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12. (a,b) Answers will vary.
(d) Answers will vary.
13. (a) b
λ= 0.25616
(b–d) Answers will vary.
Supplementary Exercises for Chapter 4
1. Let Xbe the number of people out of 105 who appear for the flight.
2. (a) Let Xbe the number of cracks in a 500 meter length of pavement.
SUPPLEMENTARY EXERCISES FOR CHAPTER 4 223
(b) Let Ybe the number of cracks in a 100 meter length of pavement.
3. (a) Let Xbe the number of plants out of 10 that have green seeds. Then X∼Bin(10,0.25).
(c) Let Ybe the number of plants out of 100 that have green seeds.
Then Y∼Bin(100,0.25) so Yis approximately normal with mean µY= 100(0.25) = 25 and standard
Page 223
224 CHAPTER 4
(e) Fewer than 80 have yellow seeds if more than 20 have green seeds.
4. (a) True. If ln X1, ..., ln Xncome from an approximately normal population, then X1, ..., Xncome from
(d) True. ln X1, ..., ln Xncome from an approximately normal population.
5. Let Xdenote the number of devices that fail. Then X∼Bin(10,0.01).
= 0.00427
(c) Let pbe the required probability. Then X∼Bin(10, p).
6. The concentrations are not approximately normally distributed. There are no concentrations lower
Page 224
7. (a) The probability that a normal random variable is within one standard deviation of its mean is the
8. Let X1, ..., X5be the thicknesses of the five layers. The thickness of the plywood is S=X1+···+X5.
9. (a) The z-score of 215 is (215 −200)/10 = 1.5.
The area to the right of z= 1.5 is 1 −0.9332 = 0.0668.
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10. (a) No, we do not know the distribution of the population of beam stiffnesses. In particular, we do not
11. (a) Let Xbe the number of components in a sample of 250 that are defective.
Then X∼Bin(250,0.07), so Xis approximately normal with mean µX= 250(0.07) = 17.5 and
standard deviation σX=p250(0.07)(0.93) = 4.0342.
(c) Let pbe the required probability, and let Xrepresent the number of components in a sample of 250
that are defective.
12. (a) The mean concentration of flaws is 0.4 per m2, so the mean number of flaws in a 15 m2area is
15(0.4) = 6.0.
Page 226
SUPPLEMENTARY EXERCISES FOR CHAPTER 4 227
= 0.5543
(b) The mean number of flaws in a 6 m2area is 6(0.4) = 2.4.
(c) The total area of the 50 surfaces is 50(18) = 900 m2.
13. (a) Let λbe the true concentration. Then b
λ= 56/2 = 28.
14. (a) X∼NB(10,0.9).
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228 CHAPTER 4
(c) σX=p10(1 −0.9)/0.92= 1.1111
15. (a) Let X1, X2, X3be the three thicknesses. Let S=X1+X2+X3be the thickness of the stack. Then
Sis normally distributed with mean µS= 3(1.5) = 4.5 and standard deviation 0.2√3 = 0.34641.
16. Let Tbe the lifetime of a microprocessor.
17. (a) Let Trepresent the lifetime of a bearing.
SUPPLEMENTARY EXERCISES FOR CHAPTER 4 229
18. Let X1, ..., X16 be the times needed to complete 16 oil changes.
19. (a) Sis approximately normal with mean µS= 75(12.2) = 915 and σS= 0.1√75 = 0.86603.
The z-score of 914.8 is (914.8−915)/0.86603 = −0.23.
20. (a) The mean number of hits in five hours is 100, so X∼Poisson(100). Therefore Xis approximately
normal with mean µX= 100 and σX=√100 = 10.
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230 CHAPTER 4
21. (a) P(X≤0) = F(0) = e−e−0=e−1
22. (a) The cumulative distribution function of Xis F(x) = Zx
−∞
f(t)dt.
If x < θ,F(x) = Zx
−∞
0dt = 0.
rθr
x
Page 230
SUPPLEMENTARY EXERCISES FOR CHAPTER 4 231
=rθrx2−r
2−r
∞
r−12
23. (a) fX(x) = F′(x) = e−(x−α)/β
β[1 + e−(x−α)/β ]2
24. (a) µX=Z∞
x[pλ1e−λ1x+ (1 −p)λ2e−λ2x]dx
−∞
Page 231
232 CHAPTER 4
If x≥0, then
25. (a) P(X > s) = P(First strials are failures) = (1 −p)s
(b) P(X > s +t|X > s) = P(X > s +tand X > s)/P (X > s)
26. Let Ube the length of the part of the stick used as the length of the rectangle. Then the width is
Page 232
27. (a) FY(y) = P(Y≤y) = P(7X≤y) = P(X≤y/7).
28. (a) P(X=x)
P(X=x−1) =
n!
x!(n−x)!px(1 −p)n−x
n!
29. (a) P(X=x)
X(x) = (1/σ)Φ′(x−µ
σ) = (1/σ)φ(x−µ
σ) = 1
σ√2πe−(x−µ)2
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234 CHAPTER 4
X(x) = (1/σ)Φ′(x−µ
−σ) = (1/σ)φ(x−µ
−σ) = 1
σ√2πe−(x−µ)2
Page 234
SECTION 5.1 235
Chapter 5
Section 5.1
1. (a) 1.96
2. (a) 95%
3. The level is the proportion of samples for which the confidence interval will cover the true value.
4. (a) X= 654.1, s= 311.7, n= 50, z.025 = 1.96.
(b) X= 654.1, s= 311.7, n= 50, z.01 = 2.33.
(c) X= 654.1, s= 311.7, n= 50, so the upper confidence bound 726.6 satisfies 726.6 = 654.1 +
Page 235
