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INSTRUCTOR’S

SOLUTION MANUAL

KEYING YE AND SHARON MYERS

for

PROBABILITY & STATISTICS

FOR ENGINEERS & SCIENTISTS

EIGHTH EDITION

WALPOLE, MYERS, MYERS, YE

iv CONTENTS

17 Statistical Quality Control 273

18 Bayesian Statistics 277

Chapter 1

Introduction to Statistics and Data

Analysis

1.1 (a) 15. (b) ¯x = 151 (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.

(e) After trimming total 40% of the data (20% highest and 20% lowest), the data becomes: 2.9 3.0 3.3 3.4 3. 3.7 4.0 4.4 4.

So. the trimmed mean is

x¯tr20 =

1.2 (a) Mean=20.768 and Median=20.610. (b) ¯xtr10 = 20.743. (c) A dot plot is shown below.

18 19 20 21 22 23

Solutions for Exercises in Chapter 1 3

(d) It also seems that the variation of the tensile strength gets larger when the cure temperature is increased.

1.7 s^2 = (^151) − 1 [(3. 4 − 3 .787)^2 +(2. 5 − 3 .787)^2 +(4. 8 − 3 .787)^2 +· · ·+(4. 8 − 3 .787)^2 ] = 0.94284; s =

s^2 =

1.8 s^2 = (^201) − 1 [(18. 71 − 20 .768)^2 + (21. 41 − 20 .768)^2 + · · · + (21. 12 − 20 .768)^2 ] = 2.5345; s =

1.9 s^2 No Aging = (^101) − 1 [(227 − 222 .10)^2 + (222 − 222 .10)^2 + · · · + (221 − 222 .10)^2 ] = 42.12; sNo Aging =

s^2 Aging = (^101) − 1 [(219 − 209 .90)^2 + (214 − 209 .90)^2 + · · · + (205 − 209 .90)^2 ] = 23.62; sAging =

1.10 For company A: s^2 A = 1.2078 and sA =

For company B: s^2 B = 0.3249 and sB =

1.11 For the control group: s^2 Control = 69.39 and sControl = 8.33. For the treatment group: s^2 Treatment = 128.14 and sTreatment = 11.32.

1.12 For the cure temperature at 20◦C: s^220 ◦C = 0.005 and s 20 ◦C = 0.071. For the cure temperature at 45◦C: s^245 ◦C = 0.0413 and s 45 ◦C = 0.2032. The variation of the tensile strength is influenced by the increase of cure temperature.

1.13 (a) Mean = X¯ = 124.3 and median = X˜ = 120; (b) 175 is an extreme observation.

1.14 (a) Mean = X¯ = 570.5 and median = X˜ = 571; (b) Variance = s^2 = 10; standard deviation= s = 3.162; range=10; (c) Variation of the diameters seems too big.

1.15 Yes. The value 0.03125 is actually a P -value and a small value of this quantity 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

i=

xi − nx¯ =

∑^ n

i=

xi −

∑^ n

i=

xi = 0,

which is the term on the right side.

1.17 (a) X¯smokers = 43.70 and X¯nonsmokers = 30.32; (b) ssmokers = 16.93 and snonsmokers = 7.13;

4 Chapter 1 Introduction to Statistics and Data Analysis

(c) A dot plot is shown below.

10 20 30 40 50 60 70 In the figure, “×” represents the nonsmoker group and “◦” represents the 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. Stem Leaf Frequency 1 057 3 2 35 2 3 246 3 4 1138 4 5 22457 5 6 00123445779 11 7 01244456678899 14 8 00011223445589 14 9 0258 4

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

(c) A histogram plot is given below.

14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94. Final Exam Grades

Relative Frequency

6 Chapter 1 Introduction to Statistics and Data Analysis

Relative Frequency Distribution of Fruit Fly Lives Class Interval Class Midpoint Frequency, f Relative Frequency 0 − 4 5 − 9 10 − 14 15 − 19 20 − 24 25 − 29 30 − 34

(c) A histogram plot is shown next.

2 7 12 17 22 27 32 Fruit fly lives (seconds)

Relative Frequency

(d) X˜ = 10.50.

1.21 (a) X¯ = 1.7743 and X˜ = 1.7700; (b) s = 0.3905.

1.22 (a) X¯ = 6.7261 and X˜ = 0.0536. (b) A histogram plot is shown next.

6.62 6.66 6.7 6.74 6.78 6. Relative Frequency Histogram for Diameter

(c) The data appear to be skewed to the left.

1.23 (a) A dot plot is shown next.

0 100 200 300 400 500 600 700 800 900 1000

160.15 395.

(b) X¯ 1980 = 395.1 and X¯ 1990 = 160.2.

Solutions for Exercises in Chapter 1 7

(c) The sample mean for 1980 is over twice as large as that of 1990. The 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.

1.24 (a) X¯ = 2.8973 and s = 0.5415. (b) A histogram plot is shown next.

1.8 2.1 2.4 2.7 3 3.3 3.6 3. Salaries

Relative Frequency

(c) Use the double-stem-and-leaf plot, we have the following.

Stem Leaf Frequency 1 (84) 1 2* (05)(10)(14)(37)(44)(45) 6 2 (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) 11 3* (10)(13)(14)(22)(36)(37) 6 3 (51)(54)(57)(71)(79)(85) 6

1.25 (a) X¯ = 33.31; (b) X˜ = 26.35; (c) A histogram plot is shown next.

10 20 30 40 50 60 70 80 90 Percentage of the families

Relative Frequency

Solutions for Exercises in Chapter 1 9

(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.

1.29 (a) A dot plot is shown next.

76 79 82 85 88 91 94

Low High

In the figure, “×” represents the low-injection-velocity group and “◦” 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 tem- perature and high mold temperature, the apparent interactions between the mold temperature and injection velocity are significant.

1.30 An interaction plot is shown next.

Low high^

injection velocity

low mold temp

high mold temp

mean shrinkage value

It is quite obvious to find the interaction between the two variables. Since in this exper- imental data, those two variables can be controlled each at two levels, the interaction can be investigated. 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.

12 Chapter 2 Probability

(b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6)}.

(c) C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

(d) A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}. (e) A ∩ B = φ.

(f) B ∩ C = {(5, 2), (6, 2)}.

(g) A Venn diagram is shown next.

A

A C

B

B C

C

S

∩

∩

2.9 (a) A = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T }.

(b) B = { 1 T T, 3 T T, 5 T T }.

(c) A′^ = { 3 HH, 3 HT, 3 T H, 3 T T, 4 H, 4 T, 5 HH, 5 HT, 5 T H, 5 T T, 6 H, 6 T }. (d) A′^ ∩ B = { 3 T T, 5 T T }.

(e) A ∪ B = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T, 3 T T, 5 T T }.

2.10 (a) S = {F F F, F F N, F NF, NF F, F NN, NF N, NNF, NNN}.

(b) E = {F F F, F F N, F NF, NF F }.

(c) The second river was safe for fishing.

2.11 (a) S = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 , F 1 M 1 , F 1 M 2 , F 1 F 2 , F 2 M 1 , F 2 M 2 , F 2 F 1 }.

(b) A = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 }.

(c) B = {M 1 F 1 , M 1 F 2 , M 2 F 1 , M 2 F 2 , F 1 M 1 , F 1 M 2 , F 2 M 1 , F 2 M 2 }. (d) C = {F 1 F 2 , F 2 F 1 }.

(e) A ∩ B = {M 1 F 1 , M 1 F 2 , M 2 F 1 , M 2 F 2 }.

(f) A ∪ C = {M 1 M 2 , M 1 F 1 , M 1 F 2 , M 2 M 1 , M 2 F 1 , M 2 F 2 , F 1 F 2 , F 2 F 1 }.

Solutions for Exercises in Chapter 2 13

(g)

A

A B

B

C

S

∩

2.12 (a) S = {ZY F, ZNF, W Y F, W NF, SY F, SNF, ZY M}. (b) A ∪ B = {ZY F, ZNF, W Y F, W NF, SY F, SNF } = A. (c) A ∩ B = {W Y F, SY F }.

2.13 A Venn diagram is shown next.

S

P

F

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}.

Solutions for Exercises in Chapter 2 15

(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, 6, 8.

2.21 With n 1 = 6 sightseeing tours each available on n 2 = 3 different days, the multiplication rule gives n 1 n 2 = (6)(3) = 18 ways for a person to arrange a tour.

2.22 With n 1 = 8 blood types and n 2 = 3 classifications of blood pressure, the multiplication rule gives n 1 n 2 = (8)(3) = 24 classifications.

2.23 Since the die can land in n 1 = 6 ways and a letter can be selected in n 2 = 26 ways, the multiplication rule gives n 1 n 2 = (6)(26) = 156 points in S.

2.24 Since a student may be classified according to n 1 = 4 class standing and n 2 = 2 gender classifications, the multiplication rule gives n 1 n 2 = (4)(2) = 8 possible classifications for the students.

2.25 With n 1 = 5 different shoe styles in n 2 = 4 different colors, the multiplication rule gives n 1 n 2 = (5)(4) = 20 different pairs of shoes.

2.26 Using Theorem 2.8, we obtain the followings.

(a) There are

5

= 21 ways. (b) There are

3

= 10 ways.

2.27 Using the generalized multiplication rule, there are n 1 ×n 2 ×n 3 ×n 4 = (4)(3)(2)(2) = 48 different house plans available.

2.28 With n 1 = 5 different manufacturers, n 2 = 3 different preparations, and n 3 = 2 different strengths, the generalized multiplication rule yields n 1 n 2 n 3 = (5)(3)(2) = 30 different ways to prescribe a drug for asthma.

2.29 With n 1 = 3 race cars, n 2 = 5 brands of gasoline, n 3 = 7 test sites, and n 4 = 2 drivers, the generalized multiplication rule yields (3)(5)(7)(2) = 210 test runs.

2.30 With n 1 = 2 choices for the first question, n 2 = 2 choices for the second question, and so forth, the generalized multiplication rule yields n 1 n 2 · · · n 9 = 2^9 = 512 ways to answer the test.

16 Chapter 2 Probability

2.31 (a) With n 1 = 4 possible answers for the first question, n 2 = 4 possible answers for the second question, and so forth, the generalized multiplication rule yields 45 = 1024 ways to answer the test. (b) With n 1 = 3 wrong answers for the first question, n 2 = 3 wrong answers for the second question, and so forth, the generalized multiplication rule yields

n 1 n 2 n 3 n 4 n 5 = (3)(3)(3)(3)(3) = 3^5 = 243

ways to answer the test and get all questions wrong.

2.32 (a) By Theorem 2.3, 7! = 5040. (b) Since the first letter must be m, the remaining 6 letters can be arranged in 6! = 720 ways.

2.33 Since the first digit is a 5, there are n 1 = 9 possibilities for the second digit and then n 2 = 8 possibilities for the third digit. Therefore, by the multiplication rule there are n 1 n 2 = (9)(8) = 72 registrations to be checked.

2.34 (a) By Theorem 2.3, there are 6! = 720 ways. (b) A certain 3 persons can follow each other in a line of 6 people in a specified order 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.

2.35 The first house can be placed on any of the n 1 = 9 lots, the second house on any of the remaining n 2 = 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)(5)(5) = 150 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.