Solucionario de estadistica y probabilidad de walpole edicion 8

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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

Contents
1 Introduction to Statistics and Data Analysis

1

2 Probability

11

3 Random Variables and Probability Distributions

29

4 Mathematical Expectation

45

5 Some Discrete Probability Distributions59

6 Some Continuous Probability Distributions

71

7 Functions of Random Variables

85

8 Fundamental Sampling Distributions and Data Descriptions

91

9 One- and Two-Sample Estimation Problems

103

10 One- and Two-Sample Tests of Hypotheses

121

11 Simple Linear Regression and Correlation

149

12 Multiple Linear Regression and Certain Nonlinear Regression Models171

13 One-Factor Experiments: General

185

14 Factorial Experiments (Two or More Factors)

213

15 2k Factorial Experiments and Fractions

237

16 Nonparametric Statistics

257
iii

iv

CONTENTS

17 Statistical Quality Control

273

18 Bayesian Statistics

277

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

1
(3.4
15

+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 3.3 3.4 3.6
4.0 4.4 4.8

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.

18

19

20

21

1

22

23

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) 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.
¯
˜
1.4 (a) XA = 7.950 and XA = 8.250;
¯
˜
XB = 10.260 and XB = 10.150.
(b) A dot plot is shown below.
7.5

6.5

8.5

9.5

10.5

11.5In 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) 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 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) 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.

3

Solutions for Exercises in Chapter 1

(d) It alsoseems that the variation of the tensile strength gets larger when the cure
temperature is increased.
1.7 s2 = 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 = s2 = 0.9428 = 0.971.
1.8 s2 = 201 1 [(18.71 − 20.768)2 + (21.41 − 20.768)2 + · · · + (21.12 − 20.768)2 ] = 2.5345;
√−
s = 2.5345 = 1.592.
1.9 s2 Aging = 101 1 [(227 − 222.10)2 +...
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