bayesian Statistics

251

iii

Chapter 1

Introduction to Statistics and Data Analysis 1.1 (a) 15. (b) x ¯=

1 15 (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 data becomes: 2.9 3.7

3.0 4.0

3.3 4.4

3.4 4.8

3.6

So. the trimmed mean is 1 x ¯tr20 = (2.9 + 3.0 + · · · + 4.8) = 3.678. 9 (f) They are about the same. 1.2 (a) Mean=20.7675 and Median=20.610. (b) x ¯tr10 = 20.743. (c) A dot plot is shown below.

Chapter 1 Introduction to Statistics and Data Analysis

2 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 “Aging” group. (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 for each group are similar to each other. ˜ A = 8.250; ¯ A = 7.950 and X 1.4 (a) X ˜ B = 10.150. ¯ B = 10.260 and X X (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. The steel rods made by company B show more ﬂexibility. 1.5 (a) A dot plot is shown below.

−10

In the ﬁgure, “×” represents the control group and “◦” represents the treatment group. ¯ Control = 5.60, X ˜ Control = 5.00, and X ¯ tr(10);Control = 5.13; (b) X ¯ Treatment = 7.60, X ˜ Treatment = 4.50, and X ¯ tr(10);Treatment = 5.63. X (c) The diﬀerence of the means is 2.0 and the diﬀerences of the medians and the trimmed means are 0.5, which are much smaller. The possible cause of this might be 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 20◦ C group and “◦” represents the 45◦ C group. ¯ 45◦ C = 2.2350. ¯ 20◦ C = 2.1075, and X (b) X (c) Based on the plot, it seems that high temperature yields more high values of tensile strength, along with a few low values of tensile strength. Overall, the temperature does have an inﬂuence on the tensile strength. (d) It also seems that the variation of the tensile strength gets larger when the cure temperature is increased. 1 [(3.4 − 3.787)2 + (2.5 − 3.787)2 + (4.8 − 3.787)2 + · · · + (4.8 − 3.787)2 ] = 0.94284; 1.7 s2 = 15−1 √ √ s = s2 = 0.9428 = 0.971.

Solutions for Exercises in Chapter 1 1 [(18.71 − 20.7675)2 + (21.41 − 20.7675)2 + · · · + (21.12 − 20.7675)2 ] = 2.5329; 1.8 s2 = 20−1 √ s = 2.5345 = 1.5915. 1 [(227 − 222.10)2 + (222 − 222.10)2 + · · · + (221 − 222.10)2 ] = 23.66; 1.9 (a) s2No Aging = 10−1 √ sNo Aging = 23.62 = 4.86. 1 [(219 − 209.90)2 + (214 − 209.90)2 + · · · + (205 − 209.90)2 ] = 42.10; s2Aging = 10−1 √ sAging = 42.12 = 6.49.

(b) Based on the numbers in (a), the variation in “Aging” is smaller that the variation in “No Aging” although the diﬀerence is not so apparent in the plot. √ 1.10 For company A: s2A = 1.2078 and sA = √1.2072 = 1.099. For company B: s2B = 0.3249 and sB = 0.3249 = 0.570. 1.11 For the control group: s2Control = 69.38 and sControl = 8.33. For the treatment group: s2Treatment = 128.04 and sTreatment = 11.32. 1.12 For the cure temperature at 20◦ C: s220◦ C = 0.005 and s20◦ C = 0.071. For the cure temperature at 45◦ C: s245◦ C = 0.0413 and s45◦ C = 0.2032. The variation of the tensile strength is inﬂuenced by the increase of cure temperature. ¯ = 124.3 and median = X ˜ = 120; 1.13 (a) Mean = X (b) 175 is an extreme observation. ¯ = 570.5 and median = X ˜ = 571; 1.14 (a) Mean = X (b) Variance = s2 = 10; standard deviation= s = 3.162; range=10; (c) Variation of the diameters seems too big so the quality is questionable. 1.15 Yes. The value 0.03125 is actually a P -value and a small value of this quantity means that 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 n

xi − n¯ x=

i=1

xi −

i=1

xi = 0,

i=1

which is the term on the right side. ¯ nonsmokers = 30.32; ¯ smokers = 43.70 and X 1.17 (a) X (b) ssmokers = 16.93 and snonsmokers = 7.13; (c) A dot plot is shown below. 10

In the ﬁgure, “×” represents the nonsmoker group and “◦” represents the smoker group. (d) Smokers appear to take longer time to fall asleep and the time to fall asleep for smoker group is more variable. 1.18 (a) A stem-and-leaf plot is shown below. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 1 Introduction to Statistics and Data Analysis

4 Stem 1 2 3 4 5 6 7 8 9

Leaf 057 35 246 1138 22457 00123445779 01244456678899 00011223445589 0258

Frequency 3 2 3 4 5 11 14 14 4

(b) The following is the relative frequency distribution table. Class Interval 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 90 − 99

Relative Frequency Distribution of Grades Class Midpoint Frequency, f Relative Frequency 14.5 3 0.05 24.5 2 0.03 34.5 3 0.05 44.5 4 0.07 54.5 5 0.08 64.5 11 0.18 74.5 14 0.23 84.5 14 0.23 94.5 4 0.07

Relative Frequency

14.5

24.5

34.5

44.5 54.5 64.5 Final Exam Grades

74.5

84.5

94.5

The distribution skews to the left. ¯ = 65.48, X ˜ = 71.50 and s = 21.13. (d) X 1.19 (a) A stem-and-leaf plot is shown below. Stem 0 1 2 3 4 5 6

Leaf 22233457 023558 035 03 057 0569 0005

Frequency 8 6 3 2 3 4 4

Solutions for Exercises in Chapter 1

(b) The following is the relative frequency distribution table. Class Interval 0.0 − 0.9 1.0 − 1.9 2.0 − 2.9 3.0 − 3.9 4.0 − 4.9 5.0 − 5.9 6.0 − 6.9

Relative Frequency Distribution of Years Class Midpoint Frequency, f Relative Frequency 0.45 8 0.267 1.45 6 0.200 2.45 3 0.100 3.45 2 0.067 4.45 3 0.100 5.45 4 0.133 6.45 4 0.133

¯ = 2.797, s = 2.227 and Sample range is 6.5 − 0.2 = 6.3. (c) X 1.20 (a) A stem-and-leaf plot is shown next. Stem 0* 0 1* 1 2* 2 3*

Leaf 34 56667777777889999 0000001223333344 5566788899 034 7 2

Frequency 2 17 16 10 3 1 1

(b) The relative frequency distribution table is shown next. Relative Frequency Distribution of Fruit Fly Lives Class Interval Class Midpoint Frequency, f Relative Frequency 0−4 2 2 0.04 5−9 7 17 0.34 10 − 14 12 16 0.32 15 − 19 17 10 0.20 20 − 24 22 3 0.06 25 − 29 27 1 0.02 30 − 34 32 1 0.02

Relative Frequency

12 17 22 Fruit fly lives (seconds)

Chapter 1 Introduction to Statistics and Data Analysis

6 ¯ = 74.02 and X ˜ = 78; 1.21 (a) X (b) s = 39.26. ¯ = 6.7261 and X ˜ = 0.0536. 1.22 (a) X

(b) A histogram plot is shown next.

6.62

6.66 6.7 6.74 6.78 Relative Frequency Histogram for Diameter

6.82

100

395.10

200

300

400

500

600

700

800

900

1000

¯ 1980 = 395.1 and X ¯ 1990 = 160.2. (b) X (c) The sample mean for 1980 is over twice as large as that of 1990. The variability for 1990 decreased also as seen by looking at the picture in (a). The gap represents an increase of over 400 ppm. It appears from the data that hydrocarbon emissions decreased considerably between 1980 and 1990 and that the extreme large emission (over 500 ppm) were no longer in evidence. ¯ = 2.8973 and s = 0.5415. 1.24 (a) X

Relative Frequency

(b) A histogram plot is shown next.

1.8

2.1

2.4

2.7

3 Salaries

3.3

3.6

3.9

Solutions for Exercises in Chapter 1

Leaf (84) (05)(10)(14)(37)(44)(45) (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) (10)(13)(14)(22)(36)(37) (51)(54)(57)(71)(79)(85)

Frequency 1 6 11 6 6

¯ = 33.31; 1.25 (a) X ˜ = 26.35; (b) X

Relative Frequency

40 50 60 70 Percentage of the families

¯ tr(10) = 30.97. This trimmed mean is in the middle of the mean and median using the (d) X full amount of data. Due to the skewness of the data to the right (see plot 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 likely that the model would be wrong due to several extreme values of the data. Actually if a scatter plot of these two data sets is made, it is easy to see that some outlier would inﬂuence the trend.

300 250

wear

350

1.27 (a) The averages of the wear are plotted here.

700

800

900

1000

1100

1200

1300

load

Chapter 1 Introduction to Statistics and Data Analysis

500 100

300

wear

700

800

900

1000

1100

1200

1300

load

(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. High 71.45

71.65

Low 71.85

72.05

72.25

72.45

72.65

72.85

73.05

In the ﬁgure, “×” represents the low-injection-velocity group and “◦” represents the high-injection-velocity group. (b) It appears that shrinkage values for the low-injection-velocity group is higher than those for the high-injection-velocity group. Also, the variation of the shrinkage is a little larger for the low injection velocity than that for the high injection velocity.

2.0

2.5

3.0

3.5

1.29 A box plot is shown next.

Solutions for Exercises in Chapter 1

700

800

900

1000

1100

1200

1300

1.30 A box plot plot is shown next.

1.31 (a) A dot plot is shown next. High

Low 76

In the ﬁgure, “×” represents the low-injection-velocity group and “◦” represents the high-injection-velocity group. (b) In this time, the shrinkage values are much higher for the high-injection-velocity group than those for the low-injection-velocity group. Also, the variation for the former group is much higher as well. (c) Since the shrinkage eﬀects change in diﬀerent direction between low mode temperature and high mold temperature, the apparent interactions between the mold temperature and injection velocity are signiﬁcant. 1.32 An interaction plot is shown next. mean shrinkage value high mold temp

Low

low mold temp injection velocity

high

It is quite obvious to ﬁnd the interaction between the two variables. Since in this experimental data, those two variables can be controlled each at two levels, the interaction can be invesc Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 1 Introduction to Statistics and Data Analysis

tigated. However, if the data are from an observational studies, in which the variable values cannot be controlled, it would be diﬃcult to study the interactions among 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 = φ. 2.2 S = {(x, y) | x2 + y 2 < 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, 1T H, 1T T, 2H, 2T, 3HH, 3HT, 3T H, 3T T, 4H, 4T, 5HH, 5HT, 5T H, 5T T, 6H, 6T }. 2.6 S = {A1 A2 , A1 A3 , A1 A4 , A2 A3 , A2 A4 , A3 A4 }. 2.7 S1 = {M M M M, M M M F, M M F M, M F M M, F M M M, M M F F, M F M F, M F F M, F M F M, F F M M, F M M F, M F F F, F M F F, F F M F, F F F M, F F F F }. 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)}. (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 Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 11

Chapter 2 Probability

12 (c) (d) (e) (f) (g)

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)}. A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}. A ∩ B = φ. B ∩ C = {(5, 2), (6, 2)}. A Venn diagram is shown next. S B

A B ∩C

A ∩C

2.9 (a) (b) (c) (d) (e)

A = {1HH, 1HT, 1T H, 1T T, 2H, 2T }. B = {1T T, 3T T, 5T T }. A = {3HH, 3HT, 3T H, 3T T, 4H, 4T, 5HH, 5HT, 5T H, 5T T, 6H, 6T }. A ∩ B = {3T T, 5T T }. A ∪ B = {1HH, 1HT, 1T H, 1T T, 2H, 2T, 3T T, 5T T }.

2.10 (a) S = {F F F, F F N, F N F, N F F, F N N, N F N, N N F, N N N }. (b) E = {F F F, F F N, F N F, N F F }. (c) The second river was safe for ﬁshing. 2.11 (a) S = {M1 M2 , M1 F1 , M1 F2 , M2 M1 , M2 F1 , M2 F2 , F1 M1 , F1 M2 , F1 F2 , F2 M1 , F2 M2 , F2 F1 }. (b) A = {M1 M2 , M1 F1 , M1 F2 , M2 M1 , M2 F1 , M2 F2 }. (c) B = {M1 F1 , M1 F2 , M2 F1 , M2 F2 , F1 M1 , F1 M2 , F2 M1 , F2 M2 }. (d) C = {F1 F2 , F2 F1 }. (e) A ∩ B = {M1 F1 , M1 F2 , M2 F1 , M2 F2 }. (f) A ∪ C = {M1 M2 , M1 F1 , M1 F2 , M2 M1 , M2 F1 , M2 F2 , F1 F2 , F2 F1 }. S A A ∩B

Solutions for Exercises in Chapter 2

2.12 (a) S = {ZY F, ZN F, W Y F, W N F, SY F, SN F, ZY M }. (b) A ∪ B = {ZY F, ZN F, W Y F, W N F, SY F, SN F } = A. (c) A ∩ B = {W Y F, SY F }. 2.13 A Venn diagram is shown next. Sample Space

P S

2.14 (a) A ∪ C = {0, 2, 3, 4, 5, 6, 8}. (b) A ∩ B = φ. (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}. (d) B ∩ C = {copper, uranium, zinc}. (e) A ∩ B ∩ C = φ. (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. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 2 Probability

14 S

A 1

B 3

(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. S

A 5 2

7 3

(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 for traﬃc 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 vacancies but will not experience mechanical problems. (c) The family will experience mechanical problems and will arrive at a campsite that has no vacancies. (d) The family will receive a traﬃc ticket but will not arrive at a campsite that has no vacancies. (e) The family will not experience mechanical problems. 2.20 (a) 6; (b) 2; (c) 2, 5, 6; (d) 4, 5, 7, 8. 2.21 With n1 = 6 sightseeing tours each available on n2 = 3 diﬀerent days, the multiplication rule gives n1 n2 = (6)(3) = 18 ways for a person to arrange a tour. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 2

2.22 With n1 = 8 blood types and n2 = 3 classiﬁcations of blood pressure, the multiplication rule gives n1 n2 = (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, the multiplication rule gives n1 n2 = (6)(26) = 156 points in S. 2.24 Since a student may be classiﬁed according to n1 = 4 class standing and n2 = 2 gender classiﬁcations, the multiplication rule gives n1 n2 = (4)(2) = 8 possible classiﬁcations for the students. 2.25 With n1 = 5 diﬀerent shoe styles in n2 = 4 diﬀerent colors, the multiplication rule gives n1 n2 = (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 n1 × n2 × n3 × n4 = (4)(3)(2)(2) = 48 diﬀerent house plans available. 2.28 With n1 = 5 diﬀerent manufacturers, n2 = 3 diﬀerent preparations, and n3 = 2 diﬀerent strengths, the generalized multiplication rule yields n1 n2 n3 = (5)(3)(2) = 30 diﬀerent 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 n1 n2 · · · n9 = 29 = 512 ways to answer the test. 2.31 Since the ﬁrst digit is a 5, there are n1 = 9 possibilities for the second digit and then n2 = 8 possibilities for the third digit. Therefore, by the multiplication rule there are n1 n2 = (9)(8) = 72 registrations to be checked. 2.32 (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 is 4 ways or in (4)(3!) = 24 ways with regard to order. The other 3 persons can then be placed in line in 3! = 6 ways. By Theorem 2.1, there are total (24)(6) = 144 ways 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 each other 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. 2.33 (a) With n1 = 4 possible answers for the ﬁrst question, n2 = 4 possible answers for the second question, and so forth, the generalized multiplication rule yields 45 = 1024 ways to answer the test. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 2 Probability

(b) With n1 = 3 wrong answers for the ﬁrst question, n2 = 3 wrong answers for the second question, and so forth, the generalized multiplication rule yields n1 n2 n3 n4 n5 = (3)(3)(3)(3)(3) = 35 = 243 ways to answer the test and get all questions wrong. 2.34 (a) By Theorem 2.3, 7! = 5040. (b) Since the ﬁrst letter must be m, the remaining 6 letters can be arranged in 6! = 720 ways. 2.35 The ﬁrst house can be placed on any of the n1 = 9 lots, the second house on any of the remaining n2 = 8 lots, and so forth. Therefore, there are 9! = 362, 880 ways to place the 9 homes on the 9 lots. 2.36 (a) Any of the 6 nonzero digits can be chosen for the hundreds position, and of the remaining 6 digits for the tens position, leaving 5 digits for the units position. So, there are (6)(6)(5) = 180 three digit numbers. (b) The units position can be ﬁlled using any of the 3 odd digits. Any of the remaining 5 nonzero digits can be chosen for the hundreds position, leaving a choice of 5 digits for the tens position. By Theorem 2.2, there are (3)(5)(5) = 75 three digit odd 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 three digit 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 units position. In this case, there are (1)(3)(5) = 15 three digit number begin with a 3. So, the total number of three digit numbers that are greater than 330 is 90 + 15 = 105. 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 be interchanged resulting in 24 (4!) = 384 ways. (c) By Theorem 2.3, the members of each gender can be seated in 4! ways. Then using 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 ﬁnalists can then come in second, and so forth. By Theorem 2.3, there 8! = 40320 possible orders in which 8 ﬁnalists may ﬁnish the spelling bee. (b) The possible orders for the ﬁrst three positions are 8 P3 = 2.40 By Theorem 2.4, 8 P5 =

8! 3!

= 6720.

2.41 By Theorem 2.4, 6 P4 =

6! 2!

= 360.

2.42 By Theorem 2.4,

40 P3

40! 37!

8! 5!

= 336.

= 59, 280.

Solutions for Exercises in Chapter 2

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!

2.46 By Theorem 2.6, there are

9! 3!4!2!

2.47 By Theorem 2.8, there are

8 3

= 3360. = 1260 ways.

= 56 ways.

2.48 Assume February 29th as March 1st for the leap year. There are total 365 days in a year. The number of ways that all these 60 students will have diﬀerent birth dates (i.e, arranging 60 from 365) is 365 P60 . This is a very large number. 2.49 (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. 2.50 Assuming equal weights (a) P (A) = (b) P (C) =

5 18 ; 1 3;

7 36 .

2.51 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. 2.52 (a) P (S ∩ D ) = 88/500 = 22/125. (b) P (E ∩ D ∩ S ) = 31/500. (c) P (S ∩ E ) = 171/500. 2.53 Consider the events S: 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. 2.54 Consider the events B: 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. 2.55 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 vowel and end with n 10 = 117 . an even digit. Therefore, N c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 2 Probability

2.56 (a) Let A = Defect in brake system; B = Defect in fuel system; P (A ∪ B) = 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. 2.57 (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) 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 is then 5/36. (b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a total of at most is 10/36=5/18. (43)(48 94 2) = 54145 . (52 5) (13)(13) 143 (b) 4 52 1 = 39984 . (5)

2.59 (a)

(11)(82) = 13 . (93) (5)(3) 5 (b) 2 9 1 = 14 . (3)

2.60 (a)

2.61 (a) P (M ∪ H) = 88/100 = 22/25; (b) P (M ∩ H ) = 12/100 = 3/25; (c) P (H ∩ M ) = 34/100 = 17/50. 2.62 (a) 9; (b) 1/9. 2.63 (a) 0.32; (b) 0.68; (c) oﬃce or den. 2.64 (a) 1 − 0.42 = 0.58; (b) 1 − 0.04 = 0.96. 2.65 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. 2.66 (a) 0.02 + 0.30 = 0.32 = 32%; (b) 0.32 + 0.25 + 0.30 = 0.87 = 87%; (c) 0.05 + 0.06 + 0.02 = 0.13 = 13%; (d) 1 − 0.05 − 0.32 = 0.63 = 63%. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 2

2.67 (a) 0.12 + 0.19 = 0.31; (b) 1 − 0.07 = 0.93; (c) 0.12 + 0.19 = 0.31. 2.68 (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 each type is purchased is 1 − 0.111 = 0.889. 2.69 (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. 2.70 (a) ($4.50 − $4.00) × 50, 000 = $25, 000; (b) Since the probability of underﬁlling is 0.001, we would expect 50, 000 × 0.001 = 50 boxes to be underﬁlled. So, instead of having ($4.50 − $4.00) × 50 = $25 proﬁt for those 50 boxes, there are a loss of $4.00×50 = $200 due to the cost. So, the loss in proﬁt expected due to underﬁlling is $25 + $200 = $250. 2.71 (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. 2.72 P (A ∩ B ) = 1 − P (A ∪ B) = 1 − (P (A) + P (B) − P (A ∩ B) = 1 + P (A ∩ B) − P (A) − P (B). 2.73 (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 armed robbery. 2.74 P (S | A) = 10/18 = 5/9. 2.75 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. 2.76 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. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 2 Probability

20 2.77 (a) P (M ∩ P ∩ H) = (b) P (H ∩ M | P ) =

10 68

5 34 ;

P (H∩M ∩P ) P (P )

22−10 100−68

12 32

= 38 .

2.78 (a) (0.90)(0.08) = 0.072; (b) (0.90)(0.92)(0.12) = 0.099. 2.79 (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+0.046 0.175+0.134

= 0.479.

2.80 Consider the events: C: an oil change is needed, F : an oil ﬁlter is needed. (a) P (F | C) =

P (F ∩C) P (C)

0.14 0.25

= 0.56.

(b) P (C | F ) =

P (C∩F ) P (F )

0.14 0.40

= 0.35.

2.81 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 (W ∩H) 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. 2.82 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) = (c) P (H | W ) =

P (H∩W ) P (H)

P (H∩W ) P (W )

0.15 0.21

= 57 .

0.06 0.72

1 12 .

2.83 Consider the events: A: the vehicle is a camper, B: the vehicle has Canadian license plates. (a) P (B | A) =

P (A∩B) P (A)

0.09 0.28

(b) P (A | B) =

P (A∩B) P (B)

0.09 0.12

= 34 .

9 28 .

Solutions for Exercises in Chapter 2

2.84 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. 2.85 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. 2.86 (a) 0.43; (b) (0.53)(0.22) = 0.12; (c) 1 − (0.47)(0.22) = 0.90. 2.87 Consider the events: A: the house is open, B: the correct key is selected.

(11)(72) = 38 = 0.375. (83) So, P [A ∪ (A ∩ B)] = P (A) + P (A )P (B) = 0.4 + (0.6)(0.375) = 0.625. P (A) = 0.4, P (A ) = 0.6, and P (B) =

2.88 Consider the events: F : failed the test, P : passed the test. (a) P (failed at least one tests) = 1 − P (P1 P2 P3 P4 ) = 1 − (0.99)(0.97)(0.98)(0.99) = 1 − 0.93 = 0.07, (b) P (failed 2 or 3) = 1 − P (P2 P3 ) = 1 − (0.97)(0.98) = 0.0494. (c) 100 × 0.07 = 7. (d) 0.25. 2.89 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. 2.90 (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 | A ∩ B )P (B | A )P (A ) = (0.80)(1 − 0.75)(0.3) + (0.90)(1 − 0.20)(1 − 0.3) = 0.564. (c) Use similar argument as in (a) and (b), P (C) = P (A ∩ B ∩ C) + P (A ∩ B ∩ 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. 2.91 (a) P (Q1 ∩ Q2 ∩ Q3 ∩ Q4 ) = P (Q1 )P (Q2 | Q1 )P (Q3 | Q1 ∩ Q2 )P (Q4 | Q1 ∩ Q2 ∩ Q3 ) = (15/20)(14/19)(13/18)(12/17) = 91/323. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 2 Probability

(b) Let A be the event that 4 good quarts of milk are selected. Then 15 91 4 = . P (A) = 20 323 4 2.92 P = (0.95)[1 − (1 − 0.7)(1 − 0.8)](0.9) = 0.8037. 2.93 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 ∩C∩D∩E) P system works

(0.3)(0.8)(0.8)(0.8) 0.75112

= 0.2045.

2.94 Deﬁne S: the system works. ∩S ) P (A )(1−P (C∩D∩E)) = P (A | S ) = P (A P (S ) = 1−P (S)

(0.3)[1−(0.8)(0.8)(0.8)] 1−0.75112

= 0.588.

2.95 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. 2.96 Let S1 , S2 , S3 , and S4 represent the events that a person is speeding as he passes through the respective locations and let R represent the event that the radar traps is operating resulting in a speeding ticket. Then the probability that he receives a speeding ticket: 4 P (R | Si )P (Si ) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27. P (R) = i=1

2.97 P (C | D) =

P (C∩D) P (D)

2.98 P (S2 | R) =

P (R∩ S2 ) P (R)

0.039 0.096

0.03 0.27

= 0.40625. = 1/9.

2.99 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, (0.005)(0.20) P (B1 | A) = (0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05) = 0.1124. 2.100 Consider the events E: a malfunction by other human errors, A: station A, B: station B, and C: station C. (E | C)P (C) P (C | E) = P (E | A)P (A)+PP (E | B)P (B)+P (E | C)P (C) = 0.1163 0.4419 = 0.2632.

(5/10)(10/43) (7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43)

2.101 Consider the events: A: a customer purchases latex paint, A : a customer purchases semigloss paint, B: a customer purchases rollers. P (B | A)P (A) (0.60)(0.75) P (A | B) = P (B | A)P (A)+P (B | A )P (A ) = (0.60)(0.75)+(0.25)(0.30) = 0.857. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 2

2.102 If we use the assumptions that the host would not open the door you picked nor the door with the prize behind it, we can use Bayes rule to solve the problem. Denote by events A, B, and C, that the prize is behind doors A, B, and C, respectively. Of course P (A) = P (B) = P (C) = 1/3. Denote by H the event that you picked door A and the host opened door B, while there is no prize behind the door B. Then P (H|B)P (B) P (H|A)P (A) + P (H|B)P (B) + P (H|C)P (C) 1/2 1 P (H|B) = = . = P (H|A) + P (H|B) + P (H|C) 0 + 1/2 + 1 3

P (A|H) =

Hence you should switch door. 2.103 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 (g | I)P (I) P (I | g) = P (g | G)P (G)+P (g | I)P (I) =

(0.01)(0.95) (0.05)(0.90)+(0.01)(0.95)

= 0.1743.

2.104 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 (A1 ∩ A2 ∩ A3 ∩ A4 ) = P (A1 )P (A2 )P (A3 )P (A4 )+P (A1 )P (A2 )P (A3 )P (A4 )+P (A1 )P (A2 )P (A3 )P (A4 )+ P (A1 )P (A2 )P (A3 )P (A4 ) = (4)(1/2)4 = 1/4.

(b) P (A1 ∩ A2 ∩ A3 ∩ A4 ) = P (A1 )P (A2 )P (A3 )P (A4 ) = (1/2)4 = 1/16. 2.105 No solution necessary. 2.106 (a) 0.28 + 0.10 + 0.17 = 0.55. (b) 1 − 0.17 = 0.83. (c) 0.10 + 0.17 = 0.27. 2.107 The number of hands =

13131313 4

2.108 (a) P (M1 ∩ M2 ∩ M3 ∩ M4 ) = (0.1)4 = 0.0001, where Mi represents that ith person make a mistake. (b) P (J ∩ C ∩ R ∩ W ) = (0.1)(0.1)(0.9)(0.9) = 0.0081. 2.109 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 event that 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 .

Chapter 2 Probability

(a) P (R1 ∩R2 ∩R3 )+P (R1 ∩R2 ∩R3 )P (R1 ∩R2 ∩R3 ) = P (R1 )P (R2 )P (R3 )+P (R1 )P (R2 )P (R3 )+ P (R1 )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. 2.111 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. 2.112 There are 43 53 63 = 800 possible selections. 2.113 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 (B1 ∩ B2 ∩ B3 ) + P (G1 ∩ G2 ∩ G3 ) = (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. 2.114 The number of ways to receive 2 or 3 defective sets among 5 that are purchased is total3 3 9 9 2 3 + 3 2 = 288. 2.115 Consider the events: O: overrun, A: consulting ﬁrm A, B: consulting ﬁrm B, C: consulting ﬁrm C. (a) P (C | O) = P (O 0.0375 0.0680 = 0.5515. (b) P (A | O) =

P (O | C)P (C) | A)P (A)+P (O | B)P (B)+P (O | C)P (C)

(0.05)(0.40) 0.0680

(0.15)(0.25) (0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25)

= 0.2941.

2.116 (a) 36; (b) 12; (c) order is not important. 1 = 0.0016; (36 2) (12)(24) (b) 1 36 1 = 288 630 = 0.4571. (2)

2.117 (a)

2.118 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 (P | C)P (C) (0.1)(0.07) P (C | P ) = P (P | C)P (C)+P (P | C )P (C ) = (0.1)(0.07)+(1−0.05)(1−0.07) =

0.007 0.8905

= 0.00786.

2.119 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, c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 2

(19 2) 20 = (2) (18 2) = T : a lot contains two defective components,P (T ) = 0.1, P (A | T ) = 20 (2)

O: a lot contains one defective component, P (O) = 0.3, P (A | O) =

P (A | N )P (N ) (a) P (N | A) = P (A | N )P (N )+P (A | O)P (O)+P (A 0.6 = 0.9505 = 0.6312;

| T )P (T )

9 10 , 153 190 .

(1)(0.6) (1)(0.6)+(9/10)(0.3)+(153/190)(0.1)

= 0.2841; (b) P (O | A) = (9/10)(0.3) 0.9505 (c) P (T | A) = 1 − 0.6312 − 0.2841 = 0.0847. 2.120 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 (P | D)P (D) (0.95)(1/500) P (D | P ) = P (P | D)P (D)+P (P | D )P (D ) = (0.95)(1/500)+(0.01)(1−1/500) = 0.1599. 2.121 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 (E | 1)P (1) (0.02)(0.7) P (1 | E) = P (E | 1)P (1)+P (E | 2)P (2) = (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. 2.122 Consider the events: D: an item is defective (a) P (D1 D2 D3 ) = 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. 2.123 Let A be the event that an injured worker is admitted to the hospital and N be the event that an injured worker is back to work the next day. P (A) = 0.10, P (N ) = 0.15 and P (A ∩ N ) = 0.02. So, P (A ∪ N ) = P (A) + P (N ) − P (A ∩ N ) = 0.1 + 0.15 − 0.02 = 0.23. 2.124 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. | T )P (T ) (0.9)(0.5) P (T | M ) = P (M | T P)P(M (T )+P (M | T )P (T ) = (0.9)(0.5)+(0.65)(0.5) = 0.581. 2.125 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. D)P (D) So, P (D | A) = P (A P| (A) = (0.5)(0.1) = 0.25. 0.2 2.126 (a) P (Union member | New company (same ﬁeld)) = (b) P (Unemployed | Union member) =

2 40+13+4+2

13 13 13+10 = 23 2 59 = 0.034.

= 0.5652.

2.127 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 | D1 D2 D3 ) =

P (D1 D2 D3

P (D1 D2 D3 | C)P (C) | C)P (C)+P (D1 D2 D3 | C )P (C )

(0.5)3 (0.5) (0.5)3 (0.5)+1(0.5)

= 19 .

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 NNN NNB N BN BN N N BB BN B BBN BBB

x 0 1 1 1 2 2 2 3

3.3 A table of sample space and assigned values of the random variable is shown next. Sample Space HHH HHT HT H T HH HT T T HT TTH TTT

w 3 1 1 1 −1 −1 −1 −3

3.4 S = {HHH, T HHH, HT HHH, T T HHH, T T T HHH, HT T HHH, T HT HHH, HHT HHH, . . . }; The sample space is discrete containing as many elements as there are positive integers. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 27

Chapter 3 Random Variables and Probability Distributions

28 3

3.5 (a) c = 1/30 since 1 =

c(x2 + 4) = 30c.

x=0

(b) c = 1/10 since 2 2 3 2 3 2 3 2 3 c =c + + = 10c. 1= x 3−x 0 3 1 2 2 1 x=0

3.6 (a) P (X > 200) =

∞

20000 200 (x+100)3

(b) P (80 < X < 200) =

120 80

dx = −

∞ 10000 (x+100)2 200

= 19 . 120 10000 dx = − (x+100) = 2

20000 (x+100)3

1 x2 (2 − x) dx = 2 0 + 2x − 0 1

1 2 1 (b) P (0.5 < X < 1) = 0.5 x dx = x2 = 0.375.

3.7 (a) P (X < 1.2) =

x dx +

1.2

1000 9801 x2 2

= 0.1020.

1.2 = 0.68. 1

0.5

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 have P (W = −3) = P (T T T ) = (1/3)3 = 1/27; P (W = −1) = P (HT T ) + P (T HT ) + P (T T H) = 3(2/3)(1/3)2 = 2/9; P (W = 1) = P (HHT ) + P (HT H) + P (T HH) = 3(2/3)2 (1/3) = 4/9; P (W = 3) = P (HHH) = (2/3)3 = 8/27; The probability distribution for W is then w P (W = w)

3.9 (a) P (0 < X < 1) =

2(x+2) 5 0

(b) P (1/4 < X < 1/2) =

dx =

1/2

2(x+2) 5 1/4

−3 1/27

1 (x+2)2 5 0 dx =

−1 2/9

1 4/9

3 8/27

= 1.

1/2 (x+2)2 5 1/4

= 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. 5 ways. A random 3.11 We can select x defective sets from 2, and 3 − x good sets from 5 in x2 3−x 7 selection of 3 from 7 sets can be made in 3 ways. Therefore, 2 f (x) =

5 3−x 7 3

x = 0, 1, 2.

In tabular form x f (x)

0 2/7

1 4/7

2 1/7

Solutions for Exercises in Chapter 3

The following is a probability histogram: 4/7

f(x)

3/7

2/7

1/7

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. (d) P (T ≤ 5|T ≥ 2) = 3.13 The c.d.f. of X ⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎪ ⎪ 0.41, ⎪ ⎪ ⎪ ⎨0.78, F (x) = ⎪ 0.94, ⎪ ⎪ ⎪ ⎪ ⎪ 0.99, ⎪ ⎪ ⎪ ⎩1,

P (2≤T ≤5) P (T ≥2)

3/4−1/4 1−1/4

= 23 .

is for for for for for for

x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 2 ≤ x < 3, 3 ≤ x < 4, x ≥ 4.

3.14 (a) P (X < 0.2) = F (0.2) = 1 − e−1.6 = 0.7981; (b) f (x) = F (x) = 8e−8x . Therefore, P (X < 0.2) = 8 3.15 The c.d.f. of X ⎧ ⎪ 0, ⎪ ⎪ ⎪ ⎨2/7, F (x) = ⎪6/7, ⎪ ⎪ ⎪ ⎩1,

0.2 0

0.2 e−8x dx = −e−8x 0 = 0.7981.

is for for for for

x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 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. 3.16 A graph of the c.d.f. is shown next. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 3 Random Variables and Probability Distributions

30 1 6/7

F(x)

5/7 4/7 3/7 2/7 1/7 0

3 (1/2) dx = x2 1 = 1. 2.5

2.5 P (2 < X < 2.5) 2 (1/2) dx = x2 2 = 14 . 1.6

1.6 P (X ≤ 1.6) = 1 (1/2) dx = x2 1 = 0.3. 4

4 (1+x)2 dx = P (X < 4) = 2 2(1+x) 27 27 2 = 16/27. 4

4 (1+x)2 P (3 ≤ X < 4) = 3 2(1+x) dx = 27 27 = 1/3.

3.17 (a) Area = (b) (c) 3.18 (a) (b)

3 1

3.19 For 1 ≤ x < 3, F (x) = Hence,

x 1

(1/2) dt =

x−1 2 ,

⎧ ⎪ ⎨0, F (x) =

x−1 , ⎪ 2

⎩

x 26) = 26 5 dx = 5 (26.25 − 26) = 0.1. It is not extremely rare.

∞ −3 ∞ 3.29 (a) f (x) ≥ 0 and 1 3x−4 dx = −3 x 3 = 1. So, this is a density function. 1

x −4 (b) For x ≥ 1, F (x) = 1 3t dt = 1 − x−3 . So, 0, x < 1, F (x) = −3 1 − x , x ≥ 1. (c) P (X > 4) = 1 − F (4) = 4−3 = 0.0156.

1 3 1 3 3.30 (a) 1 = k −1 (3 − x2 ) dx = k 3x − x3 = 16 3 k. So, k = 16 . −1

x 3 1 3 x 2 (b) For −1 ≤ x < 1, F (x) = 16 = −1 (3 − t ) dt = 3t − 3 t −1 1 9 1 1 1 1 3 99 So, P X < 2 = 2 − 16 2 − 16 2 = 128 .

1 2

9 16 x

−

x3 16 .

(c) P (|X| < 0.8) = P (X < −0.8) > 0.8) = F (−0.8) + 1 − F (0.8) + P (X 9 1 9 1 0.8 + 16 0.83 − 12 + 16 0.8 − 16 0.83 = 0.164. = 1 + 12 − 16 c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 3

y 3.31 (a) For y ≥ 0, F (y) = 14 0 e−t/4 dy = 1 − ey/4 . So, P (Y > 6) = e−6/4 = 0.2231. This probability certainly cannot be considered as “unlikely.” (b) P (Y ≤ 1) = 1 − e−1/4 = 0.2212, which is not so small either. 1

1 3.32 (a) f (y) ≥ 0 and 0 5(1 − y)4 dy = − (1 − y)5 0 = 1. So, this is a density function. 0.1 (b) P (Y < 0.1) = − (1 − y)5 0 = 1 − (1 − 0.1)5 = 0.4095. (c) P (Y > 0.5) = (1 − 0.5)5 = 0.03125. (b) For 0 ≤ y < 1, F (y) = 0.3633.

56y 5 (1 − Y

4 3 0 y (1 − y) dy, we obtain )3 + 28y 6 (1 − y)2 + 8y 7 (1 − y) + y 8 .

3.33 (a) Using integral by parts and setting 1 = k

k = 280. So, P (Y ≤ 0.5) =

(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 speciﬁcation (M ) and 5 − y (M ) does not. The probability for one combination of such a situation is 5! (0.99)y (1 − 0.99)5−y if we assume independence among the tubes. Since there are y!(5−y)! permutations of getting y M s and 5 − y M s, the probability 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) = 9.8 × 10−6 which is extremely small. So, the conjecture is false. 0

8 x 1 8 e−6 6x! = e−6 60! + 61! + · · · + 68! = 0.1528. 3.35 (a) P (X > 8) = 1 − P (X ≤ 8) = x=0

2 e−6 62!

= 0.0446. x

x 3.36 For 0 < x < 1, F (x) = 2 0 (1 − t) dt = − (1 − t)2 0 = 1 − (1 − x)2 . (b) P (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) = 3.37 (a)

3 3 x=0 y=0

(b)

3 3

f (x, y) = c

Chapter 3 Random Variables and Probability Distributions

(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 select x oranges from 3, y apples from 2, and 4 − x − y bananas from 8 3 in 32can 3 ways. A random selection of 4 pieces of fruit can be made in x y 4−x−y 4 ways. Therefore, 32 3 f (x, y) =

4−x−y

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.

1 3.40 (a) g(x) = 32 0 (x + 2y) dy = 23 (x + 1), for 0 ≤ x ≤ 1.

1 (b) h(y) = 23 0 (x + 2y) dx = 13 (1 + 4y), for 0 ≤ y ≤ 1.

1/2 5 (c) P (X < 1/2) = 23 0 (x + 1) dx = 12 . 2

1/2 1/2−y

1/2 1 1 3.41 (a) P (X + Y ≤ 1/2) = 0 24xy dx dy = 12 0 2 − y y dy = 16 . 0

1−x (b) g(x) = 0 24xy dy = 12x(1 − x)2 , for 0 ≤ x < 1. 24xy 12x(1−x)2

2y , (1−x)2

for 0 ≤ y ≤ 1 − x.

1/8 Therefore, P (Y < 1/8 | X = 3/4) = 32 0 y dy = 1/4.

∞ 3.42 Since h(y) = e−y 0 e−x dx = e−y , for y > 0, then f (x|y) = f (x, y)/h(y) = e−x , for x > 0.

1 So, P (0 < X < 1 | Y = 2) = 0 e−x dx = 0.6321.

1/2 1/2

1/2 3.43 (a) P (0 ≤ X ≤ 1/2, 1/4 ≤ Y ≤ 1/2) = 0 x dx = 3/64. 1/4 4xy dy dx = 3/8 0

1 3

1 y (b) P (X < Y ) = 0 0 4xy dx dy = 2 0 y dy = 1/2.

50 50 50 4 3.44 (a) 1 = k 30 30 (x2 + y 2 ) dx dy = k(50 − 30) 30 x2 dx + 30 y 2 dy = 392k 3 · 10 . So, (c) f (y|x) =

3 392

· 10−4 .

40 50 3 · 10−4 30 40 (x2 + y 2 ) dy dx (b) P (30 ≤ X ≤ 40, 40 ≤ Y ≤ 50) = 392

50 3 3 3 3 3 3 · 10−3 ( 30 x2 dx + 40 y 2 dy) = 392 · 10−3 40 −30 + 50 −40 = = 392 3 3

40 40 3 (c) P (30 ≤ X ≤ 40, 30 ≤ Y ≤ 40) = 392 · 10−4 30 30 (x2 + y 2 ) dx dy

40 3 3 3 3 37 = 2 392 · 10−4 (40 − 30) 30 x2 dx = 196 · 10−3 40 −30 = 196 . 3

1/4 1/2−x 1 3.45 P (X + Y > 1/2) = 1 − P (X + Y < 1/2) = 1 − 0 y dy dx x 1 1 1/4

1/4 1 ln 2 − x − ln x dx = 1 + 2 − x ln 2 − x − x ln x 0 =1− 0 = 1 + 41 ln 14 = 0.6534.

49 196 .

3.46 (a) From the column totals of Exercise 3.38, we have x g(x)

0 1/10

1 1/5

2 3/10

3 2/5

Solutions for Exercises in Chapter 3

(b) From the row totals of Exercise 3.38, we have y h(y)

0 1/5

1 1/3

2 7/15

1 3.47 (a) g(x) = 2 x dy = 2(1 − x) for 0 < x < 1; y h(y) = 2 0 dx = 2y, for 0 < y < 1. Since f (x, y) = 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.

1/2 Therefore, P (1/4 < X < 1/2 | Y = 3/4) = 43 1/4 dx = 13 . 3.48 (a) g(2) =

f (2, y) = f (2, 0) + f (2, 1) + f (2, 2) = 9/70 + 18/70 + 3/70 = 3/7. So,

y=0

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. In tabular form, y f (y|2)

0 3/10

1 3/5

2 1/10

(b) P (Y = 0 | X = 2) = f (0|2) = 3/10. 3.49 (a)

x g(x)

1 0.10

2 0.35

3 0.55

(b)

y h(y)

1 0.20

2 0.50

3 0.30

0.1 0.05+0.10+0.20

= 0.2857.

x 3.50

f (x, y) 1 3 5 g(x)

2 0.10 0.20 0.10 0.40

4 0.15 0.30 0.15 0.60

(a)

x g(x)

2 0.40

4 0.60

(b)

y h(y)

1 0.25

3 0.50

h(y) 0.25 0.50 0.25

5 0.25

3.51 (a) If (x, y) represents the selection of x kings and y jacks in 3 draws, we must have x = 0, 1, 2, 3; y = 0, 1, 2, 3; and 0 ≤ x + y ≤ 3. Therefore, (1, 2) represents the selection of 1 king and 2 jacks which will occur with probability 44 6 f (1, 2) = 1122 = . 55 3 Proceeding in a similar fashion for the other possibilities, we arrive at the following joint probability distribution:

Chapter 3 Random Variables and Probability Distributions

f (x, y) 0 1 2 3

0 1/55 6/55 6/55 1/55

1 6/55 16/55 6/55

2 6/55 6/55

3 1/55

(b) P [(X, Y ) ∈ A] = P (X + Y ≥ 2) = 1 − P (X + Y < 2) = 1 − 1/55 − 6/55 − 6/55 = 42/55. 3.52 (a) P (H) = 0.4, P (T ) = 0.6, and S = {HH, HT, T H, T T }. Let (W, Z) represent a typical outcome of the experiment. The particular outcome (1, 0) indicating a total of 1 head and no heads on the ﬁrst toss corresponds to the event T H. Therefore, f (1, 0) = P (W = 1, Z = 0) = P (T H) = 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 following joint probability distribution:

f (w, z) 0 1

0 0.36

w 1 0.24 0.24

2 0.16

(b) Summing the columns, the marginal distribution of W is w g(w)

0 0.36

1 0.48

2 0.16

0 0.60

1 0.40

(d) P (W ≥ 1) = f (1, 0) + f (1, 1) + f (2, 1) = 0.24 + 0.24 + 0.16 = 0.64.

4 3.53 g(x) = 18 2 (6 − x − y) dy = 3−x 4 , for 0 < x < 2. f (x,y) 6−x−y So, f (y|x) = g(x) = 2(3−x) , for 2 < y < 4,

3 and P (1 < Y < 3 | X = 1) = 14 2 (5 − y) dy = 58 . 3.54 Since f (1, 1) = g(1)h(1), the variables are not independent. 3.55 X and Y are independent since f (x, y) = g(x)h(y) for all (x, y).

1−y (x,y) 2x 3.56 (a) h(y) = 6 0 x dx = 3(1 − y)2 , for 0 < y < 1. Since f (x|y) = fh(y) = (1−y) 2 , for 0 < x < 1 − y, involves the variable y, X and Y are not independent.

0.5 (b) P (X > 0.3 | Y = 0.5) = 8 0.3 x dx = 0.64.

2 1 1

1 1 k 3.57 (a) 1 = k 0 0 0 xy 2 z dx dy dz = 2k 0 0 y 2 z dy dz = 2k 3 0 z dz = 3 . So, k = 3.

2 1 1/4

1/4 1 2 (b) P X < 14 , Y > 12 , 1 < Z < 2 = 3 1 1/2 0 xy 2 z dx dy dz = 92 0 1/2 y z dy dz

Solutions for Exercises in Chapter 3

1 3.58 g(x) = 4 0 xy dy = 2x, for 0 < x < 1; h(y) = 4 0 xy dx = 2y, for 0 < y < 1. Since f (x, y) = g(x)h(y) for all (x, y), X and Y are independent.

50 2 + y 2 ) dy = k x2 y + y 3 (x = k 20x2 + 3 30 30

h(y) = k 20y 2 + 98,000 . 3 Since f (x, y) = g(x)h(y), X and Y are not independent.

3.59 g(x) = k

3.60 (a) g(y, z) =

4 9

1 0

98,000 3

, and

xyz 2 dx = 29 yz 2 , for 0 < y < 1 and 0 < z < 3.

3 (b) h(y) = 29 0 yz 2 dz = 2y, for 0 < y < 1.

2 1 1/2 7 (c) P 14 < X < 12 , Y > 13 , Z < 2 = 49 1 1/3 1/4 xyz 2 dx dy dz = 162 . (x,y,z) (d) Since f (x|y, z) = fg(y,z) = 2x, for 0 < x < 1, P 0 < X < 12 | Y = 14 , Z = 2 =

1/2 2 0 x dx = 14 .

3.61 g(x) = 24

1−x 0

xy dy = 12x(1 − x)2 , for 0 < x < 1.

1 (a) P (X ≥ 0.5) = 12 0.5 x(1 − x)2 dx = 0.5 (12x − 24x2 + 12x3 ) dx =

1−y (b) h(y) = 24 0 xy dx = 12y(1 − y)2 , for 0 < y < 1.

5 16

= 0.3125.

(x,y) 24xy 2x = 12y(1−y) (c) f (x|y) = fh(y) 2 = (1−y)2 , for 0 < x < 1 − y.

1/8 1/8 2x dx = 32 0 x = 0.25. So, P X < 18 | Y = 34 = 0 1/16

3.62 (a)

x f (x)

1 0.4

3 0.2

5 0.2

7 0.2

(b) P (4 < X ≤ 7) = P (X ≤ 7) − P (X ≤ 4) = F (7) − F (4) = 1 − 0.6 = 0.4. ∞

∞ −y(1+x)

∞ 1 1 ye−y(1+x) + 1+x dy 3.63 (a) g(x) = 0 ye−y(1+x) dy = − 1+x 0 e 0 ∞ 1 −y(1+x) =− 2 e (1+x)

1 = (1+x) 2 , for x > 0.

∞ ∞ h(y) = ye−y 0 e−yx dx = −e−y e−yx |0 = e−y , for y > 0.

∞ ∞

∞

∞ ∞ (b) P (X ≥ 2, Y ≥ 2) = 2 2 ye−y(1+x) dx dy = − 2 e−y e−yx |2 dy = 2 e−3y dy ∞ = − 13 e−3y 2 = 3e16 .

1/2 1/2 2 2 3.64 (a) P X ≤ 12 , Y ≤ 12 = 32 0 0 (x + y ) dydx =

1/2 2 1 1 x + 12 dx = 16 . = 34 0

1 53 (b) P X ≥ 34 = 32 3/4 x2 + 13 dx = 128 . 3.65 (a)

x f (x)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

3 2

4 0.0902

1/2 2 x y+ 0

5 0.0361

y3 3

1/2 dx 0

6 0.0120

Chapter 3 Random Variables and Probability Distributions

38 0.3

f(x)

0.2

0.1

0.0

(c)

x F (x)

3.66 (a) g(x) =

0 0.1353

1 0

1 0.4060

2 0.6767

3 0.8571

4 0.9473

5 0.9834

(x + y) dy = x + 12 , for 0 < x < 1, and h(y) = y +

6 0.9954

1 2

for 0 < y < 1.

1 2

1 1 dy (b) P (X > 0.5, Y > 0.5) = 0.5 0.5 (x + y) dx dy = 0.5 x2 + xy 0.5 1 y

1 1 = 0.5 2 + y − 8 + 2 dy = 38 .

3.67 f (x) =

5 x

(0.1)x (1 − 0.1)5−x , for x = 0, 1, 2, 3, 4, 5.

3.68 (a) g(x) = h(y) =

2 3x−y 9 1

3 3x−y 9 1

dy =

2 3xy−y 2 /2 9 1

dx =

4 3

x 3

− 16 , for 1 < x < 3, and

− 29 y, for 1 < y < 2.

(b) No, since g(x)h(y) = f (x, y). (c) P (X > 2) =

3.69 (a) f (x) =

3 x 2

d dx F (x)

−

1 6

dx =

1 −x/50 , 50 e

x2 6

−

x 6

3 = 23 . 2

for x > 0.

(b) P (X > 70) = 1 − P (X ≤ 70) = 1 − F (70) = 1 − (1 − e−70/50 ) = 0.2466.

3.70 (a) f (x) =

1 10 ,

for x = 1, 2, . . . , 10.

Solutions for Exercises in Chapter 3

1.0 0.9 0.8

F(x)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

∞

e−y/2 = e−3/2 = 0.2231.

10 1 3.72 (a) f (x) ≥ 0 and 0 10 dx = 1. This is a continuous uniform distribution.

7 1 (b) P (X ≤ 7) = 10 0 dx = 0.7.

1 10 1 = 1. 3.73 (a) f (y) ≥ 0 and 0 f (y) dy = 10 0 (1 − y)9 dy = − 10 10 (1 − y) 0 1

1 10 10 (b) P (Y > 0.6) = 0.6 f (y) dy = − (1 − y) 0.6 = (1 − 0.6) = 0.0001. ∞

∞ −z/10 1 3.74 (a) P (Z > 20) = 10 dz = − e−z/10 20 = e−20/10 = 0.1353. 20 e 10 (b) P (Z ≤ 10) = − e−z/10 0 = 1 − e−10/10 = 0.6321.

1 3.75 (a) g(x1 ) = x1 2 dx2 = 2(1 − x1 ), for 0 < x1 < 1.

x (b) h(x2 ) = 0 2 2 dx1 = 2x2 , for 0 < x2 < 1.

1 0.2 (c) P (X1 < 0.2, X2 > 0, 5) = 0.5 0 2 dx1 dx2 = 2(1 − 0.5)(0.2 − 0) = 0.2.

3.71 P (X ≥ 3) =

1 2

(d) fX1 |X2 (x1 |x2 ) =

f (x1 ,x2 ) h(x2 )

2 2x2

1 x2 ,

for 0 < x1 < x2 .

1 3.76 (a) fX1 (x1 ) = 0 1 6x2 dx2 = 3x21 , for 0 < x1 < 1. Apparently, fX1 (x1 ) ≥ 0 and 0 fX1 (x1 ) dx1 =

1 2 0 3x1 dx1 = 1. So, fX1 (x1 ) is a density function. (b) fX2 |X1 (x2 |x1 ) =

f (x1 ,x2 ) fX1 (x1 )

6x2 3x21

= 2 xx22 , for 0 < x2 < x1 .

10.5 So, P (X2 < 0.5 | X1 = 0.7) = 0.72 2 0 x2 dx2 = 25 49 .

3.77 (a) g(x) =

9 (16)4y

∞ x=0

1 4x

9 1 (16)4y 1−1/4

3 4

1 4x ,

for x = 0, 1, 2, . . . ; similarly, h(y) =

3 4

1 4y ,

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) + f9(1, 2) +2f (2,30) + 4 9 1 1 1 1 1 1 1 1 1 f (2, 1)+f (3, 0) = 16 1 + 4 + 42 + 43 + 4 + 42 + 43 + 42 + 43 + 43 = 16 1 + 4 + 42 + 43 = 63 64 . 3.78 P (the system works) = P (all components work) = (0.95)(0.99)(0.92) = 0.86526. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 3 Random Variables and Probability Distributions

3.79 P (the system does not fail) = P (at least one of the components works) = 1 − P (all components fail) = 1 − (1 − 0.95)(1 − 0.94)(1 − 0.90)(1 − 0.97) = 0.999991. 3.80 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) = 5 3 (1 − 0.92)2 + 5 (0.92)4 (1 − 0.92) + 5 (0.92)5 = 0.9955. (0.92) 3 4 5

Chapter 4

Mathematical Expectation 4.1 μ = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88. 4.2 E(X) =

x f (x) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 3/4.

x=0

4.3 μ = 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) = 3/4 and P (T ) = 1/4. The sample space for the experiment is S = {HH, HT, T H, T T }. 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 (T H) = (2)(3/4)(1/4) = 3/8, P (X = 2) = P (T T ) = (1/4)(1/4) = 1/16. The probability distribution for X is then x f (x)

0 9/16

1 3/8

2 1/16

from which we get μ = E(X) = (0)(9/16) + (1)(3/8) + (2)(1/16) = 1/2. 4.5 Let c = amount to play the game and Y = amount won. y f (y)

5−c 2/13

3−c 2/13

−c 9/13

E(Y ) = (5 − c)(2/13) + (3 − c)(2/13) + (−c)(9/13) = 0. So, 13c = 16 which implies c = $1.23. 4.6 μ = 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. c Copyrightf 2012 Pearson Education, Inc. Publishing as Prentice Hall. 41

Chapter 4 Mathematical Expectation

42 4.8 Let X = proﬁt. Then

μ = E(X) = (250)(0.22) + (150)(0.36) + (0)(0.28) + (−150)(0.14) = $88. 4.9 For the insurance of $200,000 pilot, the distribution of the claim the insurance company would have is as follows: $200,000 0.002

Claim Amount f (x)

$100,000 0.01

$50,000 0.1

0 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. 4.10 μX = xg(x) = (1)(0.17) + (2)(0.5) + (3)(0.33) = 2.16, yh(y) = (1)(0.23) + (2)(0.5) + (3)(0.27) = 2.04. μY =

1 x ln 4 4.11 E(X) = π4 0 1+x 2 dx = π . 4.12 E(X) =

1 0

2x(1 − x) dx = 1/3. So, (1/3)($5, 000) = $1, 667.67.

2 4.13 E(X) = 0 x2 dx + 1 x(2 − x) dx = 1. Therefore, the average number of hours per year is (1)(100) = 100 hours.

1 8 dx = 15 . 4.14 E(X) = 0 2x(x+2) 5 √ 2 2 2 2

a a2 −y2 a −y − a −y dy = 0. 4.15 E(X) = πa1 2 −a √ 2 2 x dx dy = πa1 2 2 2 −

a −y

4.16 P (X1 + X2 ≥ 1) = 1 − P (X1 = 0, X2 = 0) (980)(20) = 1 − 21000 0 = 1 − 0.9604 = 0.040. ( 2 ) 4.17 The probability density function is, x f (x) g(x)

−3 1/6 25

6 1/2 169

9 1/3 361

μg(X) = E[(2X + 1)2 ] = (25)(1/6) + (169)(1/2) + (361)(1/3) = 209. 4.18 E(X 2 ) = (0)(27/64) + (1)(27/64) + (4)(9/64) + (9)(1/64) = 9/8. 4.19 Let Y = 1200X − 50X 2 be the amount spent. x f (x) y = g(x)

0 1/10 0

1 3/10 1150

2 2/5 2200

3 1/5 3150

Solutions for Exercises in Chapter 4

μY = E(1200X − 50X 2 ) = (0)(1/10) + (1150)(3/10) + (2200)(2/5) + (3150)(1/5) = $1, 855.

∞

∞ 4.20 E[g(X)] = E(e2X/3 ) = 0 e2x/3 e−x dx = 0 e−x/3 dx = 3.

1 4.21 E(X 2 ) = 0 2x2 (1 − x) dx = 16 . Therefore, the average proﬁt per new automobile is (1/6)($5000.00) = $833.33.

∞ 1 4.22 E(Y ) = E(X + 4) = 0 32(x + 4) (x+4) 3 dx = 8 days. 4.23 (a) E[g(X, Y )] = E(XY 2 ) =

xy 2 f (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) μX = E(X) = (2)(0.40) + (4)(0.60) = 3.20, μY = E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3.00. 4.24 (a) E(X 2 Y − 2XY ) =

3 2

(x2 y − 2xy)f (x, y) = (1 − 2)(18/70) + (4 − 4)(18/70) + · · · +

x=0 y=0

(8 − 8)(3/70) = −3/7. (b)

y 0 1 2 3 0 1 2 x g(x) 5/70 30/70 30/70 5/70 h(y) 15/70 40/70 15/70 μX = E(X) = (0)(5/70) + (1)(30/70) + (2)(30/70) + (3)(5/70) = 3/2, μY = E(Y ) = (0)(15/70) + (1)(40/70) + (2)(15/70) = 1. Hence μX − μY = 3/2 − 1 = 1/2.

4.25 μX+Y = E(X +Y ) =

3 3

(x+y)f (x, y) = (0+0)(1/55)+(1+0)(6/55)+· · ·+(0+3)(1/55) =

x=0 y=0

2. √

1 1

1 4.26 E(Z) = E( X 2 + Y 2 ) = 0 0 4xy x2 + y 2 dx dy = 43 0 [y(1 + y 2 )3/2 − y 4 ] dy = 8(23/2 − 1)/15 = 0.9752.

∞

∞ 1 4.27 E(X) = 2000 0 x exp(−x/2000) dx = 2000 0 y exp(−y) dy = 2000. 4.28 (a) The density function is shown next.

f(x)

2/5

23.75

(b) E(X) =

2 26.25 5

23.75

26.25

x dx = 15 (26.252 − 23.752 ) = 25.

Chapter 4 Mathematical Expectation

44 4.29 (a) The density function is shown next 3

f(x)

1.5

(b) μ = E(X) = 4.30 E(Y ) =

1 4

∞ 0

∞ 1

2.5

3.5

3x−3 dx = 32 .

ye−y/4 dy = 4.

∞ y d(1 − y)5 = 0 (1 − y)5 dy = 16 . 1

1 (b) P (Y > 1/6) = 1/6 5(1 − y)4 dy = − (1 − y)5 1/6 = (1 − 1/6)5 = 0.4019.

4.31 (a) μ = E(Y ) = 5

1 0

y(1 − y)4 dy = −

1 0

4.32 (a) A histogram is shown next. 0.4

f(x)

0.3

0.2

0.1

(b) μ = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88. (c) E(X 2 ) = (0)2 (0.41) + (1)2 (0.37) + (2)2 (0.16) + (3)2 (0.05) + (4)2 (0.01) = 1.62. 4.33 μ = $500. So, σ 2 = E[(X − μ)2 ] = (x − μ)2 f (x) = (−1500)2 (0.7) + (3500)2 (0.3) = x

$5, 250, 000. 4.34 μ = (−2)(0.3) + (3)(0.2) + (5)(0.5) = 2.5 and E(X 2 ) = (−2)2 (0.3) + (3)2 (0.2) + (5)2 (0.5) = 15.5. So, σ 2 = E(X 2 ) − μ2 = 9.25 and σ = 3.041. 4.35 μ = (2)(0.01) + (3)(0.25) + (4)(0.4) + (5)(0.3) + (6)(0.04) = 4.11, E(X 2 ) = (2)2 (0.01) + (3)2 (0.25) + (4)2 (0.4) + (5)2 (0.3) + (6)2 (0.04) = 17.63. So, σ 2 = 17.63 − 4.112 = 0.74. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 4

4.36 μ = (0)(0.4) + (1)(0.3) + (2)(0.2) + (3)(0.1) = 1.0, and E(X 2 ) = (0)2 (0.4) + (1)2 (0.3) + (2)2 (0.2) + (3)2 (0.1) = 2.0. So, σ 2 = 2.0 − 1.02 = 1.0. 4.37 It is know μ = 1/3. 1 So, E(X 2 ) = 0 2x2 (1 − x) dx = 1/6 and σ 2 = 1/6 − (1/3)2 = 1/18. So, in the actual proﬁt, 1 (5000)2 . the variance is 18 4.38 It is known μ = 8/15. 1 Since E(X 2 ) = 0 25 x2 (x + 2) dx =

11 30 ,

then σ 2 = 11/30 − (8/15)2 = 37/450.

4.39 It is known μ = 1.

2 1 Since E(X 2 ) = 0 x2 · x dx + 1 x2 (2 − x) dx = 7/6, then σ 2 = 7/6 − (1)2 = 1/6.

1 dx = 15 0 (6x3 + 12x2 + 8x + 16) dx = 5.1. 4.40 μg(X) = E[g(X)] = 0 (3x2 + 4) 2x+4 5

1 So, σ 2 = E[g(X) − μ]2 = 0 (3x2 + 4 − 5.1)2 2x+4 dx 5 2x+4

1 4 2 = 0 (9x − 6.6x + 1.21) 5 dx = 0.83. 4.41 It is known μg(X) = E[(2X + 1)2 ] = 209. Hence 2 = [(2X + 1)2 − 209]2 g(x) σg(X) x

= (25 − 209)2 (1/6) + (169 − 209)2 (1/2) + (361 − 209)2 (1/3) = 14, 144. √ So, σg(X) = 14, 144 = 118.9. 4.42 It is known μg(X) = E(X 2 ) = 1/6. Hence 2

1 2 σg(X) = 0 2 x2 − 16 (1 − x) dx = 7/180.

∞ 4.43 μY = E(3X − 2) = 14 0 (3x − 2)e−x/4 dx = 10. So

∞ σY2 = E{[(3X − 2) − 10]2 } = 94 0 (x − 4)2 e−x/4 dx = 144. 4.44 E(XY ) = xyf (x, y) = (1)(1)(18/70) + (2)(1)(18/70) x

+ (3)(1)(2/70) + (1)(2)(9/70) + (2)(2)(3/70) = 9/7; xf (x, y) = (0)f (0, 1) + (0)f (0, 2) + (1)f (1, 0) + · · · + (3)f (3, 1) = 3/2, μX = x

and μY = 1. So, σXY = E(XY ) − μX μY = 9/7 − (3/2)(1) = −3/14. 4.45 μX = xg(x) = 2.45, μY = yh(y) = 3.20, and x y xyf (x, y) = (1)(0.05) + (2)(0.05) + (3)(0.10) + (2)(0.05) E(XY ) = x

+ (4)(0.10) + (6)(0.35) + (3)(0) + (6)(0.20) + (9)(0.10) = 7.85. So, σXY = 7.85 − (2.45)(3.20) = 0.01. 3 −4 10 , and g(x) = k 20x2 + 98000 , with 4.46 From previous exercise, k = 392 3

50 98000 3 μX = E(X) = 30 xg(x) dx = k 30 20x + 3 x dx = 40.8163. Similarly, μY = 40.8163. On the other hand, 50 50 E(XY ) = k 30 30 xy(x2 + y 2 ) dy dx = 1665.3061. Hence, σXY = E(XY ) − μX μY = 1665.3061 − (40.8163)2 = −0.6642. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 4 Mathematical Expectation

1 4.47 g(x) = 23 0 (x + 2y) dy = 23 (x + 1), for 0 < x < 1, so μX = 23 0 x(x + 1) dx = 59 ;

1 h(y) = 23 0 (x + 2y) dx = 23 12 + 2y , so μY = 23 0 y 12 + 2y dy = 11 18 ; and

2 1 1 1 E(XY ) = 3 0 0 xy(x + 2y) dy dx= 3. So, σXY = E(XY ) − μX μY = 13 − 59 11 18 = −0.0062. 2 and σ 2 = b2 σ 2 , then 4.48 Since σXY = Cov(a + bX, X) = bσX Y X 2 bσ b ρ = σσXY σY = √ 2 X2 2 = |b| = sign of b. σX b σX

Hence ρ = 1 if b > 0 and ρ = −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 + (3)2 (0.05) + (4)2 (0.01) = 1.62. and E(X 2 ) = (0)2 (0.41) + (1)2 (0.37) + (2)2 (0.16) √ So, V ar(X) = 1.62 − 0.882 = 0.8456 and σ = 0.8456 = 0.9196. 2

1 3 1 4.50 E(X) = 2 0 x(1 − x) dx = 2 x2 − x3 = 13 and 0 3

1 2 x x4 2 E(X ) = 2 0 x (1 − x) dx = 2 3 − 4 = 16 . Hence, 0 2 1 , and σ = 1/18 = 0.2357. V ar(X) = 16 − 13 = 18 4.51 The joint and marginal probability mass functions are given in the following table.

Hence, E(X) = ρXY = − √15 .

3 2,

0 1 2 3 fX (x)

y 1

0 0

1 35 9 35 9 35 1 35 4 7

3 70 9 70 3 70 3 14

E(Y ) = 1, E(XY ) =

9 7,

fX (x)

3 70 9 70 3 70

1 14 3 7 3 7 1 14

0 3 14

V ar(X) =

15 28 ,

and V ar(Y ) =

3 7.

Finally,

4.52 Since fX (x) = 2(1 − x), for 0 < x < 1, and fY (y) = 2y, for 0 < y < 1, we obtain E(X) = 13 , 1 , and E(XY ) = 14 . Hence, ρXY = 12 . E(Y ) = 23 , V ar(X) = V ar(Y ) = 18 4.53 Previously we found μ = 4.11 and σ 2 = 0.74, Therefore, μg(X) = E(3X − 2) = 3μ − 2 = (3)(4.11) − 2 = 10.33 and σg(X) = 9σ 2 = 6.66. 4.54 Previously we found μ = 1 and σ 2 = 1. Therefore, μg(X) = E(5X + 3) = 5μ + 3 = (5)(1) + 3 = 8 and σg(X) = 25σ 2 = 25. 4.55 Let X = number of cartons sold and Y = proﬁt. We can write Y = 1.65X + (0.90)(5 − X) − 6 = 0.75X − 1.50. Now E(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.

∞ 4.56 μX = E(X) = 14 0 xe−x/4 dx = 4. − 2) = 3E(X) − 2 = (3)(4) − 2 = 10. Therefore, μY = E(3X

Solutions for Exercises in Chapter 4

4.57 E(X) = (−3)(1/6) + (6)(1/2) + (9)(1/3) = 11/2, E(X 2 ) = (−3)2 (1/6) + (6)2 (1/2) + (9)2 (1/3) = 93/2. So, E[(2X + 1)2 ] = 4E(X 2 ) + 4E(X) + 1 = (4)(93/2) + (4)(11/2) + 1 = 209.

2 4.58 Since E(X) = 0 x2 dx + 1 x(2 − x) dx = 1, and

1 2 E(X 2 ) = 0 x3 2 dx + 1 x2 (2 − x) dx = 7/6, then E(Y ) = 60E(X 2 ) + 39E(X) = (60)(7/6) + (39)(1) = 109 kilowatt hours. 4.59 The equations E[(X − 1)2 ] = 10 and E[(X − 2)2 ] = 6 may be written in the form: E(X 2 ) − 2E(X) = 9,

E(X 2 ) − 4E(X) = 2.

Solving these two equations simultaneously we obtain E(X) = 7/2,

and

E(X 2 ) = 16.

Hence μ = 7/2 and σ 2 = 16 − (7/2)2 = 15/4. 4.60 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. 4.61 E(2XY 2 − X 2 Y ) = 2E(XY 2 ) − E(X 2 Y ). Now, 2 2 xy 2 f (x, y) = (1)(1)2 (3/14) = 3/14, and E(XY 2 ) = E(X 2 Y ) =

x=0 y=0 2 2

x2 yf (x, y) = (1)2 (1)(3/14) = 3/14.

x=0 y=0

Therefore, E(2XY 2 − X 2 Y ) = (2)(3/14) − (3/14) = 3/14. 2 2 2 4.62 σZ2 = σ−2X+4Y −3 = 4σX + 16σY = (4)(5) + (16)(3) = 68. 2 2 2 4.63 σZ2 = σ−2X+4Y −3 = 4σX + 16σY − 16σXY = (4)(5) + (16)(3) − (16)(1) = 52.

4.64 E(Z) = E(XY ) = E(X)E(Y ) =

1 ∞ 0

16xy(y/x3 ) dx dy = 8/3.

4.65 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. (c) E(XY ) = E(X)E(Y ) = (3.5)(3.5) = 12.25. 2 = σ 2 = [(1)2 + (2)2 + · · · + (6)2 ](1/6) − (3.5)2 = 4.66 μX = μY = 3.5. σX Y

35 12 .

175 12 ; 9σY2 = 175 6 .

2 + σ2 = (a) σ2X−Y = 4σX Y 2 + (b) σX+3Y −5 = σX

Chapter 4 Mathematical Expectation

4.67 E[g(X, Y )] = E(X/Y 3 + X 2 Y ) = E(X/Y 3 ) + E(X 2 Y ).

2 1 E(X/Y 3 ) = 1 0 2x(x+2y) dx dy = 27 1 3y13 + y12 dy = 15 84 ; 7y 3

2 2 1 2 E(X 2 Y ) = 1 0 2x y(x+2y) dx dy = 27 1 y 14 + 2y dy = 139 7 3 252 . Hence, E[g(X, Y )] =

15 84

139 252

46 63 .

4.68 P = I 2 R with R = 50, μI = E(I) = 15 and σI2 = V ar(I) = 0.03. E(P ) = E(I 2 R) = 50E(I 2 ) = 50[V ar(I) + μ2I ] = 50(0.03 + 152 ) = 11251.5. If we use the approximation formula, with g(I) = I 2 , g (I) = 2I and g (I) = 2, we obtain, σI2 = 50(152 + 0.03) = 11251.5. E(P ) ≈ 50 g(μI ) + 2 2 Since V ar[g(I)] ≈

2 ∂g(i) ∂i i=μ I

σI2 , we obtain

V ar(P ) = 502 V ar(I 2 ) = 502 (2μI )2 σI2 = 502 (30)2 (0.03) = 67500. 4.69 For 0 < a < 1, since g(a) = g (a) =

∞

ax =

x=0

x(x − 1)ax−2 =

x=2

(a) E(X) = (3/4)

∞

1 1−a ,

g (a) =

x(1/4)x = (3/4)(1/4)

x=1

xax−1 =

1 (1−a)2

and

∞

x(1/4)x−1 = (3/16)[1/(1 − 1/4)2 ]

x=1

E(X 2 ) − E(X) = E[X(X − 1)] = (3/4) ∞

x=1

2 . (1−a)3

= 1/3, and E(Y ) = E(X) = 1/3.

= (3/4)(1/4)2

∞

x(x − 1)(1/4)x

x=2

x(x − 1)(1/4)x−2 = (3/43 )[2/(1 − 1/4)3 ] = 2/9.

x=2

So, V ar(X) = E(X 2 ) − [E(X)]2 = [E(X 2 ) − E(X)] + E(X) − [E(X)]2 2/9 + 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, and V ar(Z) = V ar(X + Y ) = V ar(X) + V ar(Y ) = (4/9) + (4/9) = 8/9, since X and Y are independent (from Exercise 3.79).

1 4.70 (a) g(x) = 23 0 (x2 + y 2 ) dy = 12 (3x2 + 1) for 0 < x < 1 and h(y) = 12 (3y 2 + 1) for 0 < y < 1. Since f (x, y) = g(x)h(y), X and Y are not independent.

1 2 (b) E(X + Y ) = E(X) + E(Y ) = 2E(X) = 0 x(3x + 1)

dx = 3/4 + 1/2 = 5/4.

y2 3 1 1 3 1 1 2 2 E(XY ) = 2 0 0 xy(x + y ) dx dy = 2 0 y 4 + 2 dy = 32 14 12 + 12 14 = 38 . 2

1 7 73 = 15 − 25 (c) V ar(X) = E(X 2 ) − [E(X)]2 = 12 0 x2 (3x2 + 1) dx − 58 64 = 960 , and 2 73 1 . Also, Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 38 − 58 = − 64 . V ar(Y ) = 960 73 1 (d) V ar(X + Y ) = V ar(X) + V ar(Y ) + 2Cov(X, Y ) = 2 960 − 2 64 =

∞ 4.71 (a) E(Y ) = 0 ye−y/4 dy = 4.

29 240 .

Solutions for Exercises in Chapter 4

(b) E(Y 2 ) =

∞ 0

y 2 e−y/4 dy = 32 and V ar(Y ) = 32 − 42 = 16.

4.72 (a) The density function is shown next.

f(x)

8 x

8 (b) E(Y ) = 7 y dy = 12 [82 − 72 ] = 15 2 = 7.5,

8 2 1 3 2 3 E(Y ) = 7 y dy = 3 [8 − 7 ] = 169 3 , and V ar(Y ) =

169 3

−

15 2 2

1 12 .

4.73 Using the exact formula, we have

E(e ) = 7

ey dy = ey |87 = 1884.32.

Using the approximation, since g(y) = ey , so g (y) = ey . Hence, using the approximation formula, E(e ) ≈ e Y

μY

σY2 = 2

1 1+ 24

e7.5 = 1883.38.

The approximation is very close to the true value. 8

8 4.74 Using the exact formula, E(Z 2 ) = 7 e2y dy = 12 e2y 7 = 3841753.12. Hence, V ar(Z) = E(Z 2 ) − [E(Z)]2 = 291091.3. Using the approximation formula, we have V ar(eY ) = (eμY )2 V ar(Y ) =

e(2)(7.5) = 272418.11. 12

The approximation is not so close to each other. One reason is that the ﬁrst order approximation may not always be good enough. 4.75 μ = 900 hours and σ = 50 hours. Solving μ − kσ = 700 we obtain k = 4. So, using Chebyshev’s theorem with P (μ − 4σ < X < μ + 4σ) ≥ 1 − 1/42 = 0.9375, we obtain P (700 < X < 1100) ≥ 0.9375. Therefore, P (X ≤ 700) ≤ 0.03125. 4.76 Using μ = 60 and σ = 6 and Chebyshev’s theorem P (μ − kσ < X < μ + kσ) ≥ 1 −

1 , k2

Chapter 4 Mathematical Expectation

50 since from μ + kσ = 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 as 84 or higher. Since there will be 70 positions, the applicant will have the job. 4.77 (a) P (|X − 10| ≥ 3) = 1 − P (|X − 10| < 3) = 1 − P [10 − (3/2)(2) < X < 10 + (3/2)(2)] ≤ 1 − 1 − (b) P (|X − 10| < 3) = 1 − P (|X − 10| ≥ 3) ≥ 1 −

4 9

1 (3/2)2

= 49 .

= 59 .

1 (5/2)2

21 25 .

(d) P (|X − 10| ≥ c) ≤ 0.04 implies that P (|X − 10| < c) ≥ 1 − 0.04 = 0.96. Solving 0.96 = 1 − k12 we obtain k = 5. So, c = kσ = (5)(2) = 10.

1 4.78 μ = E(X) = 6 0 x2 (1 − x) dx = 0.5, E(X 2 ) = 6 0 x3 (1 − x) dx = 0.3, which imply σ 2 = 0.3 − (0.5)2 = 0.05 and σ = 0.2236. Hence, P (μ − 2σ < X < μ + 2σ) = P (0.5 − 0.4472 < X < 0.5 + 0.4472) 0.9472 = P (0.0528 < X < 0.9472) = 6 x(1 − x) dx = 0.9839, 0.0528

compared to a probability of at least 0.75 given by Chebyshev’s theorem. 4.79 Deﬁne I1 = {xi | |xi − μ| < kσ} and I2 = {xi | |xi − μ| ≥ kσ}. Then (x − μ)2 f (x) = (xi − μ)2 f (xi ) + (xi − μ)2 f (xi ) σ 2 = E[(X − μ)2 ] = xi ∈I1

≥

xi ∈I2

(xi − μ) f (xi ) ≥ k σ 2

2 2

xi ∈I2

f (xi ) = k σ P (|X − μ| ≥ kσ), 2 2

xi ∈I2

which implies P (|X − μ| ≥ kσ) ≤

1 . k2

Hence, P (|X − μ| < kσ) ≥ 1 − k12 .

1 1

1 1 7 7 and E(Y ) = 12 . 4.80 E(XY ) = 0 0 xy(x + y) dx dy = 13 , E(X) = 0 0 x(x + y) dx dy = 12 1 7 2 1 Therefore, σXY = E(XY ) − μX μY = 3 − 12 = − 144 .

1 y 4.81 E(Y − X) = 0 0 2(y − x) dx dy = 0 y 2 dy = 13 . Therefore, the average amount of kerosene left in the tank at the end of each day is (1/3)(1000) = 333 liters.

∞ 4.82 (a) E(X) = 0 x5 e−x/5 dx = 5.

∞ 2 (b) E(X 2 ) = 0 x5 e−x/5 dx = 50, so V ar(X) = 50 − 52 = 25, and σ = 5. (c) E[(X + 5)2 ] = E{[(X − 5) + 10]2 } = E[(X − 5)2 ] + 102 + 20E(X − 5) = V ar(X) + 100 = 125. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 4

1 1−y 2 , 4.83 E(XY ) = 24 0 0 x2 y 2 dx dy = 8 0 y 2 (1 − y)3 dy = 15

1 1−y 2

1 1−y 2 μX = 24 0 0 x y dx dy = 5 and μY = 24 0 0 xy 2 dx dy = 2 2 2 − 25 = − 75 . E(XY ) − μX μY = 15

1 1−y 4.84 E(X + Y ) = 0 0 24(x + y)xy dx dy = 45 .

∞ x −x/900 e dx = 900 hours. 4.85 (a) E(X) = 0 900

2 ∞ x e−x/900 dx = 1620000 hours2 . (b) E(X 2 ) = 0 900

2 5.

Therefore, σXY =

(c) V ar(X) = E(X 2 ) − [E(X)]2 = 810000 hours2 and σ = 900 hours. 4.86 It is known g(x) = 23 (x + 1), for 0 < x < 1, and h(y) = 13 (1 + 4y), for 0 < y < 1.

1 (a) μX = 0 32 x(x + 1) dx = 59 and μY = 0 13 y(1 + 4y) dy = 11 18 . (b) E[(X + Y )/2] = 12 [E(X) + E(Y )] =

7 12 .

4.87 Cov(aX, bY ) = E[(aX − aμX )(bY − bμY )] = abE[(X − μX )(Y − μY )] = abCov(X, Y ). 4.88 It is known μ = 900 and σ = 900. For k = 2, P (μ − 2σ < X < μ + 2σ) = P (−900 < X < 2700) ≥ 0.75 using Chebyshev’s theorem. On the other hand, P (μ − 2σ < X < μ + 2σ) = P (−900 < X < 2700) = 1 − e−3 = 0.9502. For k = 3, Chebyshev’s theorem yields P (μ − 3σ < X < μ + 3σ) = P (−1800 < X < 3600) ≥ 0.8889, while P (−1800 < X < 3600) = 1 − e−4 = 0.9817. ∞

∞

1 dy = x83 , for x > 2, with μX = 2 x82 dx = − x8 2 = 4, 4.89 g(x) = 0 16y x3 ∞

∞

1 2 2 8y h(y) = 2 16y dx = − = 2y, for 0 < y < 1, with μ = Y 3 2 0 2y = 3 , and x x 2

∞ 1 16y2 ∞ 1 8 E(XY ) = 2 0 x2 dy dx = 16 3 2 x2 dx = 3 . Hence, 8 2 σXY = E(XY ) − μX μY = 3 − (4) 3 = 0. 2 = 5 and σ 2 = 3, we have ρ = 4.90 Since σXY = 1, σX Y

σXY σX σY

=√

1 (5)(3)

= 0.2582.

4.91 (a) From Exercise 4.37, we have σ 2 = 1/18, so σ = 0.2357. (b) Also, μX = 1/3 from Exercise 4.12. So, P (μ − 2σ < X < μ + 2σ) = P [1/3 − (2)(0.2357) < X < 1/3 + (2)(0.2357)] 0.8047 = P (0 < X < 0.8047) = 2(1 − x) dx = 0.9619. 0

Using Chebyshev’s theorem, the probability of this event should be larger than 0.75, which is true.

Chapter 4 Mathematical Expectation

4.92 Since g(0)h(0) = (0.17)(0.23) = 0.10 = f (0, 0), X and Y are not independent. 4.93 E(X) = (−5000)(0.2) + (10000)(0.5) + (30000)(0.3) = $13, 000. 4.94 (a) f (x) = x3 (0.15)x (0.85)3−x , for x = 0, 1, 2, 3. x f (x)

0 0.614125

1 0.325125

2 0.057375

3 0.003375

(b) E(X) = 0.45. (c) E(X 2 ) = 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. 4.95 (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(X 2 ) = 2, 697, 500, 000 dollars2 . So, σ = E(X 2 ) − [E(X)]2 = $39.689k.

50 3 2 2 4.96 (a) E(X) = 4×50 3 −50 x(50 − x ) dx = 0.

50 2 2 3 2 (b) E(X 2 ) = 4×50 3 −50 x (50 − x ) dx = 500. √ (c) σ = E(X 2 ) − [E(X)]2 = 500 − 0 = 22.36. 4.97 (a) The marginal density of X is x1 fX1 (x1 )

0 0.13

1 0.21

2 0.31

3 0.23

4 0.12

0 0.10

1 0.30

2 0.39

3 0.15

4 0.06

(b) The marginal density of Y is x2 fX2 (x2 )

7 15

1 5

1 15

1 5

1 15

(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. 1 1 18 7 (f) E(X1 |X2 = 3) = (0) 15 + (1) 15 + (2) 15 + (3) 15 + (4) 15 = 15 = (g)

E(X12 )

(0)2 (0.13)

(1)2 (0.21)

(2)2 (0.31)

(3)2 (0.23)

(4)2 (0.12)

= + +√ + √ + So, σX1 = E(X12 ) − [E(X1 )]2 = 5.44 − 22 = 1.44 = 1.2.

6 5

= 5.44.

4.98 (a) The marginal densities of X and Y are, respectively, x g(x)

0 0.2

1 0.32

2 0.48

y h(y)

0 0.26

1 0.35

= 1.2.

2 0.39

Solutions for Exercises in Chapter 4

x fX|Y =2 (x|2)

4 39

5 39

30 39

(b) E(X) = (0)(0.2) + (1)(0.32) + (2)(0.48) = 1.28, E(X 2 ) = (0)2 (0.2) + (1)2 (0.32) + (2)2 (0.48) = 2.24, and V ar(X) = 2.24 − 1.282 = 0.6016. 5 65 2 2 5 2 30 (c) E(X|Y = 2) = (1) 39 + (2) 30 39 = 39 and E(X |Y = 2) = (1) 39 + (2) 39 = 65 2 650 50 = 1521 = 117 . V ar(X) = 125 39 − 39

125 39 .

So,

4.99 The proﬁt is 8X + 3Y − 10 for each trip. So, we need to calculate the average of this quantity. The marginal densities of X and Y are, respectively, x g(x)

0 0.34

1 0.32

2 0.34

y h(y)

0 0.05

1 0.18

2 0.15

3 0.27

4 0.19

5 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. k ∂h(x1 ,x2 ,...,xk ) 2 4.100 Using the approximation formula, V ar(Y ) ≈ i=1 σi2 , we have ∂xi xi =μi , 1≤i≤k

2 b0 +b1 k1 +b2 k2 2 ∂e V ar(Yˆ ) ≈ ∂bi i=0

4.101 (a) E(Y ) = 10

1 0

σb2i = e2(β0 +k1 β1 +k2 β2 ) (σ02 + k12 σ12 + k22 σ22 ). bi =βi , 0≤i≤2

1 1 y(1 − y)9 dy = − y(1 − y)10 0 + 0 (1 − y)10 dy =

(b) E(1 − Y ) = 1 − E(Y ) =

1 11 .

10 11 .

10 112 ×12

= 0.006887.

Chapter 5

Some Discrete Probability Distributions 5.1 By deﬁnition, we have μ = E(X) =

k i=1

and σ2 =

k 1 1 = xi , k k i=1

k 1 (xi − μ)2 . k i=1

5.2 Binomial distribution with n = 12 and p = 0.5. Hence P (X = 3) = P (X ≤ 3) − P (X ≤ 2) = 0.0730 − 0.0193 = 0.0537. 5.3 This is a uniform distribution: f (x) = 3 3 f (x) = 10 . Therefore P (X < 4) =

1 10 ,

for x = 1, 2, . . . , 10, and f (x) = 0 elsewhere.

x=1

5.4 For n = 5 and p = 3/4, we have (a) P (X = 2) = 52 (3/4)2 (1/4)3 = 0.0879, 3 (b) P (X ≤ 3) = b(x; 5, 3/4) = 1 − P (X = 4) − P (X = 5) x=0 5 = 1 − 4 (3/4)4 (1/4)1 − 55 (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. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 55

Chapter 5 Some Discrete Probability Distributions

(a) P (2 ≤ X ≤ 5) = P (X ≤ 5) − P (X ≤ 1) = 0.9844 − 0.1094. (b) P (X < 3) = P (X ≤ 2) = 0.3438. 5.7 p = 0.7. (a) For n = 10, P (X < 5) = P (X ≤ 4) = 0.0474. (b) For n = 20, P (X < 10) = P (X ≤ 9) = 0.0171. 5.8 For n = 8 and p = 0.6, we have (a) P (X = 3) = b(3; 8, 0.6) = P (X ≤ 3) − P (X ≤ 2) = 0.1737 − 0.0498 = 0.1239. (b) P (X ≥ 5) = 1 − P (X ≤ 4) = 1 − 0.4059 = 0.5941. 5.9 For n = 15 and p = 0.25, we have (a) P (3 ≤ X ≤ 6) = P (X ≤ 6) − P (X ≤ 2) = 0.9434 − 0.2361 = 0.7073. (b) P (X < 4) = P (X ≤ 3) = 0.4613. (c) P (X > 5) = 1 − P (X ≤ 5) = 1 − 0.8516 = 0.1484. 5.10 From Table A.1 with n = 12 and p = 0.7, we have (a) P (7 ≤ X ≤ 9) = P (X ≤ 9) − P (X ≤ 6) = 0.7472 − 0.1178 = 0.6294. (b) P (X ≤ 5) = 0.0386. (c) P (X ≥ 8) = 1 − P (X ≤ 7) = 1 − 0.2763 = 0.7237. 5.11 From Table A.1 with n = 7 and p = 0.9, we have P (X = 5) = P (X ≤ 5) − P (X ≤ 4) = 0.1497 − 0.0257 = 0.1240. 5.12 From Table A.1 with n = 9 and p = 0.25, we have P (X < 4) = 0.8343. 5.13 From Table A.1 with n = 5 and p = 0.7, we have P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.1631 = 0.8369. 5.14 (a) n = 4, P (X = 4) = (0.9)4 = 0.6561. (b) The bulls could win in 4, 5, 6 or 7 games. So the probability is 0.6561 +

5 6 4 (0.9)4 (0.1)2 + (0.9)4 (0.1)3 = 0.9973. (0.9)4 (0.1) + 3 3 3

(c) The probability that the Bulls win is always 0.9. 5.15 p = 0.4 and n = 5. (a) P (X = 0) = 0.0778. (b) P (X < 2) = P (X ≤ 1) = 0.3370. (c) P (X > 3) = 1 − P (X ≤ 3) = 1 − 0.9130 = 0.0870. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 5

5.16 Probability of 2 or more of 4 engines operating when p = 0.6 is P (X ≥ 2) = 1 − P (X ≤ 1) = 0.8208, and the probability of 1 or more of 2 engines operating when p = 0.6 is P (X ≥ 1) = 1 − P (X = 0) = 0.8400. The 2-engine plane has a slightly higher probability for a successful ﬂight when p = 0.6. 5.17 The mean is μ = np = (5)(0.7) = 3.5 and the variance is σ 2 = npq = (5)(0.7)(0.3) = 1.05 with σ = 1.025. 5.18 (a) μ = np = (15)(0.25) = 3.75. (b) σ 2 = npq = (15)(0.25)(0.75) = 2.8125. 5.19 Let X1 = number of times encountered green light with P (Green) = 0.35, X2 = number of times encountered yellow light with P (Yellow) = 0.05, and X3 = number of times encountered red light with P (Red) = 0.60. Then n (0.35)x1 (0.05)x2 (0.60)x3 . f (x1 , x2 , x3 ) = x1 , x 2 , x 3 5.20 (a) (b) (c)

10 2 5 3 2,5,3 (0.225) (0.544) (0.231) = 10 10 0 10 (0.544) (0.456) = 0.0023. 10 0 10 = 0.0782. 0 (0.225) (0.775)

0.0749.

5.21 Using the multinomial distribution with required probability is 7 (0.02)(0.82)4 (0.1)2 = 0.0095. 0, 0, 1, 4, 2 5.22 Using the multinomial distribution, we have

8 5,2,1

(1/2)5 (1/4)2 (1/4) = 21/256.

5.23 Using the multinomial distribution, we have 9 (0.4)3 (0.2)3 (0.3)(0.1)2 = 0.0077. 3, 3, 1, 2 5.24 p = 0.40 and n = 6, so P (X = 4) = P (X ≤ 4) − P (X ≤ 3) = 0.9590 − 0.8208 = 0.1382. 5.25 n = 20 and the probability of a defective is p = 0.10. So, P (X ≤ 3) = 0.8670. 5.26 n = 8 and p = 0.60; (a) P (X = 6) = 86 (0.6)6 (0.4)2 = 0.2090. (b) P (X = 6) = P (X ≤ 6) − P (X ≤ 5) = 0.8936 − 0.6846 = 0.2090. 5.27 n = 20 and p = 0.90; c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5 Some Discrete Probability Distributions

(a) P (X = 18) = P (X ≤ 18) − P (X ≤ 17) = 0.6083 − 0.3231 = 0.2852. (b) P (X ≥ 15) = 1 − P (X ≤ 14) = 1 − 0.0113 = 0.9887. (c) P (X ≤ 18) = 0.6083. 5.28 n = 20; (a) p = 0.20, P (X ≥ x) ≤ 0.5 and P (X < x) > 0.5 yields x = 3. (b) p = 0.80, P (Y ≥ y) ≥ 0.8 and P (Y < y) < 0.2 yields y = 15. 5.29 Using the hypergeometric distribution, we get h(2; 9, 6, 4) = 5.30 P (X ≥ 1) = 1 − P (X = 0) = 1 − h(0; 15, 3, 6) = 1 −

(60)(93) = (15 3)

(42)(54) = (96)

5 14 .

53 65 .

2 ) (x4)(3−x , for x = 1, 2, 3. 6 (3) P (2 ≤ X ≤ 3) = h(2; 6, 3, 4) + h(3; 6, 3, 4) = 45 .

5.31 h(x; 6, 3, 4) =

5.32 (a) Probability that all 4 ﬁre = h(4; 10, 4, 7) = 16 . (b) Probability that at most 2 will not ﬁre =

2 x=0

h(x; 10, 4, 3) =

29 30 .

5.33 Using the hypergeometric distribution, we get 40 (12 2 )( 5 ) = 0.3246. 52 (7) (48 7) = 0.4496. (b) 1 − 52 (7)

(a)

5.34 h(2; 9, 5, 4) = 5.35 P (X ≤ 2) =

(42)(53) = (95) 2

10 21 .

h(x; 50, 5, 10) = 0.9517.

x=0

5.36 (a) P (X = 0) = h(0; 25, 3, 3) = (b) P (X = 1) = h(1; 25, 3, 1) =

77 115 . 3 25 .

5.37 (a) P (X = 0) = b(0; 3, 3/25) = 0.6815. (b) P (1 ≤ X ≤ 3) =

b(x; 3, 1/25) = 0.1153.

x=1

5.38 Using the binomial approximation of the hypergeometric distribution with p = 30/150 = 0.2, 2 the probability is 1 − b(x; 10, 0.2) = 0.3222. x=0

5.39 Using the binomial approximation of the hypergeometric with p = 0.5, the probability is 2 b(x; 10, 0.5) = 0.9453. 1− x=0

Solutions for Exercises in Chapter 5

5.40 The binomial approximation of the hypergeometric with p = 1 − 4000/10000 = 0.6 gives a 7 b(x; 15, 0.6) = 0.2131. probability of x=0

5.41 Using the binomial approximation of the hypergeometric distribution with 0.7, the probability 13 b(x; 18, 0.7) = 0.6077. is 1 − x=10

5.42 Using the extension of the hypergeometric distribution the probability is 13131313 5

523

= 0.0129.

5.43 (a) The extension of the hypergeometric distribution gives a probability 2352 4 1 1 1 1 12 = . 33 4 (b) Using the extension of the hypergeometric distribution, we have 232 232 232 8 1 1 2 1 1 2 1 12 + 2 12 + 1 12 = . 165 4 4 4 5.44 Using the extension of the hypergeometric distribution the probability is 243 243 243 17 2 1 2 9 + 2 29 1 + 2 39 0 = . 63 5 5 5 5.45 h(5; 25, 15, 10) =

15 (10 5 )(10) = 0.2315. 25 (15)

(21)(13 4) = 0.4762. (15 5) (2)(13) (b) 2 15 3 = 0.0952. (5)

5.46 (a)

(30)(17 5) = 0.3991. (20 ) 5 (3)(17) (b) 2 20 3 = 0.1316. (5)

5.47 (a)

5.48 N = 10000, n = 30 and k = 300. Using binomial approximation to the hypergeometric distribution with p = 300/10000 = 0.03, the probability of {X ≥ 1} can be determined by 1 − b(0; 30, 0.03) = 1 − (0.97)30 = 0.599. 5.49 Using the negative binomial distribution, the required probability is 9 ∗ (0.3)5 (0.7)5 = 0.0515. b (10; 5, 0.3) = 4 c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5 Some Discrete Probability Distributions

5.50 (a) Using the negative binomial distribution, we get 6 ∗ (1/2)7 = 0.1172. b (7; 3, 1/2) = 2 (b) From the geometric distribution, we have g(4; 1/2) = (1/2)(1/2)3 = 1/16. 5.51 The probability that all coins turn up the same is 1/4. Using the geometric distribution with p = 3/4 and q = 1/4, we have P (X < 4) =

g(x; 3/4) =

x=1

(3/4)(1/4)x−1 =

x=1

63 . 64

5.52 From the negative binomial distribution, we obtain 7 ∗ (1/6)2 (5/6)6 = 0.0651. b (8; 2, 1/6) = 1 5.53 (a) P (X > 5) =

∞

p(x; 5) = 1 −

x=6

p(x; 5) = 0.3840.

x=0

(b) P (X = 0) = p(0; 5) = 0.0067. 5.54 (a) Using the geometric distribution, we have g(5; 2/3) = (2/3)(1/3)4 = 2/243. (b) Using the negative binomial distribution, we have 4 16 ∗ b (5; 3, 2/3) = (2/3)3 (1/3)2 = . 81 2 5.55 Using the geometric distribution (a) P (X = 3) = g(3; 0.7) = (0.7)(0.3)2 = 0.0630. 3 3 (b) P (X < 4) = g(x; 0.7) = (0.7)(0.3)x−1 = 0.9730. x=1

x=1

5.56 (a) Using the Poisson distribution with x = 5 and μ = 3, we ﬁnd from Table A.2 that p(5; 3) =

p(x; 3) −

x=0

p(x; 3) = 0.1008.

x=0

(b) P (X < 3) = P (X ≤ 2) = 0.4232. (c) P (X ≥ 2) = 1 − P (X ≤ 1) = 0.8009. 5.57 (a) P (X ≥ 4) = 1 − P (X ≤ 3) = 0.1429. (b) P (X = 0) = p(0; 2) = 0.1353. 5.58 (a) P (X < 4) = P (X ≤ 3) = 0.1512. (b) P (6 ≤ X ≤ 8) = P (X ≤ 8) − P (X ≤ 5) = 0.4015. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 5

5.59 (a) Using the negative distribution, we obtain 5 binomial ∗ 4 2 b (6; 4, 0.8) = 3 (0.8) (0.2) = 0.1638. (b) From the geometric distribution, we have g(3; 0.8) = (0.8)(0.2)2 = 0.032. 5.60 (a) Using the Poisson distribution with μ = 12, we ﬁnd from Table A.2 that P (X < 7) = P (X ≤ 6) = 0.0458. (b) Using the binomial distribution with p = 0.0458, we get 3 b(2; 3, 0.0458) = (0.0458)2 (0.9542) = 0.0060. 2 5.61 μ = np = (10000)(0.001) = 10, so P (6 ≤ X ≤ 8) = P (X ≤ 8) − P (X ≤ 5) ≈

p(x; 10) −

x=0

p(x; 10) = 0.2657.

x=0

5.62 (a) μ = np = (1875)(0.004) = 7.5, so P (X < 5) = P (X ≤ 4) ≈ 0.1321. (b) P (8 ≤ X ≤ 10) = P (X ≤ 10) − P (X ≤ 7) ≈ 0.8622 − 0.5246 = 0.3376. 5.63 μ = 6 and σ 2 = 6. 5.64 μ = (10000)(0.001) = 10 and σ 2 = 10. 5.65 (a) P (X ≤ 3|λt = 5) = 0.2650. (b) P (X > 1|λt = 5) = 1 − 0.0404 = 0.9596. 5.66 (a) P (X = 4|λt = 6) = 0.2851 − 0.1512 = 0.1339. (b) P (X ≥ 4|λt = 6) = 1 − 0.1512 = 0.8488. 74 p(x; 74) = 0.3773. (c) P (X ≥ 75|λt = 72) = 1 − x=0

5.67 (a) P (X > 10|λt = 14) = 1 − 0.1757 = 0.8243. (b) λt = 14. 5.68 μ = np = (1875)(0.004) = 7.5. 5.69 μ = (4000)(0.001) = 4. 5.70 μ = 1 and σ 2 = 0.99. 5.71 μ = λt = (1.5)(5) = 7.5 and P (X = 0|λt = 7.5) = e−7.5 = 5.53 × 10−4 . 5.72 (a) P (X ≤ 1|λt = 2) = 0.4060. (b) μ = λt = (2)(5) = 10 and P (X ≤ 4|λt = 10) = 0.0293. 5.73 (a) P (X > 10|λt = 5) = 1 − P (X ≤ 10|λt = 5) = 1 − 0.9863 = 0.0137. (b) μ = λt = (5)(3) = 15, so P (X > 20|λt = 15) = 1 − P (X ≤ 20|λ = 15) = 1 − 0.9170 = 0.0830. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5 Some Discrete Probability Distributions

5.74 p = 0.03 with a g(x; 0.03). So, P (X = 16) = (0.03)(1 − 0.03)14 = 0.0196 and 1 − 1 = 32.33. μ = 0.03 5.75 So, Let Y = number of shifts until it fails. Then Y follows a geometric distribution with p = 0.10. So, P (Y ≤ 6) = g(1; 0.1) + g(2; 0.1) + · · · + g(6; 0.1) = (0.1)[1 + (0.9) + (0.9)2 + · · · + (0.9)5 ] = 0.4686. 5.76 (a) The number of people interviewed before the ﬁrst refusal follows a geometric distribution with p = 0.2. So. P (X ≥ 51) =

∞

(0.2)(1 − 0.2)x = (0.2)

x=51

(1 − 0.2)50 = 0.00001, 1 − (1 − 0.2)

which is a very rare event. (b) μ =

1 0.2

− 1 = 4.

5.77 n = 15 and p = 0.05. (a) P (X ≥ 2) = 1 − P (X ≤ 1) = 1 −

1 15 x=0

(b) p = 0.07. So, P (X ≤ 1) =

1 15 x

x=0

(0.05)x (1 − 0.05)15−x = 1 − 0.8290 = 0.1710.

(0.07)x (1 − 0.07)15−x = 1 − 0.7168 = 0.2832.

5.78 n = 100 and p = 0.01. (a) P (X > 3) = 1 − P (X ≤ 3) = 1 −

3 100 x=0

(b) For p = 0.05, P (X ≤ 3) =

3 100 x

x=0

(0.01)x (1 − 0.01)100−x = 1 − 0.9816 = 0.0184.

(0.05)x (1 − 0.05)100−x = 0.2578.

5.79 Using the extension of the hypergeometric distribution, the probability is 57434 2

1 1 23

= 0.0308.

5.80 λ = 2.7 call/min. (a) P (X ≤ 4) =

4 x=0

(b) P (X ≤ 1) =

1 x=0

e−2.7 (2.7)x x!

= 0.8629.

e−2.7 (2.7)x x!

= 0.2487.

10 x=0

e−13.5 (13.5)x x!

= 1 − 0.2112 = 0.7888.

Solutions for Exercises in Chapter 5

(a) P (X = 5) =

15 5

(0.05)5 (1 − 0.05)10 = 0.000562.

(b) I would not believe the claim of 5% defective. 5.82 λ = 0.2, so λt = (0.2)(5) = 1. (a) P (X ≤ 1) =

1 x=0

e−1 (1)x x!

= 2e−1 = 0.7358. Hence, P (X > 1) = 1 − 0.7358 = 0.2642.

(b) λ = 0.25, so λt = 1.25. P (X|le1) =

1 x=0

5.83 (a) 1 − P (X ≤ 1) = 1 −

1 100 x

x=0

(b) P (X ≤ 1) =

1 100 x

x=0

e−1.25 (1.25)x x!

= 0.6446.

(0.01)x (0.99)100−x = 1 − 0.7358 = 0.2642.

(0.05)x (0.95)100−x = 0.0371.

5.84 (a) 100 visits/60 minutes with λt = 5 visits/3 minutes. P (X = 0) = (b) P (X > 5) = 1 −

5 x=0

e5 5x x!

e−5 50 0!

= 0.0067.

= 1 − 0.6160 = 0.3840.

5.85 (a) P (X ≥ 1) = 1 − P (X = 0) = 1 −

4 1 0 5 4

= 1 − 0.4822 = 0.5177. 0 6 6 24 1 0 35 24 = 1 − 0.5086 = 0.4914. (b) P (X ≥ 1) = 1 − P (X = 0) = 1 − 0 36 36

5.86 n = 5 and p = 0.4; P (X ≥ 3) = 1 − P (X ≤ 2) = 1 − 0.6826 = 0.3174. 5.87 (a) μ = bp = (200)(0.03) = 6. (b) σ 2 = npq = 5.82. 6

= 0.0025 (using the Poisson approximation). (c) P (X = 0) = e (6) 0! P (X = 0) = (0.97)200 = 0.0023 (using the binomial distribution). 5.88 (a) p10 q 0 = (0.99)10 = 0.9044. (b) In this case, the last 10 of 12 must be good starts and the 2nd attempt must be a bad one. However, the 1st one can be either bad or good. So, the probability is p10 q = (0.99)10 (0.01) = (0.9044)(0.01) = 0.009. 5.89 n = 75 with p = 0.999. (a) X = the number of trials, and P (X = 75) = (0.999)75 (0.001)0 = 0.9277. (b) Y = the number of trials before the ﬁrst failure (geometric distribution), and P (Y = 20) = (0.001)(0.999)19 = 0.000981. (c) 1 − P (no failures) = 1 − (0.001)0 (0.999)10 = 0.01. 9 9 5.90 (a) 10 1 pq = (10)(0.25)(0.75) = 0.1877. (b) Let X be the number of drills until the ﬁrst success. X follows a geometric distribution with p = 0.25. So, the probability of having the ﬁrst 10 drills being failure is q 10 = (0.75)10 = 0.056. So, there is a small prospects for bankruptcy. Also, the probability that the ﬁrst success appears in the 11th drill is pq 10 = 0.014 which is even smaller. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 5 Some Discrete Probability Distributions

64 5.91 It is a negative binomial distribution.

x−1

5.92 It is a negative binomial distribution.

x−1

k−1 k−1

pk q x−k =

6−1

pk q x−k =

4−1

2−1 2−1

(0.25)2 (0.75)4 = 0.0989. (0.5)2 (0.5)2 = 0.1875.

5.93 n = 1000 and p = 0.01, with μ = (1000)(0.01) = 10. P (X < 7) = P (X ≤ 6) = 0.1301. 5.94 n = 500; (a) If p = 0.01, P (X ≥ 15) = 1 − P (X ≤ 14) = 1 −

14 500 x=0

(0.01)x (0.99)500−x = 0.00021.

This is a very rare probability and thus the original claim that p = 0.01 is questionable. 3 497 = 0.1402. (b) P (X = 3) = 500 3 (0.01) (0.99) (c) For (a), if p = 0.01, μ = (500)(0.01) = 5. So P (X ≥ 15) = 1 − P (X ≤ 14) = 1 − 0.9998 = 0.0002. For (b), P (X = 3) = 0.2650 − 0.1247 = 0.1403. 5.95 N = 50 and n = 10. (20)(48 10) = 1 − 0.6367 = 0.3633. 50 (10) (b) Even though the lot contains 2 defectives, the probability of reject the lot is not very high. Perhaps more items should be sampled. (a) k = 2; P (X ≥ 1) = 1 − P (X = 0) = 1 −

(20)(48 n) =1− (50 ) n

(50−n)(49−n) (50)(49)

≥ 0.9. The

5.97 Deﬁne X = number of screens will detect. Then X ∼ b(x; 3, 0.8). (a) P (X = 0) = (1 − 0.8)3 = 0.008. (b) P (X = 1) = (3)(0.2)2 (0.8) = 0.096. (c) P (X ≥ 2) = P (X = 2) + P (X = 3) = (3)(0.8)2 (0.2) + (0.8)3 = 0.896. 5.98 (a) P (X = 0) = (1 − 0.8)n ≤ 0.0001 implies that n ≥ 6. (b) (1 − p)3 ≤ 0.0001 implies p ≥ 0.9536. 5.99 n = 10 and p =

2 50

= 0.04.

10 P (X ≥ 1) = 1 − P (X = 0) ≈ 1 − (0.04)0 (1 − 0.04)10 = 1 − 0.6648 = 0.3351. 0 The approximation is not that good due to

n N

= 0.2 is too large.

Solutions for Exercises in Chapter 5

(22)(30) = 0.1. (52) (2)(1) (b) P = 1 5 1 = 0.2. (2)

5.100 (a) P =

5.101 n = 200 with p = 0.00001. (a) P (X ≥ 5) = 1 − P (X ≤ 4) = 1 −

4 200 x=0

(0.0001)x (1 − 0.0001)200−x ≈ 0. This is a rare

event. Therefore, the claim does not seem right. (b) μ = np = (200)(0.0001) = 0.02. Using Poisson approximation, P (X ≥ 5) = 1 − P (X ≤ 4) ≈ 1 −

4 x=0

e−0.02

(0.02)x = 0. x!

Chapter 6

Some Continuous Probability Distributions 6.1 f (x) =

1 B−A

(a) μ =

for A ≤ x ≤ B.

x A B−A

(b) E(X 2 ) = So, σ 2 =

dx =

B 2 −A2 2(B−A)

A+B 2 .

B 3 −A3 x2 A B−A dx = 3(B−A) . A+B 2 4(B 2 +AB+A2 )−3(B 2 +2AB+A2 ) B 3 −A3 = 2 12 3(B−A) −

6.2 P (X > 2.5|X ≤ 4) =

P (2.5 7) =

10−8.5 3

10−7 10

(b) P (2 < X < 7) =

= 0.70.

= 0.3. 7−2 10

= 0.5.

6.5 (a) Area=0.0823. (b) Area=1 − 0.9750 = 0.0250. (c) Area=0.2578 − 0.0154 = 0.2424. (d) Area=0.9236. (e) Area=1 − 0.1867 = 0.8133. (f) Area=0.9591 − 0.3156 = 0.6435. 6.6 (a) The area to the left of z is 1 − 0.3622 = 0.6378 which is closer to the tabled value 0.6368 than to 0.6406. Therefore, we choose z = 0.35. (b) From Table A.3, z = −1.21. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 67

Chapter 6 Some Continuous Probability Distributions

(c) The total area to the left of z is 0.5000+0.4838=0.9838. Therefore, from Table A.3, z = 2.14. (d) The distribution contains an area of 0.025 to the left of −z and therefore a total area of 0.025+0.95=0.975 to the left of z. From Table A.3, z = 1.96. 6.7 (a) Since P (Z > k) = 0.2946, then P (Z < k) = 0.7054/ From Table A.3, we ﬁnd k = 0.54. (b) From Table A.3, k = −1.72. (c) The area to the left of z = −0.93 is found from Table A.3 to be 0.1762. Therefore, the total area to the left of k is 0.1762+0.7235=0.8997, and hence k = 1.28. 6.8 (a) z = (17 − 30)/6 = −2.17. Area=1 − 0.0150 = 0.9850. (b) z = (22 − 30)/6 = −1.33. Area=0.0918. (c) z1 = (32 − 30)/6 = 0.33, z2 = (41 − 30)/6 = 1.83. Area = 0.9664 − 0.6293 = 0.3371. (d) z = 0.84. Therefore, x = 30 + (6)(0.84) = 35.04. (e) z1 = −1.15, z2 = 1.15. Therefore, x1 = 30+(6)(−1.15) = 23.1 and x2 = 30+(6)(1.15) = 36.9. 6.9 (a) z = (15 − 18)/2.5 = −1.2; P (X < 15) = P (Z < −1.2) = 0.1151. (b) z = −0.76, k = (2.5)(−0.76) + 18 = 16.1. (c) z = 0.91, k = (2.5)(0.91) + 18 = 20.275. (d) z1 = (17 − 18)/2.5 = −0.4, z2 = (21 − 18)/2.5 = 1.2; P (17 < X < 21) = P (−0.4 < Z < 1.2) = 0.8849 − 0.3446 = 0.5403. 6.10 z1 = [(μ − 3σ) − μ]/σ = −3, z2 = [(μ + 3σ) − μ]/σ = 3; P (μ − 3σ < Z < μ + 3σ) = P (−3 < Z < 3) = 0.9987 − 0.0013 = 0.9974. 6.11 (a) z = (224 − 200)/15 = 1.6. Fraction of the cups containing more than 224 millimeters is P (Z > 1.6) = 0.0548. (b) z1 = (191 − 200)/15 = −0.6, Z2 = (209 − 200)/15 = 0.6; P (191 < X < 209) = P (−0.6 < Z < 0.6) = 0.7257 − 0.2743 = 0.4514. (c) z = (230−200)/15 = 2.0; P (X > 230) = P (Z > 2.0) = 0.0228. Therefore, (1000)(0.0228) = 22.8 or approximately 23 cups will overﬂow. (d) z = −0.67, x = (15)(−0.67) + 200 = 189.95 millimeters. 6.12 (a) z = (31.7 − 30)/2 = 0.85; P (X > 31.7) = P (Z > 0.85) = 0.1977. Therefore, 19.77% of the loaves are longer than 31.7 centimeters. (b) z1 = (29.3 − 30)/2 = −0.35, z2 = (33.5 − 30)/2 = 1.75; P (29.3 < X < 33.5) = P (−0.35 < Z < 1.75) = 0.9599 − 0.3632 = 0.5967. Therefore, 59.67% of the loaves are between 29.3 and 33.5 centimeters in length. (c) z = (25.5 − 30)/2 = −2.25; P (X < 25.5) = P (Z < −2.25) = 0.0122. Therefore, 1.22% of the loaves are shorter than 25.5 centimeters in length. 6.13 (a) z = (32 − 40)/6.3 = −1.27; P (X > 32) = P (Z > −1.27) = 1 − 0.1020 = 0.8980. (b) z = (28 − 40)/6.3 = −1.90, P (X < 28) = P (Z < −1.90) = 0.0287. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 6

(c) z1 = (37 − 40)/6.3 = −0.48, z2 = (49 − 40)/6.3 = 1.43; So, P (37 < X < 49) = P (−0.48 < Z < 1.43) = 0.9236 − 0.3156 = 0.6080. 6.14 (a) z = (10.075 − 10.000)/0.03 = 2.5; P (X > 10.075) = P (Z > 2.5) = 0.0062. Therefore, 0.62% of the rings have inside diameters exceeding 10.075 cm. (b) z1 = (9.97 − 10)/0.03 = −1.0, z2 = (10.03 − 10)/0.03 = 1.0; P (9.97 < X < 10.03) = P (−1.0 < Z < 1.0) = 0.8413 − 0.1587 = 0.6826. (c) z = −1.04, x = 10 + (0.03)(−1.04) = 9.969 cm. 6.15 (a) z = (30 − 24)/3.8 = 1.58; P (X > 30) = P (Z > 1.58) = 0.0571. (b) z = (15 − 24)/3.8 = −2.37; P (X > 15) = P (Z > −2.37) = 0.9911. He is late 99.11% of the time. (c) z = (25 − 24)/3.8 = 0.26; P (X > 25) = P (Z > 0.26) = 0.3974. (d) z = 1.04, x = (3.8)(1.04) + 24 = 27.952 minutes. (e) Using the binomial distribution with p = 0.0571, we get b(2; 3, 0.0571) = 32 (0.0571)2 (0.9429) = 0.0092. 6.16 μ = 99.61 and σ = 0.08. (a) P (99.5 < X < 99.7) = P (−1.375 < Z < 1.125) = 0.8697 − 0.08455 = 0.7852. (b) P (Z > −1.645) = 0.05; x = (−1.645)(0.08) + 99.61 = 99.4784. 6.17 z = −1.88, x = (2)(−1.88) + 10 = 6.24 years. 6.18 (a) z = (159.75−174.5)/6.9 = −2.14; P (X < 159.75) = P (Z < −2.14) = 0.0162. Therefore, (1000)(0.0162) = 16 students. (b) z1 = (171.25 − 174.5)/6.9 = −0.47, z2 = (182.25 − 174.5)/6.9 = 1.12. P (171.25 < X < 182.25) = P (−0.47 < Z < 1.12) = 0.8686 − 0.3192 = 0.5494. Therefore, (1000)(0.5494) = 549 students. (c) z1 = (174.75 − 174.5)/6.9 = 0.04, z2 = (175.25 − 174.5)/6.9 = 0.11. P (174.75 < X < 175.25) = P (0.04 < Z < 0.11) = 0.5438 − 0.5160 = 0.0278. Therefore, (1000)(0.0278)=28 students. (d) z = (187.75 − 174.5)/6.9 = 1.92; P (X > 187.75) = P (Z > 1.92) = 0.0274. Therefore, (1000)(0.0274) = 27 students. 6.19 μ = $15.90 and σ = $1.50. (a) 51%, since P (13.75 < X < 16.22) = P 13.745−15.9 1.645) = 0.05; x = (1.645)(1.50) + 15.90 + 0.005 = 18.37. 6.20 (a) z = (9.55 − 8)/0.9 = 1.72. Fraction of poodles weighing over 9.5 kilograms = P (X > 9.55) = P (Z > 1.72) = 0.0427. (b) z = (8.65−8)/0.9 = 0.72. Fraction of poodles weighing at most 8.6 kilograms = P (X < 8.65) = P (Z < 0.72) = 0.7642. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 6 Some Continuous Probability Distributions

(c) z1 = (7.25 − 8)/0.9 = −0.83 and z2 = (9.15 − 8)/0.9 = 1.28. Fraction of poodles weighing between 7.3 and 9.1 kilograms inclusive = P (7.25 < X < 9.15) = P (−0.83 < Z < 1.28) = 0.8997 − 0.2033 = 0.6964. 6.21 (a) z = (10, 175−10, 000)/100 = 1.75. Proportion of components exceeding 10.150 kilograms in tensile strength= P (X > 10, 175) = P (Z > 1.75) = 0.0401. (b) z1 = (9, 775 − 10, 000)/100 = −2.25 and z2 = (10, 225 − 10, 000)/100 = 2.25. Proportion of components scrapped= P (X < 9, 775) + P (X > 10, 225) = P (Z < −2.25) + P (Z > 2.25) = 2P (Z < −2.25) = 0.0244. 6.22 (a) x1 = μ+1.3σ and x2 = μ−1.3σ. Then z1 = 1.3 and z2 = −1.3. P (X > μ+1.3σ)+P (X < 1.3σ) = P (Z > 1.3) + P (Z < −1.3) = 2P (Z < −1.3) = 0.1936. Therefore, 19.36%. (b) x1 = μ + 0.52σ and x2 = μ − 0.52σ. Then z1 = 0.52 and z2 = −0.52. P (μ − 0.52σ < X < μ + 0.52σ) = P (−0.52 < Z < 0.52) = 0.6985 − 0.3015 = 0.3970. Therefore, 39.70%. 6.23 z = (94.5 − 115)/12 = −1.71; P (X < 94.5) = P (Z < −1.71) = 0.0436. (0.0436)(600) = 26 students will be rejected. √ 6.24 μ = np = (400)(1/2) = 200, σ = npq = (400)(1/2)(1/2) = 10.

Therefore,

(a) z1 = (184.5 − 200)/10 = −1.55 and z2 = (210.5 − 200)/10 = 1.05. P (184.5 < X < 210.5) = P (−1.55 < Z < 1.05) = 0.8531 − 0.0606 = 0.7925. (b) z1 = (204.5 − 200)/10 = 0.45 and z2 = (205.5 − 200)/10 = 0.55. P (204.5 < X < 205.5) = P (0.45 < Z < 0.55) = 0.7088 − 0.6736 = 0.0352. (c) z1 = (175.5 − 200)/10 = −2.45 and z2 = (227.5 − 200)/10 = 2.75. P (X < 175.5) + P (X > 227.5) = P (Z < −2.45) + P (Z > 2.75) = P (Z < −2.45) + 1 − P (Z < 2.75) = 0.0071 + 1 − 0.9970 = 0.0101. 6.25 n = 100. (a) p = 0.01 with μ = (100)(0.01) = 1 and σ = (100)(0.01)(0.99) = 0.995. So, z = (0.5 − 1)/0.995 = −0.503. P (X ≤ 0) ≈ P (Z ≤ −0.503) = 0.3085. (b) p = 0.05 with μ = (100)(0.05) = 5 and σ = (100)(0.05)(0.95) = 2.1794. So, z = (0.5 − 5)/2.1794 = −2.06. P (X ≤ 0) ≈ P (X ≤ −2.06) = 0.0197. 6.26 μ = np = (100)(0.1) = 10 and σ = (100)(0.1)(0.9) = 3. (a) z = (13.5 − 10)/3 = 1.17; P (X > 13.5) = P (Z > 1.17) = 0.1210. (b) z = (7.5 − 10)/3 = −0.83; P (X < 7.5) = P (Z < −0.83) = 0.2033. 6.27 μ = (100)(0.9) = 90 and σ = (100)(0.9)(0.1) = 3. (a) z1 = (83.5 − 90)/3 = −2.17 and z2 = (95.5 − 90)/3 = 1.83. P (83.5 < X < 95.5) = P (−2.17 < Z < 1.83) = 0.9664 − 0.0150 = 0.9514. (b) z = (85.5 − 90)/3 = −1.50; P (X < 85.5) = P (Z < −1.50) = 0.0668. 6.28 μ = (80)(3/4) = 60 and σ = (80)(3/4)(1/4) = 3.873. (a) z = (49.5 − 60)/3.873 = −2.71; P (X > 49.5) = P (Z > −2.71) = 1 − 0.0034 = 0.9966. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 6

(b) z = (56.5 − 60)/3.873 = −0.90; P (X < 56.5) = P (Z < −0.90) = 0.1841. 6.29 μ = (1000)(0.2) = 200 and σ = (1000)(0.2)(0.8) = 12.649. (a) z1 = (169.5 − 200)/12.649 = −2.41 and z2 = (185.5 − 200)/12.649 = −1.15. P (169.5 < X < 185.5) = P (−2.41 < Z < −1.15) = 0.1251 − 0.0080 = 0.1171.

6.30

6.31 6.32 6.33

(b) z1 = (209.5 − 200)/12.649 = 0.75 and z2 = (225.5 − 200)/12.649 = 2.02. P (209.5 < X < 225.5) = P (0.75 < Z < 2.02) = 0.9783 − 0.7734 = 0.2049. (a) μ = (100)(0.8) = 80 and σ = (100)(0.8)(0.2) = 4 with z = (74.5 − 80)/4 = −1.38. P (Claim is rejected when p = 0.8) = P (Z < −1.38) = 0.0838. (b) μ = (100)(0.7) = 70 and σ = (100)(0.7)(0.3) = 4.583 with z = (74.5 − 70)/4.583 = 0.98. P (Claim is accepted when p = 0.7) = P (Z > 0.98) = 1 − 0.8365 = 0.1635. μ = (180)(1/6) = 30 and σ = (180)(1/6)(5/6) = 5 with z = (35.5 − 30)/5 = 1.1. P (X > 35.5) = P (Z > 1.1) = 1 − 0.8643 = 0.1357. μ = (200)(0.05) = 10 and σ = (200)(0.05)(0.95) = 3.082 with z = (9.5 − 10)/3.082 = −0.16. P (X < 10) = P (Z < −0.16) = 0.4364. μ = (400)(1/10) = 40 and σ = (400)(1/10)(9/10) = 6. (a) z = (31.5 − 40)/6 = −1.42; P (X < 31.5) = P (Z < −1.42) = 0.0778. (b) z = (49.5 − 40)/6 = 1.58; P (X > 49.5) = P (Z > 1.58) = 1 − 0.9429 = 0.0571.

(c) z1 = (34.5 − 40)/6 = −0.92 and z2 = (46.5 − 40)/6 = 1.08; P (34.5 < X < 46.5) = P (−0.92 < Z < 1.08) = 0.8599 − 0.1788 = 0.6811. 6.34 μ = (180)(1/6) = 30 and σ = (180)(1/6)(5/6) = 5. (a) z = (24.5 − 30)/5 = −1.1; P (X > 24.5) = P (Z > −1.1) = 1 − 0.1357 = 0.8643. (b) z1 = (32.5 − 30)/5 = 0.5 and z2 = (41.5 − 30)/5 = 2.3. P (32.5 < X < 41.5) = P (0.5 < Z < 2.3) = 0.9893 − 0.6915 = 0.2978. (c) z1 = (29.5 − 30)/5 = −0.1 and z2 = (30.5 − 30)/5 = 0.1. P (29.5 < X < 30.5) = P (−0.1 < Z < 0.1) = 0.5398 − 0.4602 = 0.0796. 6.35 (a) p = 0.05, n = 100 with μ = 5 and σ = (100)(0.05)(0.95) = 2.1794. So, z = (2.5 − 5)/2.1794 = −1.147; P (X ≥ 2) ≈ P (Z ≥ −1.147) = 0.8749. (b) z = (10.5 − 5)/2.1794 = 2.524; P (X ≥ 10) ≈ P (Z > 2.52) = 0.0059. 6.36 n = 200; X = The number of no shows with p = 0.02. z = √

3−0.5−4 (200)(0.02)(0.98)

= −0.76.

Therefore, P (airline overbooks the ﬂight) = 1 − P (X ≥ 3) ≈ 1 − P (Z > −0.76) = 0.2236. = 0.0228. 6.37 (a) P (X ≥ 230) = P Z > 230−170 30 (b) Denote by Y the number of students whose serum cholesterol level exceed 230 among the 300. Then Y ∼ b(y; 300, 0.0228 with μ = (300)(0.0228) = 6.84 and σ = (300)(0.0228)(1 − 0.0228) = = 0.26 and 2.5854. So, z = 8−0.5−6.84 2.5854 P (X ≥ 8) ≈ P (Z > 0.26) = 0.3974. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 6 Some Continuous Probability Distributions

6.38 (a) Denote by X the number of failures among ∼ b(x; 0.01) and19P (X > 1) = 2020, 20 the020. X 20 1 − b(0; 20, 0.01) − b(1; 20, 0.01) = 1 − 0 (0.01) (0.99) − 1 (0.01)(0.99) = 0.01686. (b) n = 500 and p = 0.01 with μ = (500)(0.01) = 5 and σ = (500)(0.01)(0.99) = 2.2249. So, P (more than 8 failures) ≈ P (Z > (8.5 − 5)/2.2249) = P (Z > 1.57) = 1 − 0.9418 = 0.0582. 6.39 Setting α = 1/2 in the gamma distribution and integrating, we have ∞ 1 √ x−1/2 e−x/β dx = 1. βΓ(1/2) 0 Substitute x = y 2 /2, dx = y dy, to give √ ∞ ∞ √ √ 2 1 −y 2 /2β −y 2 /2β Γ(1/2) = √ e dy = 2 π √ √ e dy = π, β 0 2π β 0 since the √ quantity in parentheses represents one-half of the area under the normal curve n(y; 0, β). ∞

∞ 6.40 P (X > 9) = 19 9 xe−x/3 dx = − x3 e−x/3 − e−x/3 9 = 4e−3 = 0.1992.

2.4 2.4 6.41 P (1.8 < X < 2.4) = 1.8 xe−x dx = [−xe−x − e−x ]|1.8 = 2.8e−1.8 − 3.4e−2.4 = 0.1545. 1

1 6.42 (a) P (X < 1) = 4 0 xe−2x dx = −2xe−2x − e−2x 0 = 1 − 3e−2 = 0.5940. ∞

∞ −2x (b) P (X > 2) = 4 xe dx = −2xe−2x − e−2x = 5e−4 = 0.0916. 0

6.43 (a) μ = αβ = (2)(3) = 6 million liters; σ 2 = αβ 2 = (2)(9) = 18. (b) Water consumption on √ any given day has a probability of at least 3/4 of falling in the interval μ ± 2σ = 6 ± 2 18 or from −2.485 to 14.485. That is from 0 to 14.485 million liters. 6.44 (a) μ = αβ = 6 and σ 2 = αβ 2 = 12. Substituting α = 6/β into the variance formula we ﬁnd 6β = 12 or β = 2 and then α = 3.

∞ 2 −x/2 1 (b) P (X > 12) = 16 dx. Integrating by parts twice gives 12 x e P (X > 12) =

∞ 1 −2x2 e−x/2 − 8xe−x/2 − 16e−x/2 = 25e−6 = 0.0620. 16 12

3 6.45 P (X < 3) = 14 0 e−x/4 dx = −e−x/4 0 = 1 − e−3/4 = 0.5276. Let Y be the number of days a person is served in less than 3 minutes. Then 6 b(y; 6, 1 − e−3/4 ) = 64 (0.5276)4 (0.4724)2 + 65 (0.5276)5 (0.4724) P (Y ≥ 4) = x=4 + 66 (0.5276)6 = 0.3968. 1

1 6.46 P (X < 1) = 12 0 e−x/2 dx = −e−x/2 0 = 1−e−1/2 = 0.3935. Let Y be the number of switches that fail during the ﬁrst year. Using the normal approximation we ﬁnd μ = (100)(0.3935) = 39.35, σ = (100)(0.3935)(0.6065) = 4.885, and z = (30.5−39.35)/4.885 = −1.81. Therefore, P (Y ≤ 30) = P (Z < −1.81) = 0.0352. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 6

∞

2 ∞ dx = 6.47 (a) E(X) = 0 + 0 e−x /2 dx 0 √ √

∞ 2 = 0 + 2π · √12π 0 e−x /2 dx = 22π = π2 = 1.2533. ∞

∞ 2 2 (b) P (X > 2) = 2 xe−x /2 dx = −e−x /2 = e−2 = 0.1353.

∞

2 −xe−x /2

2 x2 e−x /2

6.48 A beta density is given by f (x) = B(α + k, β) E(X ) = B(α, β)

For k = 1, μ =

B(α+1,β) B(α,β)

α α+β ,

xα−1 (1−x)β−1 B(α,β)

, for 0 < x < 1. Hence,

B(α + k, β) Γ(α + k)Γ(β) xα+k−1 (1 − x)β−1 = = . B(α + k, β) B(α, β) Γ(α)Γ(α + β + k) and for k = 2, E(X 2 ) =

V ar(X) =

α(α+1) (α+β)(α+β+1) .

αβ (α +

β)2 (α

+ β + 1)

Hence,

6.49 The density function is f (x) = 3(1 − x)2 , for 0 < x < 1. 1 = 14 . To compute median, notice the c.d.f. is F (x) = 1 − (1 − x)3 , for 0 < x < 1. (a) μ = 1+3 Hence, solving 1 − (1 − m)3 = 12 we obtain m = 0.206.

(b) σ 2 = (1+3)(1)(3) 2 (1+3+1) = 0.0375. (c) P X > 13 = (1 − 1/3)3 = 0.2963. 6.50 The density function is f (x) = 12x2 (1 − x), for 0 < x < 1. Hence, 1 P (X > 0.8) = 12 x2 (1 − x)dx = 0.1808. 0.8

6.51 R(t) = ce−

√ 1/ t dt

√

= ce−2 t . However, R(0) = 1 and hence c = 1. Now √ √ f (t) = Z(t)R(t) = e−2 t / t,

and

∞

P (T > 4) =

t > 0,

√ √ √ ∞ e−2 t / t dt = −e−2 t = e−4 = 0.0183. 4

∞ β 6.52 μ = E(T ) = αβ 0 tβ e−αt dt. Let y = αtβ , then dy = αβtβ−1 dt and t = (y/α)1/β . Then ∞ ∞ 1/β −y −1/β (y/α) e dy = α y (1+1/β)−1 e−y dy = α−1/β Γ(1 + 1/β). μ= 0

E(T 2 ) = αβ =α

∞

0 −2/β

tβ+1 e−αt dt = β

∞

(y/α)2/β e−y dy = α−2/β

∞

y (1+2/β)−1 e−y dy

Γ(1 + 2/β).

Chapter 6 Some Continuous Probability Distributions

74 6.53 α = 5; β = 10; (a) αβ = 50.

√ (b) σ 2 = αβ 2 = 500; so σ = 500 = 22.36.

∞ α−1 −x/β 1 e dx. Using the incomplete gamma with y = x/β, (c) P (X > 30) = β α Γ(α) 30 x then 3 4 −y y e dy = 1 − 0.185 = 0.815. 1 − P (X ≤ 30) = 1 − P (Y ≤ 3) = 1 − 0 Γ(5) 6.54 αβ = 10; σ =

αβ 2 =

√

50 = 7.07.

(a) Using integration by parts, 1 P (X ≤ 50) = α β Γ(α) (b) P (X < 10) = have

1 β α Γ(α)

10 0

α−1 −x/β

1 dx = 25

xe−x/5 dx = 0.9995.

xα−1 e−x/β dx. Using the incomplete gamma with y = x/β, we

P (X < 10) = P (Y < 2) =

ye−y dy = 0.5940.

6.55 μ = 3 seconds with f (x) = 13 e−x/3 for x > 0. ∞

∞ (a) P (X > 5) = 5 13 e−x/3 dx = 13 −3e−x/3 5 = e−5/3 = 0.1889. (b) P (X > 10) = e−10/3 = 0.0357. = 1 − Φ(0.7992) = 0.2119. 6.56 P (X > 270) = 1 − Φ ln 270−4 2 6.57 μ = E(X) = e4+4/2 = e6 ; σ 2 = e8+4 (e4 − 1) = e12 (e4 − 1). 6.58 β = 1/5 and α = 10. (a) P (X > 10) = 1 − P (X ≤ 10) = 1 − 0.9863 = 0.0137. (b) P (X > 2) before 10 cars arrive.

P (X ≤ 2) = 0

1 xα−1 e−x/β dx. βα Γ(α)

Given y = x/β, then

P (X ≤ 2) = P (Y ≤ 10) = 0

y α−1 e−y dy = Γ(α)

10 0

y 10−1 e−y dy = 0.542, Γ(10)

with P (X > 2) = 1 − P (X ≤ 2) = 1 − 0.542 = 0.458. 6.59 Let T be the time between two consecutive arrivals (a) P (T > 1) = P (no arrivals in 1 minute) = P (X = 0) = e−5 = 0.0067. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 6

(b) μ = β = 1/5 = 0.2. 6.60 Assume that Z(t) = αβtβ−1 , for t > 0. Then we can write f (t) = Z(t)R(t), where R(t) = − Z(t) dt = ce− αβtβ−1 dt = ce−αtβ . From the condition that R(0) = 1, we ﬁnd that c = 1. ce β β Hence R(t) = eαt and f (t) = αβtβ−1 e−αt , for t > 0. Since Z(t) =

f (t) , R(t)

where

R(t) = 1 − F (t) = 1 −

αβx

β−1 −αxβ

dx = 1 +

de−αx = e−αt , β

then αβtβ−1 e−αt = αβtβ−1 , e−αtβ β

Z(t) =

6.61 μ = np = (1000)(0.49) = 490, σ = z1 =

√

npq =

t > 0.

(1000)(0.49)(0.51) = 15.808.

481.5 − 490 = −0.54, 15.808

z2 =

510.5 − 490 = 1.3. 15.808

P (481.5 < X < 510.5) = P (−0.54 < Z < 1.3) = 0.9032 − 0.2946 = 0.6086. ∞

∞ 6.62 P (X > 1/4) = 1/4 6e−6x dx = −e−6x 1/4 = e−1.5 = 0.223. 6.63 P (X < 1/2) = 108

1/2 0

x2 e−6x dx. Letting y = 6x and using Table A.24 we have

P (X < 1/2) = P (Y < 3) =

y 2 e−y dy = 0.577.

6.64 Manufacturer A:

P (X ≥ 10000) = P Manufacturer B:

100000 − 14000 Z≥ 2000

P (X ≥ 10000) = P

Z≥

10000 − 13000 1000

= P (Z ≥ −2) = 0.9772.

= P (Z ≥ −3) = 0.9987.

Manufacturer B will produce the fewest number of defective rivets. 6.65 Using the normal approximation to the binomial with μ = np = 650 and σ = So,

√

npq = 15.0831.

P (590 ≤ X ≤ 625) = P (−10.64 < Z < −8.92) ≈ 0. 6.66 (a) μ = β = 100 hours. (b) P (X ≥ 200) = 0.01

∞

200 e

−0.01x

dx = e−2 = 0.1353.

Chapter 6 Some Continuous Probability Distributions

6.67 (a) μ = 85 and σ = 4. So, P (X < 80) = P (Z < −1.25) = 0.1056. (b) μ = 79 and σ = 4. So, P (X ≥ 80) = P (Z > 0.25) = 0.4013. 6.68 1/β = 1/5 hours with α = 2 failures and β = 5 hours. (a) αβ = (2)(5) = 10.

∞ 1 xe−x/5 dx = (b) P (X ≥ 12) = 12 52 Γ(2) = 0.3084.

1 25

∞ 12

∞ xe−x/5 dx = − x5 e−x/5 − e−x/5 12

6.69 Denote by X the elongation. We have μ = 0.05 and σ = 0.01. (a) P (X ≥ 0.1) = P Z ≥ 0.1−0.05 = P (Z ≥ 5) ≈ 0. 0.01 (b) P (X ≤ 0.04) = P Z ≤ 0.04−0.05 = P (Z ≤ −1) = 0.1587. 0.01 (c) P (0.025 ≤ X ≤ 0.065) = P (−2.5 ≤ Z ≤ 1.5) = 0.9332 − 0.0062 = 0.9270. 6.70 Let X be the error. X ∼ n(x; 0, 4). So, P (fails) = 1 − P (−10 < X < 10) = 1 − P (−2.5 < Z < 2.5) = 2(0.0062) = 0.0124. 6.71 Let X be the time to bombing with μ = 3 and σ = 0.5. Then 4−3 1−3 ≤Z≤ = P (−4 ≤ Z ≤ 2) = 0.9772. P (1 ≤ X ≤ 4) = P 0.5 0.5 P (of an undesirable product) is 1 − 0.9772 = 0.0228. Hence a product is undesirable is 2.28% of the time.

200 6.72 α = 2 and β = 100. P (X ≤ 200) = β12 0 xe−x/β dx. Using the incomplete gamma table

2 and let y = x/β, 0 ye−y dy = 0.594. 6.73 μ = αβ = 200 hours and σ 2 = αβ 2 = 20, 000 hours. 6.74 X follows a lognormal distribution. ln 50, 000 − 5 = 1 − Φ(2.9099) = 1 − 0.9982 = 0.0018. P (X ≥ 50, 000) = 1 − Φ 2 6.75 The mean of X, which follows a lognormal distribution is μ = E(X) = eμ+σ √ 6.76 μ = 10 and σ = 50.

2 /2

= e7 .

(a) P (X ≤ 50) = P (Z ≤ 5.66) ≈ 1. (b) P (X ≤ 10) = 0.5. (c) For the larger values, such as 50, both results are similar. However, for smaller values such as 10, the normal population will give you smaller probabilities. 1

1 6.77 (a) Since f (y) ≥ 0 and 0 10(1 − y)9 dy = − (1 − y)10 0 = 1, it is a density function. 1 (b) P (Y > 0.6) = − (1 − y)10 0.6 = (0.4)10 = 0.0001. (c) α = 1 and β = 10. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 6

(d) μ =

α α+β

(e) σ 2 = 6.78 (a) μ =

1 11

= 0.0909.

αβ (α+β)2 (α+β+1)

1 10

∞ 0

(1)(10) (1+10)2 (1+10+1)

= 0.006887.

∞ ∞ ze−z/10 dz = − ze−z/10 0 + 0 e−z/10 dz = 10.

(b) Using integral by parts twice, we get ∞ 1 2 E(Z ) = z 2 e−z/10 dz = 200. 10 0 So, σ 2 = E(Z 2 ) − μ2 = 200 − (10)2 = 100. ∞ (c) P (Z > 10) = − ez/10 10 = e−1 = 0.3679. 6.79 The problem has changed to a Poisson distribution: the number of calls per hour. Since the average time between two calls in 6.76 is 10 minutes, the average number of calls per hour should be 6. Therefore, the mean and variance of the number of calls per hour should all be 6. 6.80 μ = 0.5 seconds and σ = 0.4 seconds. = P (Z > −0.5) = 0.6915. (a) P (X > 0.3) = P Z > 0.3−0.5 0.4 (b) P (Z > −1.645) = 0.95. So, −1.645 = x−0.5 0.4 yields x = −0.158 seconds. The negative number in reaction time is not reasonable. So, it means that the normal model may not be accurate enough. 6.81 (a) For an exponential distribution with parameter β, P (X > a + b | X > a) =

P (X > a + b) e−a−b = −a = e−b = P (X > b). P (X > a) e

So, P (it will breakdown in the next 21 days | it just broke down) = P (X > 21) = e−21/15 = e−1.4 = 0.2466. (b) P (X > 30) = e−30/15 = e−2 = 0.1353. 6.82 α = 2 and β = 50. So,

P (X ≤ 10) = 100

x49 e−2x

dx.

Let y = 2x50 with x = (y/2)1/50 and dx = 100 P (X ≤ 10) = 1/50 2 (50)

(2)1050 0

1 y −49/50 21/50 (50)

y 49/50 2

−49/50 −y

dy.

(2)1050

dy =

e−y dy ≈ 1.

6.83 The density function of a Weibull distribution is f (y) = αβy β−1 e−αy , β

y > 0.

Chapter 6 Some Continuous Probability Distributions

78 So, for any y ≥ 0,

f (t) dt = αβ

F (y) = 0

tβ−1 e−αt dt. β

Let z = tβ which yields t = z 1/β and dt = β1 z 1/β−1 dz. Hence,

yβ

F (y) = αβ 0

1 z 1−1/β z 1/β−1 e−αz dz = α β

yβ

e−αz dz = 1 − e−αy . β

On the other hand, since de−αy = −αβy β−1 e−αy , the above result follows immediately. β

6.84 One of the basic assumptions for the exponential distribution centers around the “lack-ofmemory” property for the associated Poisson distribution. Thus the drill bit of problem 6.80 is assumed to have no punishment through wear if the exponential distribution applies. A drill bit is a mechanical part that certainly will have signiﬁcant wear over time. Hence the exponential distribution would not apply. 6.85 The chi-squared distribution is a special case of the gamma distribution when α = v/2 and β = 2, where v is the degrees of the freedom of the chi-squared distribution. So, the mean of the chi-squared distribution, using the property from the gamma distribution, is μ = αβ = (v/2)(2) = v, and the variance of the chi-squared distribution is σ 2 = αβ 2 = (v/2)(2)2 = 2v. 6.86 Let X be the length of time in seconds. Then Y = ln(X) follows a normal distribution with μ = 1.8 and σ = 2. (a) P (X > 20) = P (Y > ln 20) = P (Z > (ln 20 − 1.8)/2) = P (Z > 0.60) = 0.2743. P (X > 60) = P (Y > ln 60) = P (Z > (ln 60 − 1.8)/2) = P (Z > 1.15) = 0.1251. (b) The mean of the underlying lognormal distribution is e1.8+4/2 = 44.70 seconds. So, P (X < 44.70) = P (Z < (ln 44.70 − 1.8)/2) = P (Z < 1) = 0.8413.

Chapter 7

Functions of Random Variables 7.1 From y = 2x − 1 we obtain x = (y + 1)/2, and given x = 1, 2, and 3, then g(y) = f [(y + 1)/2] = 1/3, 7.2 From y = x2 , x = 0, 1, 2, 3, we obtain x =

√

g(y) = f ( y) =

3 √ y

√

for y = 1, 3, 5.

√y 3−√y 3 2 , 5 5

fory = 0, 1, 4, 9.

7.3 The inverse functions of y1 = x1 +x2 and y2 = x1 −x2 are x1 = (y1 +y2 )/2 and x2 = (y1 −y2 )/2. Therefore, g(y1 , y2 ) =

(y1 +y2 )/2 (y1 −y2 )/2 2−y1 1 1 5 , y1 +y2 y1 −y2 , , 2 − y 4 3 12 1 2 2 2

where y1 = 0, 1, 2, y2 = −2, −1, 0, 1, 2, y2 ≤ y1 and y1 + y2 = 0, 2, 4. 7.4 Let W = X2 . The inverse functions of y = x1 x2 and w = x2 are x1 = y/w and x2 = w, where y/w = 1, 2. Then g(y, w) = (y/w)(w/18) = y/18,

y = 1, 2, 3, 4, 6; w = 1, 2, 3, y/w = 1, 2.

In tabular form the joint distribution g(y, w) and marginal h(y) are given by

g(y, w) 1 w 2 3 h(y)

1 1/18

1/18

2 2/18 2/18 2/9

y 3

4/18 3/18 1/6

2/9

6/18 1/3

Chapter 7 Functions of Random Variables

P (Y = 2) = f (1, 2) + f (2, 1) = 2/18 + 2/18 = 2/9, P (Y = 3) = f (1, 3) = 3/18 = 1/6, P (Y = 4) = f (2, 2) = 4/18 = 2/9, P (Y = 6) = f (2, 3) = 6/18 = 1/3. 7.5 The inverse function of y = −2 ln x is given by x = e−y/2 from which we obtain |J| = | − e−y/2 /2| = e−y/2 /2. Now, g(y) = f (ey/2 )|J| = e−y/2 /2,

y > 0,

which is a chi-squared distribution with 2 degrees of freedom. 7.6 The inverse function of y = 8x3 is x = y 1/3 /2, for 0 < y < 8 from which we obtain |J| = y −2/3 /6. Therefore, 1 g(y) = f (y 1/3 /2)|J| = 2(y 1/3 /2)(y −2/3 /6) = y −1/3 , 6 7.7 To ﬁnd k we solve the equation k dv =

x−1/2 √ 2 b

v 2 e−bv dv = 1. Let x = bv 2 , then dx = 2bv dv and

dx. Then the equation becomes k 2b3/2

Hence k =

∞

x3/2−1 e−x dx = 1,

4b3/2 Γ(1/2) .

2w/m)|J| =

Now the√inverse function of w = mv 2 /2 is v = |J| = 1/ 2mw. It follows that g(w) = f (

0 < y < 8.

kΓ(3/2) = 1. 2b3/2

2w/m, for w > 0, from which we obtain

4b3/2 1 (2w/m)e−2bw/m 2w/m = w3/2−1 e−(2b/m)w , Γ(1/2) (m/(2b))3/2 Γ(3/2)

for w > 0, which is a gamma distribution with α = 3/2 and β = m/(2b). 7.8 (a) The inverse of y = x2 is x = Therefore,

√

√ y, for 0 < y < 1, from which we obtain |J| = 1/2 y.

√ √ √ g(y) = f ( y)|J| = 2(1 − y)/2 y = y −1/2 − 1, (b) P (Y < 0.1) =

0.1 0

0 < y < 1.

0.1 (y −1/2 − 1) dy = (2y 1/2 − y) 0 = 0.5324.

7.9 (a) The inverse of y = x + 4 is x = y − 4, for y > 4, from which we obtain |J| = 1. Therefore, g(y) = f (y − 4)|J| = 32/y 3 , (b) P (Y > 8) = 32

∞ 8

y > 4.

∞ y −3 dy = − 16y −2 8 = 14 .

Solutions for Exercises in Chapter 7

7.10 (a) Let W = X. The inverse functions of z = x + y 0 < w < z, 0 < z < 1, from which we obtain ∂x ∂x 1 ∂z J = ∂w ∂y ∂y = −1 ∂w ∂z

and w = x are x = w and y = z − w, 0 = 1. 1

Then g(w, z) = f (w, z −w)|J| = 24w(z −w), for 0 < w < z and 0 < z < 1. The marginal distribution of Z is 1 f1 (z) = 24(z − w)w dw = 4z 3 , 0 < z < 1. 0

(b) P (1/2 < Z < 3/4) = 4

3/4

z 3 dz = 65/256.

1/2

7.11 The amount of kerosene left at the end of the day is Z = Y − X. Let W = Y . The inverse functions of z = y − x and w = y are x = w − z and y = w, for 0 < z < w and 0 < w < 1, from which we obtain ∂x ∂x 1 −1 ∂w ∂z = 1. J = ∂y ∂y = 1 0 ∂w ∂z Now, g(w, z) = f (w − z, w) = 2,

0 < z < w, 0 < w < 1,

and the marginal distribution of Z is 1 h(z) = 2 dw = 2(1 − z),

0 < z < 1.

7.12 Since X1 and X2 are independent, the joint probability distribution is f (x1 , x2 ) = f (x1 )f (x2 ) = e−(x1 +x2 ) ,

x1 > 0, x2 > 0.

The inverse functions of y1 = x1 +x2 and y2 = x1 /(x1 +x2 ) are x1 = y1 y2 and x2 = y1 (1−y2 ), for y1 > 0 and 0 < y2 < 1, so that ∂x1 /∂y1 ∂x1 /∂y2 y2 y1 = = −y1 . J = ∂x2 /∂y1 ∂x2 /∂y2 1 − y2 −y1 Then, g(y1 , y2 ) = f (y1 y2 , y1 (1 − y2 ))|J| = y1 e−y1 , for y1 > 0 and 0 < y2 < 1. Therefore,

g(y1 ) =

y1 e−y1 dy2 = y1 e−y1 ,

y1 > 0,

and

∞

g(y2 ) =

y1 e−y1 dy1 = Γ(2) = 1,

0 < y2 < 1.

Chapter 7 Functions of Random Variables

7.13 Since I and R are independent, the joint probability distribution is f (i, r) = 12ri(1 − i),

0 < i < 1, 0 < r < 1.

Let V = R. The inverse functions of w = i2 r and v = r are i = w < v < 1 and 0 < w < 1, from which we obtain ∂i/∂w ∂i/∂v = √1 . J = ∂r/∂w ∂r/∂v 2 vw

w/v and r = v, for

Then, g(w, v) = f (

w/v, v)|J| = 12v

w/v(1 −

1 = 6(1 − w/v), w/v) √ 2 vw

for w < v < 1 and 0 < w < 1, and the marginal distribution of W is 1 v=1 √ √ (1 − w/v) dv = 6 (v − 2 wv) v=w = 6 + 6w − 12 w, h(w) = 6

0 < w < 1.

√ √ 7.14 The inverse functions of y = x2 are given by x1 = y and x2 = − y from which we obtain √ √ J1 = 1/2 y and J2 = 1/2 y. Therefore, √ √ 1− y 1+ y 1 1 √ √ √ g(y) = f ( y)|J1 | + f (− y)|J2 | = · √ + · √ = 1/2 y, 2 2 y 2 2 y for 0 < y < 1. √ √ √ 7.15 The inverse functions of y = x2 are x1 = y, x2 = − y for 0 < y < 1 and x3 = y for √ 0 < y < 4. Now |J1 | = |J2 | = |J3 | = 1/2 y, from which we get √ √ 2(− y + 1) 2( y + 1) 1 1 2 √ √ · √ + · √ = √ , g(y) = f ( y)|J1 | + f (− y)|J2 | = 9 2 y 9 2 y 9 y for 0 < y < 1 and √ √ y+1 2( y + 1) 1 √ g(y) = f ( y)|J3 | = · √ = √ , 9 2 y 9 y

for 0 < y < 4.

7.16 Using the formula we obtain ∞ α−1 e−x/β β α+r Γ(α + r) ∞ xα+r−1 e−x/β r r x dx = dx μr = E(X ) = x · β α Γ(α) β α Γ(α) β α+r Γ(α + r) 0 0 β r Γ(α + r) = , Γ(α) since the second integrand is a gamma density with parameters α + r and β. 7.17 The moment-generating function of X is MX (t) = E(etX ) =

k 1 tx et (1 − ekt ) , e = k k(1 − et ) x=1

Solutions for Exercises in Chapter 7

7.18 The moment-generating function of X is MX (t) = E(etX ) = p

∞

etx q x−1 =

x=1

p t x pet (e q) = , q 1 − qet x=1

by summing an inﬁnite geometric series. To ﬁnd out the moments, we use (1 − qet )pet + pqe2t (1 − q)p + pq 1 μ = MX (0) = = = , t 2 2 (1 − qe ) (1 − q) p t=0 and

(1 − qet )2 pet + 2pqe2t (1 − qet ) 2−p μ2 = MX (0) = = . t 4 (1 − qe ) p2 t=0

So, σ 2 = μ2 − μ2 =

q . p2

7.19 The moment-generating function of a Poisson random variable is MX (t) = E(etX ) =

∞ tx −μ x e e μ

x=0

= e−μ

x=0

So,

μ = MX (0) = μ eμ(e

∞ (μet )x

t −1)+t

= e−μ eμe = eμ(e t

t −1)

= μ, t μ2 = MX (0) = μeμ(e −1)+t (μet + 1) t=0

t=0

= μ(μ + 1),

and

σ 2 = μ2 − μ2 = μ(μ + 1) − μ2 = μ. 7.20 From MX (t) = e4(e

t −1)

we obtain μ = 4, σ 2 = 4, and σ = 2. Therefore,

P (μ − 2σ < X < μ + 2σ) = P (0 < X < 8) =

p(x; 4) = 0.9489 − 0.0183 = 0.9306.

x=1

7.21 The moment generating function of a chi-squared distribution with v degrees of freedom is ∞ 1 tX MX (t) = E(e ) = xv/2−1 e−(1/2−t)x dx Γ(v/2)2v/2 0 (1/2 − t)−v/2 ∞ xv/2−1 e−(1/2−t)x = dx = (1 − 2t)−v/2 , 2v/2 Γ(v/2)(1/2 − t)−v/2 0 since the last integrand is another gamma distribution with parameters v/2 and (1/2 − t)−1 . 7.22 Using the moment-generating function of the chi-squared distribution, we obtain = v, μ = MX (0) = v(1 − 2t)−v/2−1 t=0 μ2 = MX (0) = v(v + 2) (1 − 2t)−v/2−2 = v(v + 2). t=0

Chapter 7 Functions of Random Variables

7.23 The joint distribution of X and Y is fX,Y (x, y) = e−x−y for x > 0 and y > 0. The inverse v u = functions of u = x + y and v = x/(x + y) are x = uv and y = u(1 − v) with J = 1 − v −u −u for u > 0 and 0 < v < 1. Hence, J = u. The joint distribution of U and V is fU,V (u, v) = ue−uv · e−u(1−v) = ue−u , for u > 0 and 0 < v < 1.

1 (a) fU (u) = 0 ue−u dv = ue−u for u > 0, which is a gamma distribution with parameters 2 and 1.

∞ (b) fV (v) = 0 ue−u du = 1 for 0 < v < 1. This is a uniform (0,1) distribution.

Chapter 8

Fundamental Sampling Distributions and Data Descriptions 8.1 (a) Responses of all people in Richmond who have telephones. (b) Outcomes for a large or inﬁnite number of tosses of a coin. (c) Length of life of such tennis shoes when worn on the professional tour. (d) All possible time intervals for this lawyer to drive from her home to her oﬃce. 8.2 (a) x ¯ = 8.6 minutes. (b) x ˜ = 9.5 minutes. (c) Mode are 5 and 10 minutes. 8.3 (a) x ¯ = 3.2 seconds. (b) x ˜ = 3.1 seconds. 8.4 (a) Number of tickets issued by all state troopers in Montgomery County during the Memorial holiday weekend. (b) Number of tickets issued by all state troopers in South Carolina during the Memorial holiday weekend. 8.5 (a) x ¯ = 2.4. (b) x ˜ = 2. (c) m = 3. 8.6 x ¯ = 22.2 days, x ˜ = 14 days and m = 8 days. x ˜ is the best measure of the center of the data. The mean should not be used on account of the extreme value 95, and the mode is not desirable because the sample size is too small. 8.7 (a) x ¯ = 53.75. (b) Modes are 75 and 100. 8.8 (a) x ¯ = 35.7 grams. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 85

Chapter 8 Fundamental Sampling Distributions and Data Descriptions

86 (b) x ˜ = 32.5 grams.

(b)

n i=1

x2i −(

i=1

xi )2

n(n−1)

(10)(838)−862 (10)(9)

= 10.933. Taking the square root, we have s = 3.307.

8.10 (a) Range = 4.3 − 2.3 = 2.0. (b) The mean is

2.5 + 3.6 + · · · + 3.4 = 3.2. 9 Using formula (8.2.1), we calculate the sample variance as x ¯=

s2 = 8.11 (a) s2 =

1 n−1 n

(b) s2 =

(2.5 − 3.2)2 + (3.6 − 3.2)2 + · · · + (3.4 − 3.2)2 = 0.4975. 8

(xi − x ¯ )2 =

x=1

n i=1

x2i −(

i=1

xi )2

n(n−1)

1 14 [(2

− 2.4)2 + (1 − 2.4)2 + · · · + (2 − 2.4)2 ] = 2.971.

(15)(128)−362 (15)(14)

= 2.971.

8.12 (a) x ¯ = 11.69 milligrams. n

(b)

s2 n

8.13

= n

i=1

n i=1

x2i −(

i=1

xi )2

n(n−1) x2i −(

i=1

xi )2

n(n−1)

(8)(1168.21)−93.52 (8)(7)

(20)(148.55)−53.32 (20)(19)

= 10.776.

= 0.342 and hence s = 0.585.

¯ becomes X ¯ + c and 8.14 (a) Replace Xi in S 2 by Xi + c for i = 1, 2, . . . , n. Then X 1 ¯ + c)]2 = 1 ¯ 2. [(Xi + c) − (X (Xi − X) S = n−1 n−1 n

i=1

¯ becomes cX ¯ and (b) Replace Xi by cXi in S 2 for i = 1, 2, . . . , n. Then X S2 =

8.15 s2 =

n i=1

x2i −(

i=1

n(n−1)

xi )2

2 1 ¯ 2= c ¯ 2. (cXi − cX) (Xi − X) n−1 n−1 n

i=1

(6)(207)−332 (6)(5)

= 5.1.

(a) Multiplying each observation by 3 gives s2 = (9)(5.1) = 45.9. (b) Adding 5 to each observation does not change the variance. Hence s2 = 5.1. 8.16 Denote by D the diﬀerence in scores. ¯ = 25.15. (a) D c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 8

˜ = 31.00. (b) D 8.17 z1 = −1.9, z2 = −0.4. Hence, ¯ < μ ¯ − 0.4σ ¯ ) = P (−1.9 < Z < −0.4) = 0.3446 − 0.0287 = 0.3159. P (μX¯ − 1.9σX¯ < X X X 8.18 n = 36, σX¯ = 2. Hence σ = n = 100.

√

√ nσX¯ = (6)(2) = 12. If σX¯ = 1.2, then 1.2 = 12/ n and

8.19 (a) For n = 64, σX¯ = 5.6/8 = 0.7, whereas for n = 196, σX¯ = 5.6/14 = 0.4. Therefore, the variance of the sample mean is reduced from 0.49 to 0.16 when the sample size is increased from 64 to 196. (b) For n = 784, σX¯ = 5.6/28 = 0.2, whereas for n = 49, σX¯ = 5.6/7 = 0.8. Therefore, the variance of the sample mean is increased from 0.04 to 0.64 when the sample size is decreased from 784 to 49. 2 = σ 2 /n = (8/3)/54 = 4/81 with σ = 2/9. So, 8.20 n = 54, μX¯ = 4, σX ¯ ¯ X

z1 = (4.15 − 4)/(2/9) = 0.68,

and

z2 = (4.35 − 4)/(2/9) = 1.58,

and ¯ < 4.35) = P (0.68 < Z < 1.58) = 0.9429 − 0.7517 = 0.1912. P (4.15 < X √ 8.21 μX¯ = μ = 240, σX¯ = 15/ 40 = 2.372. Therefore, μX¯ ± 2σX¯ = 240 ± (2)(2.372) or from 235.256 to 244.744, which indicates that a value of x = 236 milliliters is reasonable and hence the machine need not be adjusted. √ 8.22 (a) μX¯ = μ = 174.5, σX¯ = σ/ n = 6.9/5 = 1.38. (b) z1 = (172.45 − 174.5)/1.38 = −1.49, z2 = (175.85 − 174.5)/1.38 = 0.98. So, ¯ < 175.85) = P (−1.49 < Z < 0.98) = 0.8365 − 0.0681 = 0.7684. P (172.45 < X Therefore, the number of sample means between 172.5 and 175.8 inclusive is (200)(0.7684) = 154. (c) z = (171.95 − 174.5)/1.38 = −1.85. So, ¯ < 171.95) = P (Z < −1.85) = 0.0322. P (X Therefore, about (200)(0.0322) = 6 sample means fall below 172.0 centimeters. 8.23 (a) μ = xf (x) = (4)(0.2) + (5)(0.4) + (6)(0.3) + (7)(0.1) = 5.3, and σ 2 = (x − μ)2 f (x) = (4 − 5.3)2 (0.2) + (5 − 5.3)2 (0.4) + (6 − 5.3)2 (0.3) + (7 − 5.3)2 (0.1) = 0.81. 2 = σ 2 /n = 0.81/36 = 0.0225. (b) With n = 36, μX¯ = μ = 5.3 and σX ¯

Chapter 8 Fundamental Sampling Distributions and Data Descriptions

8.24 n = 36, μX¯ = 40, σX¯ = 2/6 = 1/3 and z = (40.5 − 40)/(1/3) = 1.5. So, 36 ¯ > 40.5) = P (Z > 1.5) = 1 − 0.9332 = 0.0668. P Xi > 1458 = P (X i=1

¯ < 7.2) = P (−1.8 < Z < 0.6) = 0.6898. 8.25 (a) P (6.4 < X √ (b) z = 1.04, x ¯ = z(σ/ n) + μ = (1.04)(1/3) + 7 = 7.35. √ 8.26 n = 64, μX¯ = 3.2, σX¯ = σ/ n = 1.6/8 = 0.2. ¯ < 2.7) = P (Z < −2.5) = 0.0062. (a) z = (2.7 − 3.2)/0.2 = −2.5, P (X ¯ > 3.5) = P (Z > 1.5) = 1 − 0.9332 = 0.0668. (b) z = (3.5 − 3.2)/0.2 = 1.5, P (X (c) z1 = (3.2 − 3.2)/0.2 = 0, z2 = (3.4 − 3.2)/0.2 = 1.0, ¯ < 3.4) = P (0 < Z < 1.0) = 0.8413 − 0.5000 = 0.3413. P (3.2 < X √ 8.27 n = 50, x ¯ = 0.23 and σ = 0.1. Now, z = (0.23 − 0.2)/(0.1/ 50) = 2.12; so ¯ ≥ 0.23) = P (Z ≥ 2.12) = 0.0170. P (X Hence the probability of having such observations, given the mean μ = 0.20, is small. Therefore, the mean amount to be 0.20 is not likely to be true. 8.28 μ1 − μ2 = 80 − 75 = 5, σX¯ 1 −X¯ 2 = 25/25 + 9/36 = 1.118, z1 = (3.35 − 5)/1.118 = −1.48 and z2 = (5.85 − 5)/1.118 = 0.76. So, ¯1 − X ¯ 2 < 5.85) = P (−1.48 < Z < 0.76) = 0.7764 − 0.0694 = 0.7070. P (3.35 < X 8.29 μX¯ 1 −X¯ 2 = 72 − 28 = 44, σX¯ 1 −X¯ 2 = 100/64 + 25/100 = 1.346 and z = (44.2 − 44)/1.346 = ¯ 2 < 44.2) = P (Z < 0.15) = 0.5596. ¯1 − X 0.15. So, P (X 8.30 μ1 − μ2 = 0, σX¯ 1 −X¯ 2 = 50 1/32 + 1/50 = 11.319. (a) z1 = −20/11.319 = −1.77, z2 = 20/11.319 = 1.77, so ¯ 2 | > 20) = 2P (Z < −1.77) = (2)(0.0384) = 0.0768. ¯1 − X P (|X (b) z1 = 5/11.319 = 0.44 and z2 = 10/11.319 = 0.88. So, ¯ 2 < −5) + P (5 < X ¯1 − X ¯ 2 < 10) = 2P (5 < X ¯1 − X ¯ 2 < 10) = 2P (0.44 < ¯1 − X P (−10 < X Z < 0.88) = 2(0.8106 − 0.6700) = 0.2812. 8.31 (a) If the two population mean drying times are truly equal, the probability that the diﬀerence of the two sample means is 1.0 is 0.0013, which is very small. This means that the assumption of the equality of the population means are not reasonable. (b) If the experiment was run 10,000 times, there would be (10000)(0.0013) = 13 experiments ¯A − X ¯ B would be at least 1.0. where X 8.32 (a) n1 = n2 = 36 and z = 0.2/ 1/36 + 1/36 = 0.85. So, ¯B − X ¯ A ≥ 0.2) = P (Z ≥ 0.85) = 0.1977. P (X (b) Since the probability in (a) is not negligible, the conjecture is not true. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 8

8.33 (a) When the population equals the limit, the probability of a sample mean exceeding the limit would be 1/2 due the symmetry of the approximated normal distribution. √ ¯ ≥ 7960 | μ = 7950) = P (Z ≥ (7960 − 7950)/(100/ 25)) = P (Z ≥ 0.5) = 0.3085. (b) P (X No, this is not very strong evidence that the population mean of the process exceeds the government limit. 52 52 + 30 = 1.29 and z = 4−0 8.34 (a) σX¯ A −X¯ B = 30 1.29 = 3.10. So, ¯ B > 4 | μA = μB ) = P (Z > 3.10) = 0.0010. ¯A − X P (X Such a small probability means that the diﬀerence of 4 is not likely if the two population means are equal. (b) Yes, the data strongly support alloy A. ¯ ≤ 775 is 0.0062, given that μ = 800 is true, it suggests that this 8.35 Since the probability that X event is very rare and it is very likely that the claim of μ = 800 is not true. On the other hand, if μ is truly, say, 760, the probability √ ¯ ≤ 775 | μ = 760) = P (Z ≤ (775 − 760)/(40/ 16)) = P (Z ≤ 1.5) = 0.9332, P (X which is very high. ¯ −μW )/(σW /√n) ∼ 8.36 Deﬁne Wi = ln Xi for i = 1, 2, . . . . Using the central limit theorem, Z = (W 1 1 ¯ follows, approximately, a normal distribution when n is large. Since n(z; 0, 1). Hence W n n 1 1 1 ¯ = ln(Xi ) = ln Xi = ln(Y ), W n n n i=1

i=1

then it is easily seen that Y follows, approximately, a lognormal distribution. 8.37 (a) 27.488. (b) 18.475. (c) 36.415. 8.38 (a) 16.750. (b) 30.144. (c) 26.217. 8.39 (a) χ2α = χ20.99 = 0.297. (b) χ2α = χ20.025 = 32.852. (c) χ20.05 = 37.652. Therefore, α = 0.05 − 0.045 = 0.005. Hence, χ2α = χ20.005 = 46.928. 8.40 (a) χ2α = χ20.01 = 38.932. (b) χ2α = χ20.05 = 12.592. (c) χ20.01 = 23.209 and χ20.025 = 20.483 with α = 0.01 + 0.015 = 0.025.

Chapter 8 Fundamental Sampling Distributions and Data Descriptions

(b) P (3.462 < S 2 < 10.745) = P

(24)(3.462) 6

1.356) = 0.10. Therefore, P (−1.356 < T < 2.179) = 0.975 − 0.010 = 0.875. (d) P (T > −2.567) = 1 − P (T > 2.567) = 1 − 0.01 = 0.99. 8.46 (a) Since t0.01 leaves an area of 0.01 to the right, and −t0.005 an area of 0.005 to the left, we ﬁnd the total area to be 1 − 0.01 − 0.005 = 0.985 between −t0.005 and t0.01 . Hence, P (−t0.005 < T < t0.01 ) = 0.985. (b) Since −t0.025 leaves an area of 0.025 to the left, the desired area is 1 − 0.025 = 0.975. That is, P (T > −t0.025 ) = 0.975. 8.47 (a) From Table A.4 we note that 2.069 corresponds to t0.025 when v = 23. Therefore, −t0.025 = −2.069 which means that the total area under the curve to the left of t = k is 0.025 + 0.965 = 0.990. Hence, k = t0.01 = 2.500. (b) From Table A.4 we note that 2.807 corresponds to t0.005 when v = 23. Therefore the total area under the curve to the right of t = k is 0.095 + 0.005 = 0.10. Hence, k = t0.10 = 1.319. (c) t0.05 = 1.714 for 23 degrees of freedom. 8.48 From Table A.4 we ﬁnd t0.025 = 2.131 for v = 15 degrees of freedom. Since the value t=

27.5 − 30 = −2.00 5/4

falls between −2.131 and 2.131, the claim is valid. 8.49 t = (24 − 20)/(4.1/3) = 2.927, t0.01 = 2.896 with 8 degrees of freedom. Conclusion: no, μ > 20. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 8

8.50 x ¯ = 0.475, s2 = 0.0336 and t = (0.475 − 0.5)/0.0648 = −0.39. Hence ¯ < 0.475) = P (T < −0.39) ≈ 0.35. P (X So, the result is inconclusive. 8.51 (a) 2.71. (b) 3.51. (c) 2.92. (d) 1/2.11 = 0.47. (e) 1/2.90 = 0.34. 8.52 s21 = 10.441 and s22 = 1.846 which gives f = 5.66. Since, from Table A.6, f0.05 (9, 7) = 3.68 and f0.01 (9, 7) = 6.72, the probability of P (F > 5.66) should be between 0.01 and 0.05, which is quite small. Hence the variances may not be equal. Furthermore, if a computer software can be used, the exact probability of F > 5.66 can be found 0.0162, or if two sides are considered, P (F < 1/5.66) + P (F > 5.66) = 0.026. 8.53 s21 = 15750 and s22 = 10920 which gives f = 1.44. Since, from Table A.6, f0.05 (4, 5) = 5.19, the probability of F > 1.44 is much bigger than 0.05, which means that the two variances may be considered equal. The actual probability of F > 1.44 is 0.3442 and P (F < 1/1.44) + P (F > 1.44) = 0.7170. 8.54 The normal quantile-quantile plot is shown as

1100 1000 700

800

900

Sample Quantiles

1200

1300

Normal Q−Q Plot

−2

−1

Theoretical Quantiles

Chapter 8 Fundamental Sampling Distributions and Data Descriptions

6.75 6.70 6.65

Sample Quantiles

6.80

Normal Q−Q Plot

−2

−1

Theoretical Quantiles

8.56 The box-and-whisker plot is shown below. Box−and−Whisker Plot

The sample mean = 12.32 and the sample standard deviation = 6.08. 8.57 The moment-generating function for the gamma distribution is given by ∞ 1 tX etx xα−1 e−x/β dx MX (t) = E(e ) = α β Γ(α) 0 ∞ 1 1 1 α−1 −x β −t x e dx = α β (1/β − t)α (1/β − t)−α Γ(α) 0 ∞ α−1 −x/(1/β−t)−1 1 x e 1 dx = , = (1 − βt)α 0 [(1/β − t)−1 ]α Γ(α) (1 − βt)α c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 8

for t < 1/β, since the last integral is one due to the integrand being a gamma density function. Therefore, the moment-generating function of an exponential distribution, by substituting α to 1 and β to θ, is given by MX (t) = (1 − θt)−1 . Hence, the moment-generating function of Y can be expressed as MY (t) = MX1 (t)MX2 (t) · · · MXn (t) =

(1 − θt)−1 = (1 − θt)−n ,

i=1

which is seen to be the moment-generating function of a gamma distribution with α = n and β = θ. 8.58 The variance of the carbon monoxide contents is the same as the variance of the coded 2 measurements. That is, s2 = (15)(199.94)−39 = 7.039, which results in s = 2.653. (15)(14)

2 2

2 S S /σ 8.59 P S12 < 4.89 = P S12 /σ2 < 4.89 = P (F < 4.89) = 0.99, where F has 7 and 11 degrees of 2 2 freedom. 8.60 s2 = 2, 446, 800, 000. 8.61 Let X1 and X2 be Poisson variables with parameters λ1 = 6 and λ2 = 6 representing the number of hurricanes during the ﬁrst and second years, respectively. Then Y = X1 + X2 has a Poisson distribution with parameter λ = λ1 + λ2 = 12. e−12 1215 = 0.0724. 15! 9 e−12 12y = 0.2424. y! y=0

(a) P (Y = 15) = (b) P (Y ≤ 9) =

8.62 Dividing each observation by 1000 and then subtracting 55 yields the following data: −7, −2, −10, 6, 4, 1, 8, −6, −2, and −1. The variance of this coded data is (10)(311) − (−9)2 = 33.656. (10)(9) Hence, with c = 1000, we have s2 = (1000)2 (33.656) = 33.656 × 106 , and then s = 5801 kilometers.

Chapter 8 Fundamental Sampling Distributions and Data Descriptions

8.63 The box-and-whisker plot is shown below. Box−and−Whisker Plot

The sample mean is 2.7967 and the sample standard deviation is 2.2273.

2 2

2 S S /σ = P (F > 1.89) ≈ 0.05, where F has 24 and 30 8.64 P S12 > 1.26 = P S12 /σ12 > (15)(1.26) 10 2 2 2 degrees of freedom. 8.65 No outliers. 8.66 The value 32 is a possible outlier. 8.67 μ = 5,000 psi, σ = 400 psi, and n = 36. (a) Using approximate normal distribution (by CLT), 4800 − 5000 5200 − 5000 ¯ √ √ P (4800 < X < 5200) = P 4.0/ 52 /20 + 52 /20) ¯B − X (a) P (X √ = P (Z > 4.0/(5/ 10)) = P (Z > 2.93) = 0.0017. (b) It is extremely unlikely that μA = μB . 8.70 (a) nA = 30, x ¯A = 64.5% and σA = 5%. Hence, √ ¯ A ≤ 64.5 | μA = 65) = P (Z < (64.5 − 65)/(5/ 30)) = P (Z < −0.55) P (X = 0.2912. There is no evidence that μA is less than 65%.

¯B = 70% and σB = 5%. It turns out σX¯ B −X¯ A = (b) nB = 30, x ¯B − X ¯ A ≥ 5.5 | μA = μB ) = P P (X

5.5 Z≥ 1.29

52 30

= 1.29%. Hence,

= P (Z ≥ 4.26) ≈ 0.

It does strongly support that μB is greater than μA . ¯ B ∼ n(x; 65%, 0.9129%). (c) i) Since σX¯ B = √530 = 0.9129, X ¯A − X ¯ B ∼ n(x; 0, 1.29%). ii) X ¯

2/30

A −X B ∼ n(z; 0, 1). iii) X√

¯ B ≥ 70) = P Z ≥ 8.71 P (X

70−65 0.9129

= P (Z ≥ 5.48) ≈ 0.

8.72 It is known, from Table A.3, that P (−1.96 < Z < 1.96) = 0.95. Given μ = 20 and σ =

2 (3)(1.96) √ 3, we equate 1.96 = 20.1−20 to obtain n = = 3457.44 ≈ 3458. 0.1 3/ n 8.73 It is known that P (−2.575 < Z < 2.575) = 0.99. Hence, by equating 2.575 = 2 n = 2.575 = 2652.25 ≈ 2653. 0.05

0.05 √ , 1/ n

√

we obtain

8.74 μ = 9 and σ = 1. Then P (9 − 1.5 < X < 9 + 1.5) = P (7.5 < X < 10.5) = P (−1.5 < Z < 1.5) = 0.9332 − 0.0668 = 0.8664. Thus the proportion of defective is 1 − 0.8654 = 0.1346. To meet the speciﬁcations 99% of the time, we need to equate 2.575 = 1.5 σ , since P (−2.575 < Z < 2.575) = 0.99. Therefore, 1.5 = 0.5825. σ = 2.575 8.75 With the 39 degrees of freedom, P (S 2 ≤ 0.188 | σ 2 = 1.0) = P (χ2 ≤ (39)(0.188)) = P (χ2 ≤ 7.332) ≈ 0, which means that it is impossible to observe s2 = 0.188 with n = 40 for σ 2 = 1. Note that Table A.5 does not provide any values for the degrees of freedom to be larger than 30. However, one can deduce the conclusion based on the values in the last line of the table. Also, computer software gives the value of 0.

Chapter 9

One- and Two-Sample Estimation Problems 9.1 n = [(2.575)(5.8)/2]2 = 56 when rounded up. 9.2 n = 30, x ¯ = 780, and σ = 40. Also, z0.02 = 2.054. So, a 96% conﬁdence interval for the population mean can be calculated as √ √ 780 − (2.054)(40/ 30) < μ < 780 + (2.054)(40/ 30), or 765 < μ < 795. 9.3 n = 75, x ¯ = 0.310, σ = 0.0015, and z0.025 = 1.96. A 95% conﬁdence interval for the population mean is √ √ 0.310 − (1.96)(0.0015/ 75) < μ < 0.310 + (1.96)(0.0015/ 75), or 0.3097 < μ < 0.3103. 9.4 n = 50, x ¯ = 174.5, σ = 6.9, and z0.01 = 2.33. (a) A 98% conﬁdence interval for the population mean√is √ 174.5 − (2.33)(6.9/ 50) < μ < 174.5 + (2.33)(6.9/ 50), or 172.23 < μ < 176.77. √ (b) e < (2.33)(6.9)/ 50 = 2.27. 9.5 n = 100, x ¯ = 23, 500, σ = 3900, and z0.005 = 2.575. (a) A 99% conﬁdence interval for the population mean is 23, 500 − (2.575)(3900/10) < μ < 23, 500 + (2.575)(3900/10), or 22, 496 < μ < 24, 504. (b) e < (2.575)(3900/10) = 1004. 9.6 n = [(2.05)(40)/10]2 = 68 when rounded up. 9.7 n = [(1.96)(0.0015)/0.0005]2 = 35 when rounded up. 9.8 n = [(1.96)(40)/15]2 = 28 when rounded up. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall. 97

Chapter 9 One- and Two-Sample Estimation Problems

9.9 n = 20, x ¯ = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95% conﬁdence interval for the population mean is √ √ 11.3 − (2.093)(2.45/ 20) < μ < 11.3 + (2.093)(2.45/ 20), or 10.15 < μ < 12.45. 9.10 n = 12, x ¯ = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95% conﬁdence interval for the population mean is √ √ 79.3 − (2.201)(7.8/ 12) < μ < 79.3 + (2.201)(7.8/ 12), or 74.34 < μ < 84.26. 9.11 n = 9, x ¯ = 1.0056, s = 0.0245, and t0.005 = 3.355 with 8 degrees of freedom. A 99% conﬁdence interval for the population mean is 1.0056 − (3.355)(0.0245/3) < μ < 1.0056 + (3.355)(0.0245/3), or 0.978 < μ < 1.033. 9.12 n = 10, x ¯ = 230, s = 15, and t0.005 = 3.25 with 9 degrees of freedom. A 99% conﬁdence interval for the population mean is √ √ 230 − (3.25)(15/ 10) < μ < 230 + (3.25)(15/ 10), or 214.58 < μ < 245.42. 9.13 n = 12, x ¯ = 48.50, s = 1.5, and t0.05 = 1.796 with 11 degrees of freedom. A 90% conﬁdence interval for the population mean is √ √ 48.50 − (1.796)(1.5/ 12) < μ < 48.50 + (1.796)(1.5/ 12), or 47.722 < μ < 49.278. 9.14 n = 15, x ¯ = 3.7867, s = 0.9709, 1 − α = 95%, and t0.025 = 2.145 with 14 degrees of freedom. So, by calculating 3.7867 ± (2.145)(0.9709) 1 + 1/15 we obtain (1.6358, 5.9376) which is a 95% prediction interval the drying times of next paint. 9.15 n = 100, x ¯ = 23,500, s = 3, 900, 1−α = 0.99, and t0.005 ≈ 2.66 with 60 degrees of freedom (use table from the book) or t0.005 = 2.626 if 100 degrees of freedom is used. The prediction interval of next automobile will be driven in Virginia (using 2.66) is 23,500 ± (2.66)(3,900) 1 + 1/100 which yields 13,075 < μ < 33,925 kilometers. 9.16 n = 12, x ¯ = 79.3, s = 7.8, and t0.025 = 2.201 with 11 degrees of freedom. A 95% prediction interval for a future observation is 79.3 ± (2.201)(7.8) 1 + 1/12 = 79.3 ± 17.87, which yields (61.43, 97.17). c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Solutions for Exercises in Chapter 9

9.17 n = 20, x ¯ = 11.3, s = 2.45, and t0.025 = 2.093 with 19 degrees of freedom. A 95% prediction interval for a future observation is 11.3 ± (2.093)(2.45) 1 + 1/20 = 11.3 ± 5.25, which yields (6.05, 16.55). 9.18 n = 12, x ¯ = 48.50, s = 1.5, 1 − α = 0.90, and γ = 0.05, with k = 2.655. The tolerance interval is 48.50 ± (2.655)(1.5) which yields (44.52, 52.48). 9.19 n = 25, x ¯ = 325.05, s = 0.5, γ = 5%, and 1 − α = 90%, with k = 2.208. So, 325.05 ± (2.208)(0.5) yields (323.946, 326.151). Thus, we are 95% conﬁdent that this tolerance interval will contain 90% of the aspirin contents for this brand of buﬀered aspirin. ¯ = 1.0056, s = 0.0245, 1 − α = 0.95, and γ = 0.05, with k = 3.532. The tolerance 9.20 n = 9, x interval is 1.0056 ± (3.532)(0.0245) which yields (0.919, 1.092). 9.21 n = 15, x ¯ = 3.84, and s = 3.07. To calculate an upper 95% prediction limit, we obtain t0.05 = 1.761 with 14 degrees of freedom. So, the upper limit is 3.84 + (1.761)(3.07) 1 + 1/15 = 3.84 + 5.58 = 9.42. This means that a new observation will have a chance of 95% to fall into the interval (−∞, 9.42). To obtain an upper 95% tolerance limit, using 1 − α = 0.95 and γ = 0.05, with k = 2.566, we get 3.84 + (2.566)(3.07) = 11.72. Hence, we are 95% conﬁdent that (−∞, 11.72) will contain 95% of the orthophosphorous measurements in the river. 9.22 n = 50, x ¯ = 78.3, and s = 5.6. Since t0.05 = 1.677 with 49 degrees of freedom, the bound of a lower 95% prediction interval for a single new observation is 78.3 − (1.677)(5.6) 1 + 1/50 = 68.91. So, the interval is (68.91, ∞). On the other hand, with 1 − α = 95% and γ = 0.01, the k value for a one-sided tolerance limit is 2.269 and the bound is 78.3 − (2.269)(5.6) = 65.59. So, the tolerance interval is (65.59, ∞). 9.23 Since the manufacturer would be more interested in the mean tensile strength for future products, it is conceivable that prediction interval and tolerance interval may be more interesting than just a conﬁdence interval. 9.24 This time 1 − α = 0.99 and γ = 0.05 with k = 3.126. So, the tolerance limit is 78.3 − (3.126)(5.6) = 60.79. Since 62 exceeds the lower bound of the interval, yes, this is a cause of concern. 9.25 In Exercise 9.14, a 95% prediction interval for a new observation is calculated as (1.6358, 5.9377). Since 6.9 is in the outside range of the prediction interval, this new observation is likely to be an outlier. 9.26 n = 12, x ¯ = 48.50, s = 1.5, 1 − α = 0.95, and γ = 0.05, with k = 2.815. The lower bound of the one-sided tolerance interval is 48.50−(2.815)(1.5) = 44.278. Their claim is not necessarily correct. 9.27 (a) n = 16, x ¯ = 1.0025, s = 0.0202, 1 − α = 0.99, t0.005 = 2.947 with 15 degrees of √ freedom. A 99% conﬁdence interval for the mean diameter is 1.0025 ± (2.947)(0.0202)/ 16 which yields (0.9876, 1.0174) centimeters. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 9 One- and Two-Sample Estimation Problems

100

(b) A 99% prediction interval for the diameter of a new metal piece is 1.0025 ± (2.947)(0.0202) 1 + 1/16 which yields (0.9411, 1.0639) centimeters. (c) For n = 16, 1 − γ = 0.99 and 1 − α = 0.95, we ﬁnd k = 3.421. Hence, the tolerance limits are 1.0025 ± (3.421)(0.0202) which give (0.9334, 1.0716). ˆ − θ)2 which can be expressed as 9.28 By deﬁnition, M SE = E(Θ ˆ − E(Θ) ˆ + E(Θ) ˆ − θ]2 M SE = E[Θ ˆ − E(Θ)] ˆ 2 + E[E(Θ) ˆ − θ]2 + 2E[Θ ˆ − E(Θ)]E[E( ˆ ˆ − θ]. = E[Θ Θ) ˆ − E(Θ)] ˆ = E[Θ] ˆ − E(Θ) ˆ = 0. Hence The third term on the right hand side is zero since E[Θ the claim is valid. 9.29 From Example 9.1 on page 267, we know that E(S 2 ) = σ 2 . Therefore, n−1 n−1 2 n−1 2 2 E(S ) = E S = E(S 2 ) = σ . n n n 9.30 (a) The bias is E(S 2 ) − σ 2 = σ2 n→∞ n

(b) lim Bias = lim n→∞

n−1 2 n σ

− σ 2 = − σn .

= 0.

9.31 (a) E(X) = np; E(Pˆ ) = E(X/n) = E(X)/n = np/n = p. (b) E(P ) = √ np+ n/2 √ n→∞ n+ n

9.32 lim

√ E(X)+ n/2 √ n+ n

√ np+ n/2 √ n+ n

√ p+1/2 n √ n→∞ 1+1/ n

= lim

= p.

= p. (n−1)S 2 σ2

follows a chi-squared distribution with n − 1 2

σ 2 degrees of freedom, whose variance is 2(n − 1). So, V ar(S 2 ) = V ar n−1 X 2 = n−1 σ 4 , and 4 2 = n−1 2 V ar(S 2 ) = 2(n−1)σ . Therefore, the variance of S 2 is V ar(S 2 ) = V ar n−1 n S n n2 smaller.

9.33 Using Theorem 8.4, we know that X 2 =

9.34 Using Exercises 9.30 and 9.33, V ar(S 2 ) + [Bias(S 2 )]2 2σ 4 /(n − 1) M SE(S 2 ) = = 2 2 2 2 M SE(S ) V ar(S ) + [Bias(S )] 2(n − 1)σ 4 /n2 + σ 4 /n2 3n − 1 , =1+ 2 2n − 3n + 1 which is always larger than 1 when n is larger than 1. Hence the M SE of S 2 is usually smaller. ¯1 = 80, x ¯2 = 75, σ1 = 5, σ2 = 3, and z0.03 = 1.88. So, a 94% conﬁdence 9.35 n1 = 25, n2 = 36, x interval for μ1 − μ2 is (80 − 75) − (1.88) 25/25 + 9/36 < μ1 − μ2 < (80 − 75) + (1.88) 25/25 + 9/36, which yields 2.9 < μ1 − μ2 < 7.1. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

101

Solutions for Exercises in Chapter 9

¯A = 78.3, x ¯B = 87.2, σA = 5.6, and σB = 6.3. It is known that z0.025 = 9.36 nA = 50, nB = 50, x 1.96. So, a 95% conﬁdence interval for the diﬀerence of the population means is (87.2 − 78.3) ± 1.96 5.62 /50 + 6.32 /50 = 8.9 ± 2.34, or 6.56 < μA − μB < 11.24. ¯1 = 12.2, x ¯2 = 9.1, s1 = 1.1, and s2 = 0.9. It is known that z0.01 = 2.327. 9.37 n1 = 100, n2 = 200, x So (12.2 − 9.1) ± 2.327 1.12 /100 + 0.92 /200 = 3.1 ± 0.30, or 2.80 < μ1 − μ2 < 3.40. The treatment appears to reduce the mean amount of metal removed. 9.38 n1 = 12, n2 = 10, x ¯1 = 85, x ¯2 = 81, s1 = 4, s2 = 5, and sp = 4.478 with t0.05 = 1.725 with 20 degrees of freedom. So (85 − 81) ± (1.725)(4.478)

1/12 + 1/10 = 4 ± 3.31,

which yields 0.69 < μ1 − μ2 < 7.31. ¯1 = 84, x ¯2 = 77, s1 = 4, s2 = 6, and sp = 5.305 with t0.005 = 2.763 with 28 9.39 n1 = 12, n2 = 18, x degrees of freedom. So, (84 − 77) ± (2.763)(5.305)

1/12 + 1/18 = 7 ± 5.46,

which yields 1.54 < μ1 − μ2 < 12.46. ¯1 = 0.399, x ¯2 = 0.565, s1 = 0.07279, s2 = 0.18674, and sp = 0.14172 with 9.40 n1 = 10, n2 = 10, x t0.025 = 2.101 with 18 degrees of freedom. So, (0.565 − 0.399) ± (2.101)(0.14172)

1/10 + 1/10 = 0.166 ± 0.133,

which yields 0.033 < μ1 − μ2 < 0.299. ¯1 = 17, x ¯2 = 19, s21 = 1.5, s22 = 1.8, and sp = 1.289 with t0.005 = 2.763 with 9.41 n1 = 14, n2 = 16, x 28 degrees of freedom. So, (19 − 17) ± (2.763)(1.289)

1/16 + 1/14 = 2 ± 1.30,

which yields 0.70 < μ1 − μ2 < 3.30. ¯1 = 16, x ¯2 = 11, s1 = 1.0, s2 = 0.8, and sp = 0.915 with t0.05 = 1.725 with 9.42 n1 = 12, n2 = 10, x 20 degrees of freedom. So, (16 − 11) ± (1.725)(0.915)

1/12 + 1/10 = 5 ± 0.68,

Chapter 9 One- and Two-Sample Estimation Problems

102

¯A = 36, 300, x ¯B = 38, 100, sA = 5, 000, sB = 6, 100, and 9.43 nA = nB = 12, x v=

(50002 /12 + 61002 /12)2 (50002 /12)2 11

(61002 /12)2 11

= 21,

with t0.025 = 2.080 with 21 degrees of freedom. So, 50002 61002 (36, 300 − 38, 100) ± (2.080) + = −1, 800 ± 4, 736 12 12 which yields −6, 536 < μA − μB < 2, 936. 9.44 n = 8, d¯ = −1112.5, sd = 1454, with t0.005 = 3.499 with 7 degrees of freedom. So, 1454 −1112.5 ± (3.499) √ = −1112.5 ± 1798.7, 8 which yields −2911.2 < μD < 686.2. 9.45 n = 9, d¯ = 2.778, sd = 4.5765, with t0.025 = 2.306 with 8 degrees of freedom. So, 4.5765 = 2.778 ± 3.518, 2.778 ± (2.306) √ 9 which yields −0.74 < μD < 6.30. ¯I = 98.4, x ¯II = 110.7, sI = 8.375, and sII = 32.185, with 9.46 nI = 5, nII = 7, x v=

(8.7352 /5 + 32.1852 /7)2 (8.7352 /5)2 4

(32.1852 /7)2 6

So, t0.05 = 1.895 with 7 degrees of freedom. (110.7 − 98.4) ± 1.895 8.7352 /5 + 32.1852 /7 = 12.3 ± 24.2, which yields −11.9 < μII − μI < 36.5. 9.47 n = 10, d¯ = 14.89%, and sd = 30.4868, with t0.025 = 2.262 with 9 degrees of freedom. So, 30.4868 = 14.89 ± 21.81, 14.89 ± (2.262) √ 10 which yields −6.92 < μD < 36.70. ¯A = 32.91, x ¯B = 30.47, sA = 1.57, sB = 1.74, and Sp = 1.657. 9.48 nA = nB = 20, x (a) t0.025 ≈ 2.042 with 38 degrees of freedom. So, (32.91 − 30.47) ± (2.042)(1.657) 1/20 + 1/20 = 2.44 ± 1.07, which yields 1.37 < μA − μB < 3.51. (b) Since it is apparent that type A battery has longer life, it should be adopted. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

103

Solutions for Exercises in Chapter 9

¯A = 3.82, x ¯B = 4.94, sA = 0.7794, sB = 0.7538, and sp = 0.7667 with 9.49 nA = nB = 15, x t0.025 = 2.048 with 28 degrees of freedom. So, (4.94 − 3.82) ± (2.048)(0.7667) 1/15 + 1/15 = 1.12 ± 0.57, which yields 0.55 < μB − μA < 1.69. 9.50 n1 = 8, n2 = 13, x ¯1 = 1.98, x ¯2 = 1.30, s1 = 0.51, s2 = 0.35, and sp = 0.416. t0.025 = 2.093 with 19 degrees of freedom. So, (1.98 − 1.30) ± (2.093)(0.416) 1/8 + 1/13 = 0.68 ± 0.39, which yields 0.29 < μ1 − μ2 < 1.07. 9.51 n = 1000, pˆ =

228 1000

= 0.228, qˆ = 0.772, and z0.005 = 2.575. So, using method 1, 0.228 ± (2.575)

(0.228)(0.772) = 0.228 ± 0.034, 1000

which yields 0.194 < p < 0.262. When we use method 2, we have 2.575 0.228 + 2.5752 /2000 ± 1 + 2.5752 /1000 1 + 2.5752 /1000

2.5752 (0.228)(0.772) + = 0.2298 ± 0.0341, 1000 4(1000)2

which yields an interval of (0.1957, 0.2639). 9.52 n = 100, pˆ =

8 100

= 0.08, qˆ = 0.92, and z0.01 = 2.33. So, 0.08 ± (2.33)

(0.08)(0.92) = 0.08 ± 0.063, 100

which yields 0.017 < p < 0.143, for method 1. Now for method 2, we have 2.33 0.08 + 2.332 /200 ± 2 1 + 2.33 /100 1 + 2.332 /100

2.332 (0.08)(0.92) + = 0.1016 ± 0.0652, 100 4(100)2

which yields (0.0364, 0.1669). 9.53 (a) n = 200, pˆ = 0.57, qˆ = 0.43, and z0.02 = 2.05. So, 0.57 ± (2.05)

(0.57)(0.43) = 0.57 ± 0.072, 200

Chapter 9 One- and Two-Sample Estimation Problems

104 9.54 n = 500.ˆ p=

485 500

= 0.97, qˆ = 0.03, and z0.05 = 1.645. So, 0.97 ± (1.645)

(0.97)(0.03) = 0.97 ± 0.013, 500

which yields 0.957 < p < 0.983. 9.55 (a) n = 40, pˆ =

34 40

= 0.85, qˆ = 0.15, and z0.025 = 1.96. So, 0.85 ± (1.96)

(0.85)(0.15) = 0.85 ± 0.111, 40

which yields 0.739 < p < 0.961. (b) Since p = 0.8 falls in the conﬁdence interval, we can not conclude that the new system is better. 9.56 n = 100, pˆ =

24 100

= 0.24, qˆ = 0.76, and z0.005 = 2.575. (a) 0.24 ± (2.575) (0.24)(0.76) = 0.24 ± 0.110, which yields 0.130 < p < 0.350. 100 (b) Error ≤ (2.575) (0.24)(0.76) = 0.110. 100

9.57 n = 1600, pˆ = 23 , qˆ = 13 , and z0.025 = 1.96. 2 (a) 3 ± (1.96) (2/3)(1/3) = 23 ± 0.023, which yields 0.644 < p < 0.690. 1600 (b) Error ≤ (1.96) (2/3)(1/3) = 0.023. 1600 9.58 n =

(1.96)2 (0.32)(0.68) (0.02)2

= 2090 when round up.

9.59 n =

(2.05)2 (0.57)(0.43) (0.02)2

= 2576 when round up.

9.60 n =

(2.575)2 (0.228)(0.772) (0.05)2

9.61 n =

(2.33)2 (0.08)(0.92) (0.05)2

9.62 n =

(2.575)2 (4)(0.01)2

= 16577 when round up.

9.63 n =

(1.96)2 (4)(0.01)2

= 9604 when round up.

9.64 n =

(1.96)2 (4)(0.04)2

= 601 when round up.

= 467 when round up.

= 160 when round up.

9.65 nM = nF = 1000, pˆM = 0.250, qˆM = 0.750, pˆF = 0.275, qˆF = 0.725, and z0.025 = 1.96. So (0.250)(0.750) (0.275)(0.725) + = 0.025 ± 0.039, (0.275 − 0.250) ± (1.96) 1000 1000 which yields −0.014 < pF − pM < 0.064. c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

105

Solutions for Exercises in Chapter 9

9.66 n1 = 250, n2 = 175, pˆ1 =

80 250

= 0.32, pˆ2 =

(0.32 − 0.2286) ± (1.645)

40 175

= 0.2286, and z0.05 = 1.645. So,

(0.32)(0.68) (0.2286)(0.7714) + = 0.0914 ± 0.0713, 250 175

which yields 0.0201 < p1 − p2 < 0.1627. From this study we conclude that there is a signiﬁcantly higher proportion of women in electrical engineering than there is in chemical engineering. 9.67 n1 = n2 = 500, pˆ1 =

120 500

= 0.24, pˆ2 =

(0.24 − 0.196) ± (1.645)

98 500

= 0.196, and z0.05 = 1.645. So,

(0.24)(0.76) (0.196)(0.804) + = 0.044 ± 0.0429, 500 500

which yields 0.0011 < p1 − p2 < 0.0869. Since 0 is not in this conﬁdence interval, we conclude, at the level of 90% conﬁdence, that inoculation has an eﬀect on the incidence of the disease. 9.68 n5◦ C = n15◦ C = 20, pˆ5◦ C = 0.50, pˆ15◦ C = 0.75, and z0.025 = 1.96. So, (0.50)(0.50) (0.75)(0.25) + = −0.25 ± 0.2899, (0.5 − 0.75) ± (1.96) 20 20 which yields −0.5399 < p5◦ C − p15◦ C < 0.0399. Since this interval includes 0, the signiﬁcance of the diﬀerence cannot be shown at the conﬁdence level of 95%. 9.69 nnow = 1000, pˆnow = 0.2740, n91 = 760, pˆ91 = 0.3158, and z0.025 = 1.96. So, (0.2740)(0.7260) (0.3158)(0.6842) + = −0.0418 ± 0.0431, (0.2740 − 0.3158) ± (1.96) 1000 760 which yields −0.0849 < pnow − p91 < 0.0013. Hence, at the conﬁdence level of 95%, the signiﬁcance cannot be shown. 9.70 n90 = n94 = 20, pˆ90 = 0.337, and ˆ094 = 0.362 (a) n90 pˆ90 = (20)(0.337) ≈ 7 and n94 pˆ94 = (20)(0.362) ≈ 7. (b) Since z0.025 = 1.96, (0.337−0.362)±(1.96) (0.337)(0.663) + (0.362)(0.638) = −0.025±0.295, 20 20 which yields −0.320 < p90 − p94 < 0.270. Hence there is no evidence, at the conﬁdence level of 95%, that there is a change in the proportions. 9.71 s2 = 0.815 with v = 4 degrees of freedom. Also, χ20.025 = 11.143 and χ20.975 = 0.484. So, (4)(0.815) (4)(0.815) < σ2 < , 11.143 0.484

which yields

0.293 < σ 2 < 6.736.

Since this interval contains 1, the claim that σ 2 seems valid. 9.72 s2 = 16 with v = 19 degrees of freedom. It is known χ20.01 = 36.191 and χ20.99 = 7.633. Hence (19)(16) (19)(16) < σ2 < , or 8.400 < σ 2 < 39.827. 36.191 7.633 c Copyright 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Chapter 9 One- and Two-Sample Estimation Problems

106

9.73 s2 = 6.0025 with v = 19 degrees of freedom. Also, χ20.025 = 32.852 and χ20.975 = 8.907. Hence, (19)(6.0025) (19)(6.0025) < σ2 < , or 3.472 < σ 2 < 12.804, 32.852 8.907 9.74 s2 = 0.0006 with v = 8 degrees of freedom. Also, χ20.005 = 21.955 and χ20.995 = 1.344. Hence, (8)(0.0006) (8)(0.0006) < σ2 < , or 0.00022 < σ 2 < 0.00357. 21.955 1.344 9.75 s2 = 225 with v = 9 degrees of freedom. Also, χ20.005 = 23.589 and χ20.995 = 1.735. Hence, (9)(225) (9)(225) < σ2 < , or 85.845 < σ 2 < 1167.147, 23.589 1.735 which yields 9.27 < σ < 34.16. 9.76 s2 = 2.25 with v = 11 degrees of freedom. Also, χ20.05 = 19.675 and χ20.95 = 4.575. Hence, (11)(2.25) (11)(2.25) < σ2 < , or 1.258 < σ 2 < 5.410. 19.675 4.575 9.77 s21 = 1.00, s22 = 0.64, f0.01 (11, 9) = 5.19, and f0.01 (9, 11) = 4.63. So, σ2 σ2 1.00/0.64 < 12 < (1.00/0.64)(4.63), or 0.301 < 12 < 7.234, 5.19 σ2 σ2 which yields 0.549

Probability and statistics for engineers and scientists answers

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