Instructor s solutions manual for probability and statistics for engineers and scientists 9th edition solution notes
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Probability, Statistics, And Data
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Probability, Statistics, And Data
Instructor s solutions manual for probability and statistics for engineers and scientists 9th edition solution notes
Instructor s solutions manual for probability and statistics for engineers and scientists 9th edition solution notes
Instructor s solutions manual for probability and statistics fo...
1 Introduction z to z Statistics z and z Data z Analysis 1
2 Probability 11
3 Random z Variables z and z Probability z Distributions 27
4 Mathematical z z Expectation 41
5 Some z Discrete z Probability z Distributions 55
6 Some z Continuous z Probability z Distributions 67
7 Functions z of z Random z Variables 79
8 Fundamental z Sampling z Distributions z and z Data z Descriptions 85
9 One- zand zTwo-Sample zEstimation zProblems 97
10 One- z and z Two-Sample z Tests z of z Hypotheses 113
11 Simple z Linear z Regression z and z Correlation 139
12 Multiple z Linear z Regression z and z Certain z Nonlinear z Regression z Models 161
13 One-Factor z Experiments: z z General 175
14 Factorial z Experiments z (Two z or z More z Factors) 197
15 2k zFactorial zExperiments z and z Fractions 219
16 Nonparametric z z Statistics 233
17 Statistical z Quality z Control 247
18 Bayesian z Statistics 251
iii
,
,Chapter 1 z
Introduction to Statistics and z z z
zDataAnalysis z
1.1 (a) z 15.
(b) z z x̄ z =15z z z1 zz(3.4 + z2.5 + z4.8 + z· · · z+ z4.8) z = z 3.787.
(c) Sample z median z is z the z 8th z value, z after z the z data z is z sorted z from z smallest z to z largest: z 3.6.
(d) A z dot z plot z is z shown z below.
2.5 3.0 3.5 4.0 4.5 5.0 5.5
(e) After z trimming z total z 40% z of z the z data z (20% z highest z and z 20% z lowest), z the z data z becomes:
2.9 3.0 3.3 3.4 3.6
3.7 4.0 4.4 4.8
So. z the z trimmed z mean
z is
1
x̄tr20 z = z z (2.9 + z3.0 + z· · · z+ z4.8) z = z 3.678.
9z
(f) They z are z about z the z same.
1.2 (a) z Mean=20.7675 z and z Median=20.610.
(b) z z x̄tr10 z = z 20.743.
(c) A z dot z plot z is z shown z below.
18 19 20 21 22 23
(d) No. z They z are z all z close z to z each z other.
Copyright z Ⓧ
c z 2012 z Pearson z Education, z Inc. z zPublishing z as z Prentice z Hall.
1
, 2 Chapter z 1 z Introduction z to z Statistics z and z Data
z Analysis
1.3 (a) z A zdot zplot zis zshown zbelow.
200 205 210 215 220 225 230
In z zthe z zfigure, z z “×” z zrepresents z zthe z z“No z zaging” z zgroup z zand z z“◦” z zrepresents z
zthe z z“Aging” zgroup.
(b) Yes; z tensile z strength z is z greatly z reduced z due z to z the z aging z process.
(c) MeanAging z = z 209.90, z and z MeanNo zaging z = z 222.10.
(d) MedianAging z = z 210.00, z and z MedianNo zaging z = z 221.50. z The z means z and z medians
z for z eachzgroup z are z similar z to z each z other.
1.4 (a) X̄A z z= z 7.950 z and z X̃A z z= z 8.250;
X̄B z z= z 10.260 z and z X̃B z z= z 10.150.
(b) z A zdot zplot zis zshown zbelow.
6.5 7.5 8.5 9.5 10.5 11.5
In z the z figure, z “×” z represents z company z A z and z “◦” z represents z company z B. z zThe
z steel z rods zmade z by z company z B z show z more z flexibility.
1.5 (a) z A zdot zplot zis zshown zbelow.
−10 0 10 20 30 40
In z the z figure, z “×” z represents z the z control z group z and z “◦” z represents z the z treatment z group.
(b) X̄Control z z= z 5.60, z X̃Control z z= z 5.00, z and z X̄tr(10);Control z z= z 5.13;
X̄Treatment z z= z 7.60, z X̃Treatment z z= z 4.50, z and z X̄tr(10);Treatment z z= z 5.63.
(c) The zdifference zof zthe zmeans zis z2.0 zand zthe zdifferences zof zthe zmedians zand zthe
ztrimmed zmeans zare z0.5, zwhich zare zmuch zsmaller. zThe zpossible zcause zof zthis
zmight zbe zdue zto zthe zextreme z values z (outliers) z in z the z samples, z especially z the
z value z of z 37.
1.6 (a) z A zdot zplot zis zshown zbelow.
1.95 2.05 2.15 2.25 2.35 2.45 2.55
In z the z figure, z “×” z represents z the z 20◦ C z group z and z “◦” z represents z the z 45◦ C z group.
(b) X̄20◦ C z z= z 2.1075, z and z X̄45◦ C z z= z 2.2350.
(c) Based zon zthe zplot, zit zseems zthat zhigh ztemperature zyields zmore zhigh zvalues zof
ztensile zstrength, zalong zwith za zfew zlow zvalues zof ztensile zstrength. zOverall, zthe
ztemperature zdoes zhave z an z influence z on z the z tensile z strength.
(d) It zalso zseems zthat zthe zvariation zof zthe ztensile zstrength zgets zlarger zwhen zthe zcure
ztemper-zature z is z increased.
1.7 z z s2 z = z z z 1 z z z [(3.4 z − z3.787)2 z + z (2.5 z − z3.787)2 z + z (4.8 z − z3.787)2 z + z · · · z+ z (4.8 z − z3.787)2] z =
z 0.94284;
√15−1 √ s z= s2 z=
Copyright z Ⓧ
c z 2012 z Pearson z Education, z Inc. z zPublishing z as z Prentice
z Hall.
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