Solucionario Probabilidad Y Estadistica Para Ingenieria Y Ciencias Walpole

INSTRUCTOR’S SOLUTION MANUAL
KEYING YE AND SHARON MYERS

for PROBABILITY & STATISTICS
FOR ENGINEERS & SCIENTISTS

EIGHTH EDITION

WALPOLE, MYERS, MYERS, YE

Contents
1 Introduction to Statistics and Data Analysis 2 Probability 3 Random Variables and Probability Distributions 4 Mathematical Expectation 5 Some Discrete Probability Distributions 6 Some Continuous ProbabilityDistributions 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 11 Simple Linear Regression and Correlation 12 Multiple Linear Regression and Certain Nonlinear Regression Models 13 One-Factor Experiments: General 14 Factorial Experiments (Two or More Factors) 15 2k FactorialExperiments and Fractions 16 Nonparametric Statistics iii 1 11 29 45 59 71 85 91 103 121 149 171 185 213 237 257

iv 17 Statistical Quality Control 18 Bayesian Statistics

CONTENTS 273 277

Chapter 1 Introduction to Statistics and Data Analysis
1.1 (a) 15. (b) x = ¯
1 (3.4 15

(c) Sample median is the 8th value, after the data is sorted from smallest to largest: 3.6. (d) A dot plot is shownbelow.

+ 2.5 + 4.8 + · · · + 4.8) = 3.787.

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 So. the trimmed mean is 1 xtr20 = (2.9 + 3.0 + · · · + 4.8) = 3.678. ¯ 9 1.2 (a) Mean=20.768 and Median=20.610. (b) xtr10 = 20.743. ¯ (c) A dot plot is shown below. 3.0 3.3 3.4 3.6 4.0 4.4 4.8

18

1920

21

22

23

1

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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) MeanAging = 209.90, and MeanNo aging = 222.10. (d) MedianAging = 210.00, andMedianNo aging = 221.50. The means and medians for each group are similar to each other. ¯ ˜ 1.4 (a) XA = 7.950 and XA = 8.250; ¯ ˜ XB = 10.260 and XB = 10.150. (b) A dot plot is shown below.
6.5 7.5 8.5 9.5 10.5 11.5

In the figure, “×” represents company A and “◦” represents company B. The steel rods made by company B show more flexibility. 1.5 (a) A dot plot is shown below.

−10

0

1020

30

40

In the figure, “×” represents the control group and “◦” represents the treatment group. ¯ ˜ ¯ (b) XControl = 5.60, XControl = 5.00, and Xtr(10);Control = 5.13; ¯ ˜ ¯ XTreatment = 7.60, XTreatment = 4.50, and Xtr(10);Treatment = 5.63. (c) The difference of the means is 2.0 and the differences of the medians and the trimmed means are 0.5, which are much smaller. The possible causeof 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) X20◦ C = 2.1075, and X45◦ 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.

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 1.7 s2 = 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 = s2 = 0.9428 = 0.971.1 1.8 s2 = 20−1 [(18.71 − 20.768)2 + (21.41 − 20.768)2 + · · · + (21.12 − 20.768)2 ] = 2.5345; √ s = 2.5345 = 1.592. 1 1.9 s2 Aging = 10−1 [(227 − 222.10)2 + (222 − 222.10)2 + · · · + (221 − 222.10)2 ] = 42.12; No √ sNo Aging = 42.12 = 6.49. 1 2 2 2 s2 Aging = √ [(219 − 209.90) + (214 − 209.90) + · · · + (205 − 209.90) ] = 23.62; 10−1 sAging = 23.62 = 4.86. √ 1.10 For company A: s2 = 1.2078 and...