Homework Assignment Sets
Textbook: Probability and Statistics for the Engineering and the Sciences
7th Edition, by J. Devore (answers included for those assigned exercises having
no answer in the back of the textbook)
Problem set # 1
• Reading: Section 1.1
Problem set # 2
• Reading: Histograms (pages 13 thru 19), The Mean and The Median
(pages 24 thru 28), Categorical Data and Sample Proportions (pages 29
and 30), Sample Variance and Sample Standard Deviation (page 31 to
Boxplots)
• Exercises: Chapter 1: 33a,b,d, 47, 51, 69
Problem set # 3
• Reading: Section 2.1
• Exercises: Chapter 2: 5, 9
Problem set # 4
• Reading: Section 2.2
• Exercises: Chapter 2: 13, 18, 21, 25, 27, 28
• Answers: (18).6 25(a) .98 (b) .02
28 (a) .111 (b) .889 (c) .222
(c) .03
(d) .24
Problem set # 5
• Reading: Section 2.4 to Bayes’ Theorem
• Exercises: Chapter 2: 50, 51, 53, 58
• Answers: 50(a) .05
(f).444, .556
(b) .12
(c) .56, .44
(d) .49, .25
Problem set # 6
• Reading: Section 2.4 from Bayes’ Theorem to end of Section
• Exercises: Chapter 2: 59, 60, 64, 65
• Answers: 59(a).21 (b).455 (c).264, .462, .275
60(a).0667 (b).5091 (64).0588, .6735, .9996
(e).533
Problem set # 7
• Reading: Section 2.5
• Exercises: Chapter 2: 72, 73, 80, 84, 85, 89
• Answers: (72) A2 is independent of A3
84(a).216 (b).784 (c) .288 (d) .352
(80) .9981
(e) .231
Problem set # 8
• Reading: Sections 3.1 and 3.2
• Exercises: Chapter 3: 5, 9, 13, 17, 24
• Answers: 17(a).81 (b) .162 (c)5th is acceptable; auuua, uauua, uuaua,
uuuaa; .00324
(d) p(y) = (y − 1)(.1)y−2 (.9)2 for y = 2, 3, ...
24(a) p(1) = .3, p(3) = .10, p(4) = .05, p(6) = .15, p(12) = .40 (b) .30,
.60
Problem set # 9
• Reading: Section 3.3 to Rules of Expected Value
• Exercises: 30, 38, 103
• Answers: 30(a) .60 (b)110
(38) E[h(X)] = .408, so gamble
Problem set # 10
• Reading:Section 3.3 from Rules of Expected Value to end of Section
• Exercises: Chapter 3: 32, 33, 39
• Answers: 32(a)16.38, 272.289, 3.9936 (b)401
33(a) p (c) p (39) 2.3, .81, 88.5 lb, 20.25lb2
Problem set # 11
• Reading: Section 3.4
• Exercises: Chapter 3: 55, 59, 60, 97, 99
• Answers: 60 $40
Problem set # 12
• Reading: Section 3.6
• Exercises:Chapter 3: 82, 87, 88, 109
• Answers: 82 (a) .1637 (b) .0176 (c).6703
88(a) 2, 1.407 (b) .143 (c).0014
(c)2496
(d)13.66
Problem set # 13
• Reading:Section 4.1
• Exercises: Chapter 4: 3, 5, 7
• Answers: 3(b).5 (c).6875
(d).6328
Problem set # 14
• Reading: Section 4.2
• Exercises: Chapter 4: 11, 19, 22
−1
• Answers: 22(a) F
√(x) = 2x + 2x − 4, if 1 ≤ x ≤ 2
p+ p2 +8p
(b) η(p) = 1 +
, µ̃ = 1.64 (c)1.614, .0626
4
(d).0609
Problem set # 15
• Reading: Section 4.3 to The Normal Distribution and Discrete Populations
• Exercises: Chapter 4: 33, 39, 41, 46
• Answers: 46(a) .7938
(b)5.88
(c)7.938 (d).264
Problem set # 16
• Reading: Section 4.3 from The Normal Distribution and Discrete Populations to the end of the Section
• Exercises: 47, 54, 55
• Answers: 54(a) .9932
(b) .9875
(c).8064
Problem set # 17
• Reading: The Exponential Distribution, page 157 to the Chi-Squared
Distribution
• Exercises: 59, 61, 70
• Answers: 59(a) 1 (b)1 (c) .9817 (d) .1286
61 (a) .4493, .6988, .1481 (b) .0498, .0183
70 (a) η(p) = (−ln(1 − p))/λ (b) µ = .693/λ
Problem set # 18
• Reading: The Joint Probability Mass Function for Two Discrete Random Variables, pages 185 and 186, discrete part of Independent Random
Variables, pages 189 and 190
• Exercises: Chapter 5: 1, 3, 7, 11a,b
• Answers: 7(a) .030
(b).120
(c) .10, .30
(d).380
(e) Yes
Problem set # 19
• Reading: Discrete parts of Section 5.2 (Omit Examples 5.14 and 5.16)
• Exercises: Chapter 5: 22, 30
• Answers: 22(a) 14.1
(b) 9.6
30(a) -3 .2025
(b) -.2074
Problem set # 20
• Reading: Section 5.1 (Omit Example 5.21)
• Exercises: Chapter 5: 37, 38, 39
• Answers: 38(a)
t
0
P (T0 = t) .04
1
.20
(b) E[T0 ] = 2.2 = 2µ
2
.37
3
.30
4
.09
(c) V (T0 ) = .98 = 2σ 2
Problem set # 21
• Reading: Section 5.4 (Omit PROPOSITION on page 217)
• Exercises: Chapter 5: 49, 53, 83
Problem set # 22
• Reading: Section 5.5
• Exercises: Chapter 5: 62, 67, 72, 73
• Answers:(62).0314
(72) t=10:52.76 (rounded to 10:53)
Problem set # 23
• Reading: Section 6.1 to Some Complications (Omit Examples 6.4 and
6.6), Reporting a Point Estimate: The Standard Error (bottom of page
238 thru Example 6.10)
• Exercises: Chapter 6: 1, 4, 7
• Answers:1(a) 8.14, X̄ (b)7.7, X̃ (c)1.66, S (d).148
(e).204, S/X̄
p σ2 σ2
σ2
σ2
4(a) -.4343 (b)V (X̄ − Ȳ ) = m1 + n2 , SE of (X̄ − Ȳ ) = ( m1 + n2 ), .569
(c).789 (d)7.181
Problem set # 24
• Reading: Maximum Likelihood Estimation (bottom of page 245) to Some
Complications, omitting Examples 6.18, 6.19 and 6.21
• Exercises: Chapter 6: 20, 22b, 25, 30
• Answers: 20(a)p̃, .15 (b) Yes (c).4437
n
22(b) θ̂ = −ln(x1 ·x
− 1, 3.116
2 ···xn )
(30)
−ln(.75)
24
= 0.012
Problem set # 25
• Reading: Section 7.1 to Other Levels of Confidence, Section 7.2 thru
Example 7.6
• Exercises: Chapter 7: 12, 24, 24(b): How many additional observations
are needed to obtain a 95% CI for µ having length approximately .50?
• Answers: (12).81±.084 sec
(24) 8.17 ± .37 percent, None
24(b) 68
Problem set # 26
• Reading: No reading for this set. Use the Large-Sample CI for p having
the form:q
p̂ ± zα/2 p̂q̂
n , whenever n > 40, np̂ ≥ 10, and nq̂ ≥ 10.
• Exercises: Chapter 7: 19, 19(b): Use the data in exercise 19 as a pilot
study to find the sample size needed to obtain a 90% CI for p having
length ≈ 0.08.
19(c): What is the smallest sample size that is guaranteed to produce a
99% CI for p having length at most .10 regardless of p̂?
• Answers: (19) .565 ± .0515
19(b): 416
19(c)664
Problem set # 27
• Reading: Section 7.3 thru Example 7.11 (Omit one-sided confidence intervals)
• Exercises: Chapter 7: 32, 33b, c
• Answers: (32) 30.2 ± 2.6
33(b) Yes, because A-D Normality Test gives P-value = .418
(You will understand the meaning of this statement by the time you have
completed Minitab Assignment 2)
(c) 4383 ± 7.8, Yes, No
Problem set # 28
• Reading: Section 8.1 to Example 8.1, Case II: Large Sample Tests (page
299 to the bottom of page 300)
Section 8.4 thru Example 8.17, Section 1 of the Supplement for Statistics
260
• Exercises: Chapter 8: 26, 28, 53
• Answers: (26) Ha : µ > 50, Zobs = 3.77, pv < 0.0002, very strong
evidence that µ > 50, the estimated value of µ is 52.7 mils with ese = .72
mils, the conduit should not be used
(28) Ha : µ < 20, Zobs = −1.13, pv = .1292, little or no evidence that
µ < 20, the estimated value of µ is 18.86 min with ese = 1.01 min, retain
H0 at level .10
(53) Ha : µ 6= 5, Zobs = −3.71, pv < 2(.0002) = .0004, very strong
evidence that µ 6= 5, the estimated value of µ is 4.87 grains with ese =
.035 grains, reject H0 at level α = .01
Problem set # 29
• Reading: Section 8.3 thru Example 8.11, Case III: A Normal Population
Distribution, page 300, P-values for t Tests, bottom of page 315 thru page
317
• Exercises: Chapter 8: 35, 57(b): Is there any evidence against the null
hypothesis that the population distribution of sprinkler activation time is
normal?
• Answers: (35) Ha : p 6= .7, Zobs = −2.47, pv = 2(.0068) = .0136, strong
evidence that p 6= .7, the estimated value of p is .62 with ese = .034, reject
H0 at α = .05
55 Ha : µ > 25, Tobs = 1.88, df = 12, .041 < pv < .049, strong evidence
that µ > 25, the estimated value of µ is 27.92 sec with ese = 1.56 sec,
reject H0 at α = .05
57(b) There is little or no evidence (A-D Normality Test P-value = .145)
against the null hypothesis that the population distribution of sprinkler
activation times is normal. Therefore, the t-test results should be valid.
Problem set # 30
• Reading: Section 9.1 to Test Procedures for Normal Populations with
Known Variances, and from Large Sample Tests, page 331 to end of Section, Section 9.4 to A Large Sample Test Procedure, and from A Large
Sample Confidence interval for p1 − p2 , middle of page 357 to end of Section.
Use the test statistic Z =
qpˆ1 −pˆ2 −∆0
p̂1 q̂1
m
+
p̂2 q̂2
n
for testing H0 : pˆ1 − pˆ2 = ∆0
regardless of whether or not the specified value of ∆0 is 0. Accurate pvalue calculations will be obtained when m > 40, mpˆ1 ≥ 10, mqˆ1 ≥ 10
and n > 40, npˆ2 ≥ 10, nqˆ2 ≥ 10.
• Exercises: Chapter 9: 8a, 12, 50, 51a
• Answers: 8(a) Ha : µ1 − µ2 < −10, Zobs = −28.6, pv = .0000, very
strong evidence that µ1 − µ2 < −10, the estimated value of µ1 − µ2 is
-16.0 kg/mm2 with ese = .21 kg/mm2
(12) -8.77 ± 2.46 N/mm2
(50) 95% CI for p1 − p2 : .093 ± .077
Problem set # 31
• Reading: Section 9.2 to Type II Error Probabilities.
Use Pooled-t Procedures whenever the two population distributions are
near normal and
larger of s1 , s2
≤ 1.4 (suggesting thatσ1 ≈ σ2 )
smaller ofs1 , s2
Pooled-t Test Statistic:
T =q
X̄ − Ȳ − ∆0
(m−1)S12 +(n−1)S22 1
(m
m+n−2
+ n1 )
m + n − 2 degrees of freedom when H0 : µ1 − µ2 = ∆0 is true, and the
two population distributions are normal with σ1 = σ2 .
100(1 − α)% Pooled-t CI:
s
x̄ − ȳ ± tα/2,m+n−2
(m − 1)s21 + (n − 1)s22 1
1
( + )
m+n−2
m n
• Exercises: Chapter 9: 22, 22(b): Construct a 90% CI for µ1 − µ2 .
33
• Answers: (22) Ha : µ1 −µ2 6= 0, Tobs = −3.16, df = 14, .006 < pv < .008,
very strong evidence that µ1 − µ2 6= 0, the estimated value of µ1 − µ2 is
-3.93 N/mm2 with ese = 1.244 N/mm2 , reject H0 at level α = .05
22(b) -3.93 ± 2.19 N/mm2
Problem set # 32
• Reading: Section 9.3
• Exercises: Chapter 9: 37a, 39
• Answers: 37(a) -.424 ± .137 nanograms/m3
39(a) There is no evidence (A-D Normality Test P-Value=.683) against the
null hypothesis that the population distribution of differences is normal.
(b) Ha : µD 6= 0, Tobs = 2.74, df = 13, .016 < pv < .018, strong evidence
that µD 6= 0, the estimated value of µD is 167.2 with ese = 61.0, reject
H0 at level α = .05