148 CHAPTER 4
4. (a) If Xand Ycannot both be equal to 1, the possible values for Zare 0 and 1, so Zis a Bernoulli
random variable.
5. (a) pX= 1/2
6. (a) pX= 1/6
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SECTION 4.2 149
7. (a) Since the possible values of Xand Yare 0 and 1, the possible values of the product Z=XY are also
0 and 1. Therefore Zis a Bernoulli random variable.
Section 4.2
1. (a) P(X= 1) = 7!
1!(7 −1)!(0.3)1(1 −0.3)7−1= 0.2471
2. (a) P(X > 6) = P(X= 7) + P(X= 8) + P(X= 9)
=9!
7!(9 −7)!(0.4)7(1 −0.4)9−7+9!
8!(9 −8)!(0.4)8(1 −0.4)9−8+9!
9!(9 −9)!(0.4)9(1 −0.4)9−9
= 0.0250
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150 CHAPTER 4
(d) P(2 < X ≤5) = P(X= 3) + P(X= 4) + P(X= 5)
3. (a) P(X= 2) = 4!
2!(4 −2)!(0.6)2(1 −0.6)4−2= 0.3456
= 0.2031
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SECTION 4.2 151
4. Let Xbe the number of flights in the sample that are on time. Then X∼Bin(10,0.75).
5. Let Xbe the number of automobiles that violate the standard. Then X∼Bin(12,0.1).
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152 CHAPTER 4
6. Let Xbe the number of times the die comes up 6, and let Ybe the number of times the die comes
up an odd number. Then X∼Bin(8,1/6) and Y∼Bin(8,1/2).
7. Let Xbe the number of failures that occur in the base metal. Then X∼Bin(20,0.15).
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SECTION 4.2 153
8. Let Xbe the number of contracts, out of the 20 chosen, that have a cost overrun. Then X∼
Bin(20,0.2).
9. Let Xbe the number of women among the five winners. Then X∼Bin(5,0.6).
10. Let Xbe the number of rods in the sample that meet specifications, and let pbe the proportion of
the day’s production that meets specifications. Then X∼Bin(100, p). The observed value of Xis 92.
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154 CHAPTER 4
(b) We must find nso that rp(1 −p)
11. Let Xbe the number of defective parts in the sample from vendor A, and let Ybe the number of
defective parts in the sample from vendor B. Let pAbe the probability that a part from vendor A is
defective, and let pBbe the probability that a part from vendor B is defective. Then X∼Bin(100, pA)
and Y∼Bin(200, pB). The observed value of Xis 12 and the observed value of Yis 10.
(a) bpA=X/100 = 12/100 = 0.12, σbpA=rpA(1 −pA)
100
bpA
bpB
12. (a) Let Ddenote the event that the item is defective and let Rdenote the even that the item can
be repaired. Then P(D) = 0.2 and P(R|D) = 0.6. Then P(D∩Rc) = P(Rc|D)P(D) = [1 −
P(R|D)]P(D) = (1 −0.6)(0.2) = 0.08.
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SECTION 4.2 155
13. (a) P(can be used) = P(meets spec) + P(long) = 0.90 + 0.06 = 0.96.
14. (a) Let Xbe the number of the ten gears that are scrap. Then X∼Bin(10,0.05).
(b) Eight or more are not scrap if and only if two or fewer are scrap.
(c) The probability that a gear is either degraded or scrap is 0.15 + 0.05 = 0.20. Let Ybe the number of
gears that are either degraded or scrap. Then Y∼Bin(10,0.2).
(d) The probability that a gear is either conforming or degraded is 0.80 + 0.15 = 0.95. Let Ybe the
number of gears that are either conforming or degraded. Then Y∼Bin(10,0.95).
15. (a) Let Ybe the number of lights that are green. Then Y∼Bin(3,0.6).
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156 CHAPTER 4
16. (a) Let Xbe the number of components out of 10 that are defective. Then X∼Bin(10,0.1).
(b) Let Xbe the number of components out of 10 that are defective. Then X∼Bin(10,0.2).
(c) Let Xbe the number of components out of 10 that are defective. Then X∼Bin(10,0.02).
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SECTION 4.2 157
17. (a) Let Xbe the number of components that function. Then X∼Bin(5,0.9).
(b) We need to find the smallest value of nso that P(X≤2) <0.10 when X∼Bin(n, 0.9). Consulting
Table A.1, we find that if n= 3, P(X≤2) = 0.271, and if n= 4, P(X≤2) = 0.052. The smallest
value of nis therefore n= 4.
18. (a) On a rainy day, the number Xof components that function is distributed Bin(6,0.7).
19. (a) X∼Bin(10,0.15).
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158 CHAPTER 4
20. (a) X∼Bin(8,0.80).
(b) Yes, if the claim is true, then only about 8 or 9 out of every 100,000 samples would have 1 or fewer
(d) P(X≤6) = 1 −P(X > 6)
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SECTION 4.2 159
21. (a) Let Xbe the number of bits that are reversed. Then X∼Bin(5,0.3). The correct value is assigned
23. (a) Y= 10X+ 3(100 −X) = 7X+ 300
24. Let Xbe the number of components of the two in the first design that work. Let Ybe the number
of components of the four in the second design that work. Then X∼Bin(2,0.9) and Y∼Bin(4,0.9).
The probability that the design with two components works is
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160 CHAPTER 4
The probability that the design with four components works is
25. Let pbe the probability that a tire has no flaw. The results of Example 4.14 show that the sample
proportion is bp= 93/100 = 0.93 and σbp= 0.0255. Let Xbe the number of tires out of four that have
no flaw. Then X∼Bin(4, p). Let qbe the probability that exactly one of four tires has a flaw.
26. (a) The sample size is n= 612 and bp= 88/612 = 0.143791.
(b) Let O=p
1−pbe the odds. Then Ois estimated with bp
1−bp=88/612
1−88/612 =88
524 = 0.167939.
1. (a) P(X= 1) = e−441
1! = 0.0733
2. (a) Since X∼Poisson(3), P(X= 5) = e−335
5! = 0.1008.
(d) P(X > 1) = 1 −P(X= 0) −P(X= 1)
162 CHAPTER 4
3. Since the mean number per mile is 3, the mean number per two miles is (2)(3) = 6. Therefore
X∼Poisson(6),
(c) P(5 ≤X < 8) = P(X= 5) + P(X= 6) + P(X= 7)
=e−665
5! +e−666
6! +e−667
7!
4. (a) X∼Poisson(6), so P(X= 7) = e−667
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SECTION 4.3 163
5. Let Xbe the number of servers that fail. Then Xis the number of successes in n= 1000 Bernoulli
trials, each of which has success probability p= 0.003. The mean of Xis np = (1000)(0.003) = 3.
Since nis large and pis small, X∼Poisson(3) to a very close approximation.
= 0.5768
6. Let Xbe the number of individuals in the sample that carry the gene. Then Xis the number of suc-
cesses in
n= 1000 Bernoulli trials, each of which has success probability p= 0.0002. The mean of Xis
np = (1000)(0.0002) = 0.2. Since nis large and pis small, X∼Poisson(0.2) to a very close approxi-
mation.
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164 CHAPTER 4
7. (a) Let Xbe the number of hits in one minute. Since the mean rate is 4 messages per minute, X∼
Poisson(4).
P(X= 5) = e−445
5! = 0.1563
8. (a) Let Xbe the number of cars that arrive in a given second. Since the mean rate is 4 per second,
SECTION 4.3 165
(b) Since the mean rate is 4 per second, X∼Poisson(12)
9. (ii). Let X∼Bin(n, p) where µX=np = 3. Then σ2
X=np(1 −p), which is less than 3 because
10. Let Xrepresent the number of particles observed in 3 mL. Let λrepresent the true concentration in
11. Let Xrepresent the number of bacteria observed in 0.5 mL. Let λrepresent the true concentra-
12. (a) Let Xbe the number of plants in a two-acre region. Then X∼Poisson(20).
166 CHAPTER 4
13. (a) Let Nbe the number of defective components produced. Then N∼Poisson(20).
P(N= 15) = e−20 2015
15! = 0.0516
14. Let λbe the mean number of particles emitted in one minute. Let Xbe the number of particles
actually emitted during a given minute.
15. (a) Let Xbe the number of flaws in a 3-meter board. Since the mean number of flaws is 0.45 per meter,
X∼Poisson(1.35).
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SECTION 4.3 167
16. (a) Let Xbe the number of raisins in a randomly chosen slice.
Since the mean number of raisins per slice is 100/60 = 5/3, X∼Poisson(5/3).
P(X= 0) = e−(5/3) (5/3)0
0! = 0.1889
17. Let λ1be the mean number of chips per cookie in one of Mom’s cookies, and let λ2be the mean
number of chips per cookie in one of Grandma’s cookies. Let X1and X2be the numbers of chips in
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