Probability and Statistics for Engineering and the Sciences Solution Manual 9th Ed

222 Pages·2016·41.38 MB·English

by  Jay L. Devore

Student
Solutions
Manual

Student Solutions Manual
Probability and Statistics for
Engineering and the Sciences
NINTH EDITION
Jav 1.Devore
California Polytechnic State University
Prepared by
MattCarlton
California Polytechnic State University
*" ~ CENGAGE
Learning
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CONTENTS
Chapter I Overview and Descriptive Statistics
Chapter 2 Probability 22
Chapter 3 Discrete Random Variables and Probability 41
Distributions
Chapter 4 Continuous Random Variables and Probability 57
Distributions
Chapter 5 Joint Probability Distributions and Random Samples 79
Chapter 6 Point Estimation 94
Chapter 7 Statistical Intervals Based on a Single Sample 100
Chapter 8 Tests of Hypotheses Based on a Single Sample 108
Chapter 9 Inferences Based onTwo Samples 119
Chapter 10 The Analysis of Variance 135
Chapter II Multifactor Analysis ofVariance 142
Chapter 12 Simple Linear Regression and Correlation 157
Chapter 13 Nonlinear and Multiple Regression 174
Chapter 14 Goodness-of-Fit Tests and Categorical Data Analysis 196
Chapter 15 Distribution-Free Procedures 205
Chapter 16 Quality Control Methods 209
III
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•
CHAPTER 1
Section 1.1
1.
a. Los Angeles Times, Oberlin Tribune, Gainesville Sun, Washington Post
b. Duke Energy, Clorox, Seagate, Neiman Marcus
c. Vince Correa, Catherine Miller, Michael Cutler, Ken Lee
d. 2.97, 3.56, 2.20, 2.97
3.
a. How likely isitthat more than half ofthe sampled computers willneedor have needed
warranty service? What isthe expected number among the 100that need warranty
service? How likely is itthat the number needing warranty service willexceed the
expected number bymore than 1O?
b. Suppose that 15ofthe 100 sampled needed warranty service. How confident can we be
that the proportion of all such computers needing warranty service isbetween .08 and
.22? Does the sample provide compelling evidence for concluding that more than 10%of
all such computers need warranty service?
5.
a. No. All students taking a large statistics course who participate inan SIprogram of this
sort.
b. The advantage torandomly allocating students to the two groups isthatthe two groups
should then be fairly comparable before thestudy. Ifthe two groups perform differently
in the class, we might attribute this to the treatments (SI and control). If itwere left to
students tochoose, stronger or more dedicated students might gravitate toward SI,
confounding the results.
c. If all students were put in tbe treatment group, tbere would be nofirm basis for assessing
the effectiveness ofSI (notbing to wbich the SI scores could reasonably be compared).
7. One could generate asimple random sample ofallsingle-family homes inthe city, or a
stratified random sample by taking a simple random sample from each of the 10district
neighborhoods. Fromeach of the selected homes, values of all desired variables would be
determined. This would be an enumerative study because there exists afinite, identifiable
population of objects from wbich to sample.
102016Cengage Learning. AllRightsReserved. May notbescanned,copiedorduplicated, orposted 10apubliclyaccessible website, inwholeorinpart.
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Chapter I: Overview and Descriptive Statistics
9.
a. There could be several explanations for the variability ofthe measurements. Among
themcould be measurementerror(due to mechanical ortechnical changes across
measurements), recordingerror,differences inweatherconditions attime of
measurements, etc.
b. No, because thereisnosamplingframe.
Section 1.2
11.
31. I
3H 56678
41. 000112222234
4H 5667888 stem: tenths
51. 144 leaf: hundredths
5H 58
61. 2
6H 6678
71.
7H 5
Thestem-and-leaf display showsthat .45 is agood representative value for the data. In
addition, the display isnot symmetric and appears to bepositively skewed. The range ofthe
data is.75- .31 ~ .44, which iscomparable to the typical value of .45. This constitutes a
reasonably large amount ofvariation in the data. The data value .75 isapossible outlier.
13.
a.
12 2 stem: tens
12 445 leaf: ones
12 6667777
12 889999
13 00011111111
13 2222222222333333333333333
13 44444444444444444455555555555555555555
13 6666666666667777777777
13 888888888888999999
14 0000001111
14 2333333
14 444
14 77
Theobservations are highlyconcentrated ataround 134or 135, where the display
suggests the typical value falls.
2
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Chapter 1: Overview and Descriptive Statistics
b.
40
r-t-'
r---
I---
-
.. -
10 -
I---
r----
o ~ I
124 128 132 136 140 144 148
strength (ksi)
The histogram ofultimate strengths issymmetric and unimodal, with thepoint of
symmetry atapproximately 135 ksi. There isa moderate amount ofvariation, and there
are no gaps or outliers in the distribution.
15.
American French
8 I
755543211000 9 00234566
9432 10 2356
6630 II 1369
850 12 223558
8 13 7
14
15 8
2 16
American movie timesareunimodal strongly positively skewed, while Frenchmovie times
appear to be bimodal. A typical American movie runs about 95 minutes, while French movies
are typically either around 95 minutes oraround 125 minutes. American movies are generally
shorter than French movies and are less variable in length. Finally, both American and French
movies occasionallyrun very long (outliers at 162minutes and 158minutes, respectively, in
the samples).
3
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Chapter I: Overview and Descriptive Statistics
17. Thesample size for thisdata setisn=7 + 20+ 26+ ... +3+2 = 108.
a. "At most five hidders" means2, 3,4, or 5 hidders. Theproportion of contracts that
involved at most 5bidders is(7 + 20 +26 + 16)/108~ 69/108 ~ .639.
Similarly, the proportion ofcontracts that involved atleast 5 bidders (5 through II) is
equal to (16 + II+9+6+8+ 3 + 2)/108 ~ 55/108 ~ .509,
b. The numher ofcontracts with between 5 aod 10bidders, inclusive, is 16+ 11+9 + 6 + 8
+3 =53, so the proportion is53/108 = .491. "Strictly" between 5 and 10means6, 7, 8, or
9bidders, for aproportion equal to (11 + 9+6+8)/108 =34/1 08 ~ .315.
c. The distribution ofnumber ofbidders ispositively skewed, ranging from 2to II bidders,
WIith atrvpr.caIvaIue 0faround4-5 bi1dders.
0
•
r-
J I--
I'
..
r I--
I--Hl
, , , , , , , .. r "l
• •
"
Nu"'''oFbidd.~
19.
a. From this frequency distribution, the proportion ofwafers that contained at leastone
particle is(100-1)/1 00~ .99,or 99%. Note that itismuch easier to subtract I (which is
the number of wafers thatcontain 0 particles) from 100than it would be toaddall the
frequencies for 1,2,3, ... particles. In a similar fashion, the proportion containing at least
5particles is(100 - 1-2-3-12-11)/100 = 71/100 =.71,or,71%.
b. The proportion containing between 5 and 10particles is(15+ 18+10+12+4+5)/100~
64/100 ~ .64, or 64%. Theproportion that contain strictly between 5 and 10(meaning
strictly more than 5 andstrictly less than 10) is(18+10+12+4)/100 ~ 44/100 =.44, or
44%.
c. The following histogram wasconstructed using Minitah. The histogram isalmost
syrrunetric and unimodal; however, the distribution has afew smaller modes andhas a
ve sli ht ositive skew.
'"
r-'
s ~
~r-- r-r-
o
t-
,
t-r-
,1.-rl
Il---n,
o 1 2 J 4 5 6 7 8 9 10 II 12 13 I~
NurmerOr.onl,rrin.lirlg po";ol••
4
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7