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Probability & Statistics Solutions Manual for Engineers

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rights protections.#### 8.2 **War and Military Interventions**Military interventions, whether for
humanitarian reasons or in response to threats, pose significant ethical and legal challenges. The question of
whether military force is justified to prevent atrocities or maintain national security is hotly debated. The legal
framework governing the use of force is established under the United Nations Charter, which permits
intervention under certain conditions (e.g., self-defense or Security Council authorization).However, ethical
dilemmas arise when military actions result in civilian casualties or unintended consequences. For example,
drone strikes, though often used to target terrorist leaders, raise concerns about extrajudicial killings, civilian
harm, and sovereignty violations. The legal justification for such actions is often debated, and



INSTRUCTOR’S
SOLUTIONS MANUAL
KEYING YE SHARON L. MYERS
The University of Texas at San Antonio Radford University




PROBABILITY & STATISTICS FOR
ENGINEERS & SCIENTISTS
MYSTATLAB UPDATE
N INTH E DITION


Ronald E. Walpole
Roanoke University

Raymond H. Myers
Virginia Tech

Sharon L. Myers
Radford University

Keying Ye
ISBN-13: 978-0-13-411622-8

,ISBN-10: 0-13-411622-4




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 2k Factorial Experiments and Fractions 219

16 Nonparametric Statistics 233

17 Statistical Quality Control 247

18 Bayesian Statistics 251

,Chapter 1

Introduction to Statistics and Data
Analysis

1.1 (a) 15.

15(3.4 + 2.5 + 4.8 + · · · + 4.8)
1
(b) x̄ = = 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
1
x̄ t r 2 0 = (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) x̄ t r 1 0 = 20.743.
(c) A dot plot is shown below.


18 19 20 21 22 23

(d) No. They are all close to each other.

Copyright c⃝ 2012 Pearson Education, Inc. Publishing as Prentice Hall.

1

, 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.

1.4 (a) X̄ A = 7.950 and X˜ A = 8.250;
X̄ B = 10.260 and X̃ 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) X̄ Control = 5.60, X̃ Control = 5.00, and X̄tr(10);Control = 5.13;
X̄ Treatment = 7.60, X̃ Treatment = 4.50, and X̄tr(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) X̄ 20 ◦ C = 2.1075, and X̄ 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 s2 = 1
[(3.4 − 3.787)2 + (2.5 − 3.787)2 + (4.8 − 3.787)2 + · · · + (4.8 − 3.787)2] = 0.94284;
15−1
s= s2 = 0.9428 = 0.971.

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