Skip to content
Textbook Guides

Solution Manual For Probability And Statistics For Engineers And Scientists, 9th Edition

Struggling with textbook problems? Let Solution Manual For Probability And Statistics For Engineers And Scientists, 9th Edition guide you through with clear, step-by-step solutions.

Preview 16 of 256 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    Contents 1 Introduction to Statistics and Data Analysis 1 2 Probability 11 3 Random Variables and Probability Distributions 27 4 Mathematical Expectation 41 5 Some Discrete Probability Distributions 55 6 Some Continuous Probability Distributions 67 7 Functions of Random Variables 79 8 Fundamental Sampling Distributions and Data Descriptions 85 9 One- and Two-Sample Estimation Problems 97 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 2 k Factorial Experiments and Fractions 219 16 Nonparametric Statistics 233 17 Statistical Quality Control 247 18 Bayesian Statistics 251 iii

  2. Page 2
  3. Page 3
    Read page 3 as text

    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 3.0 3.3 3.4 3.6 3.7 4.0 4.4 4.8 So. the trimmed mean is ¯ x tr20 = 1 9 (2 . 9 + 3 . 0 + · · · + 4 . 8) = 3 . 678 . (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. 18 19 20 21 22 23 (d) No. They are all close to each other. 1

  4. Page 4
    Read page 4 as text

    2 Chapter 1 Introduction to Statistics and Data Analysis 1.3 (a) A dot plot is shown below. 200 205 210 215 220 225 230 In the figure, “ × ” represents the “No aging” group and “ ◦ ” represents the “Aging” group. (b) Yes; tensile strength is greatly reduced due to the aging process. (c) Mean Aging = 209 . 90, and Mean No aging = 222 . 10. (d) Median Aging = 210 . 00, and Median No aging = 221 . 50. The means and medians for each group are similar to each other. 1.4 (a) ¯ X A = 7 . 950 and ˜ X A = 8 . 250; ¯ X B = 10 . 260 and ˜ X B = 10 . 150. (b) A dot plot is shown below. 6.5 7.5 8.5 9.5 10.5 11.5 In the figure, “ × ” represents company A and “ ◦ ” represents company B . The steel rods made by company B show more flexibility. 1.5 (a) A dot plot is shown below. −10 0 10 20 30 40 In the figure, “ × ” represents the control group and “ ◦ ” represents the treatment group. (b) ¯ X Control = 5 . 60, ˜ X Control = 5 . 00, and ¯ X tr(10);Control = 5 . 13; ¯ X Treatment = 7 . 60, ˜ X Treatment = 4 . 50, and ¯ X tr(10);Treatment = 5 . 63. (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. (b) ¯ X 20 ◦ C = 2 . 1075, and ¯ X 45 ◦ C = 2 . 2350. (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 temper- ature is increased. 1.7 s 2 = 1 15 − 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 = √ 0 . 9428 = 0 . 971.

  5. Page 5
    Read page 5 as text

    Solutions for Exercises in Chapter 1 3 1.8 s 2 = 1 20 − 1 [(18 . 71 − 20 . 7675) 2 + (21 . 41 − 20 . 7675) 2 + · · · + (21 . 12 − 20 . 7675) 2 ] = 2 . 5329; s = √ 2 . 5345 = 1 . 5915. 1.9 (a) s 2 No Aging = 1 10 − 1 [(227 − 222 . 10) 2 + (222 − 222 . 10) 2 + · · · + (221 − 222 . 10) 2 ] = 23 . 66; s No Aging = √ 23 . 62 = 4 . 86. s 2 Aging = 1 10 − 1 [(219 − 209 . 90) 2 + (214 − 209 . 90) 2 + · · · + (205 − 209 . 90) 2 ] = 42 . 10; s Aging = √ 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 : s 2 A = 1 . 2078 and s A = √ 1 . 2072 = 1 . 099. For company B : s 2 B = 0 . 3249 and s B = √ 0 . 3249 = 0 . 570. 1.11 For the control group: s 2 Control = 69 . 38 and s Control = 8 . 33. For the treatment group: s 2 Treatment = 128 . 04 and s Treatment = 11 . 32. 1.12 For the cure temperature at 20 ◦ C: s 2 20 ◦ C = 0 . 005 and s 20 ◦ C = 0 . 071. For the cure temperature at 45 ◦ C: s 2 45 ◦ 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 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 ∑ i =1 x i − n ¯ x = n ∑ i =1 x i − n ∑ i =1 x i = 0 , which is the term on the right side. 1.17 (a) ¯ X smokers = 43 . 70 and ¯ X nonsmokers = 30 . 32; (b) s smokers = 16 . 93 and s nonsmokers = 7 . 13; (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.

  6. Page 6
    Read page 6 as text

    4 Chapter 1 Introduction to Statistics and Data Analysis 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 14 . 5 24 . 5 34 . 5 44 . 5 54 . 5 64 . 5 74 . 5 84 . 5 94 . 5 3 2 3 4 5 11 14 14 4 0 . 05 0 . 03 0 . 05 0 . 07 0 . 08 0 . 18 0 . 23 0 . 23 0 . 07 (c) A histogram plot is given below. 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 Final Exam Grades Relative Frequency The distribution skews to the left. (d) ¯ X = 65 . 48, ˜ X = 71 . 50 and s = 21 . 13. 1.19 (a) A stem-and-leaf plot is shown below. Stem Leaf Frequency 0 22233457 8 1 023558 6 2 035 3 3 03 2 4 057 3 5 0569 4 6 0005 4

  7. Page 7
    Read page 7 as text

    Solutions for Exercises in Chapter 1 5 (b) The following is the relative frequency distribution table. Relative Frequency Distribution of Years Class Interval Class Midpoint Frequency, f Relative Frequency 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 0 . 45 1 . 45 2 . 45 3 . 45 4 . 45 5 . 45 6 . 45 8 6 3 2 3 4 4 0 . 267 0 . 200 0 . 100 0 . 067 0 . 100 0 . 133 0 . 133 (c) ¯ X = 2 . 797, s = 2 . 227 and Sample range is 6 . 5 − 0 . 2 = 6 . 3. 1.20 (a) A stem-and-leaf plot is shown next. Stem Leaf Frequency 0* 34 2 0 56667777777889999 17 1* 0000001223333344 16 1 5566788899 10 2* 034 3 2 7 1 3* 2 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 5 − 9 10 − 14 15 − 19 20 − 24 25 − 29 30 − 34 2 7 12 17 22 27 32 2 17 16 10 3 1 1 0 . 04 0 . 34 0 . 32 0 . 20 0 . 06 0 . 02 0 . 02 (c) A histogram plot is shown next. 2 7 12 17 22 27 32 Fruit fly lives (seconds) Relative Frequency (d) ˜ X = 10 . 50.

  8. Page 8
    Read page 8 as text

    6 Chapter 1 Introduction to Statistics and Data Analysis 1.21 (a) ¯ X = 74 . 02 and ˜ X = 78; (b) s = 39 . 26. 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 6.82 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 395.10 160.15 (b) ¯ X 1980 = 395 . 1 and ¯ X 1990 = 160 . 2. (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 3.9 Salaries Relative Frequency

  9. Page 9
    Read page 9 as text

    Solutions for Exercises in Chapter 1 7 (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 (d) ¯ X tr(10) = 30 . 97. This trimmed mean is in the middle of the mean and median using the 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. 1.27 (a) The averages of the wear are plotted here. 700 800 900 1000 1100 1200 1300 250 300 350 load wear (b) When the load value increases, the wear value also increases. It does show certain relationship.

  10. Page 10
    Read page 10 as text

    8 Chapter 1 Introduction to Statistics and Data Analysis (c) A plot of wears is shown next. 700 800 900 1000 1100 1200 1300 100 300 500 700 load wear (d) The relationship between load and wear in (c) is not as strong as the case in (a), especially for the load at 1300. One reason is that there is an extreme value (750) which influence the mean value at the load 1300. 1.28 (a) A dot plot is shown next. 71.45 71.65 71.85 72.05 72.25 72.45 72.65 72.85 73.05 Low High 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. 1.29 A box plot is shown next. 2.0 2.5 3.0 3.5

  11. Page 11
    Read page 11 as text

    Solutions for Exercises in Chapter 1 9 1.30 A box plot plot is shown next. 700 800 900 1000 1100 1200 1300 1.31 (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 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. 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 experimental data, those two variables can be controlled each at two levels, the interaction can be inves-

  12. Page 12
    Read page 12 as text

    10 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.

  13. Page 13
    Read page 13 as text

    Chapter 2 Probability 2.1 (a) S = { 8 , 16 , 24 , 32 , 40 , 48 } . (b) For x 2 + 4 x − 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 2 x − 4 ≥ 0 gives x ≥ 2. Since we must also have x < 1, it follows that S = φ . 2.2 S = { ( x, y ) | x 2 + y 2 < 9; x ≥ 0 , y ≥ 0 } . 2.3 (a) A = { 1 , 3 } . (b) B = { 1 , 2 , 3 , 4 , 5 , 6 } . (c) C = { x | x 2 − 4 x + 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 = { 1 HH, 1 HT, 1 T H, 1 T T, 2 H, 2 T, 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 } . 2.6 S = { A 1 A 2 , A 1 A 3 , A 1 A 4 , A 2 A 3 , A 2 A 4 , A 3 A 4 } . 2.7 S 1 = { 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 } . S 2 = { 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) } . 11

  14. Page 14
    Read page 14 as text

    12 Chapter 2 Probability (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 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 = { 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 } . (g) A A B B C S ∩

  15. Page 15
    Read page 15 as text

    Solutions for Exercises in Chapter 2 13 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. S P F Sample Space 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.

  16. Page 16
    Read page 16 as text

    14 Chapter 2 Probability A B S 1 2 3 4 (a) From the above Venn diagram, ( A ∩ B ) ′ contains the regions of 1, 2 and 4. (b) ( A ∪ B ) ′ contains region 1. (c) A Venn diagram is shown next. A B C S 1 2 3 4 5 6 7 8 ( A ∩ C ) ∪ B contains the regions of 3, 4, 5, 7 and 8. 2.18 (a) Not mutually exclusive. (b) Mutually exclusive. (c) Not mutually exclusive. (d) Mutually exclusive. 2.19 (a) The family will experience mechanical problems but will receive no ticket 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 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.

240 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Statistics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.