Solucionario Probabilidad Y Estadistica En Ingles
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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