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48 CHAPTER 2
(e) The small probability in part (a) indicates that some good bottles will be scrapped. This is not so
33. Let Drepresent the event that the man actually has the disease, and let + represent the event that
the test gives a positive signal.
(a) P(D|−) = P(−|D)P(D)
P(−|D)P(D) + P(−|Dc)P(Dc)
(d) P(D|+ +) = P(+ + |D)P(D)
P(+ + |D)P(D) + P(+ + |Dc)P(Dc)
34. P(system functions) = P[(A∩B)∪(C∪D)]. Now P(A∩B) = P(A)P(B) = (1−0.10)(1−0.05) = 0.855,
and P(C∪D) = P(C) + P(D)−P(C∩D) = (1 −0.10) + (1 −0.20) −(1 −0.10)(1 −0.20) = 0.98.
35. P(system functions) = P[(A∩B)∩(C∪D)]. Now P(A∩B) = P(A)P(B) = (1 −0.05)(1 −0.03) =
0.9215, and P(C∪D) = P(C)+P(D)−P(C∩D) = (1−0.07)+(1−0.14)−(1−0.07)(1−0.14) = 0.9902.
Therefore
36. (a) P(A∩B) = P(A)P(B) = (1 −0.05)(1 −0.03) = 0.9215
37. Let Cdenote the event that component C functions, and let Ddenote the event that component D
functions.
Page 49
50 CHAPTER 2
(b) P(system functions) = 1 −P(Cc∩Dc) = 1 −p2= 0.99. Therefore p=√1−0.99 = 0.1.
38. To show that Acand Bare independent, we show that P(Ac∩B) = P(Ac)P(B). Now B= (Ac∩B)∪
Section 2.4
1. (a) Discrete
SECTION 2.4 51
2. (a) P(X≤2) = P(X= 0) + P(X= 1) + P(X= 2) = 0.4 + 0.3 + 0.15 = 0.85.
3. (a) µX= 1(0.4) + 2(0.2) + 3(0.2) + 4(0.1) + 5(0.1) = 2.3
4. (a) p3(x) is the only probability mass function, because it is the only one whose probabilities sum to 1.
Page 51
5. (a) x1 2 3 4 5
p(x) 0.70 0.15 0.10 0.03 0.02
6. (a) µX= 24(0.0825) + 25(0.0744) + 26(0.7372) + 27(0.0541) + 28(0.0518) = 25.9183
7. (a) P5
x=1 cx = 1, so c(1 + 2 + 3 + 4 + 5) = 1, so c= 1/15.
8. (a) P(X≤2) = F(2) = 0.83
Page 52
SECTION 2.4 53
9. (a)
x p1(x)
0 0.2
1 0.16
2 0.128
3 0.1024
4 0.0819
5 0.0655
x p2(x)
0 0.4
1 0.24
10. Let Adenote an acceptable chip, and Uan unacceptable one.
(a) P(A) = 0.9
54 CHAPTER 2
11. Let Adenote an acceptable chip, and Uan unacceptable one.
(a) If the first two chips are both acceptable, then Y= 2. This is the smallest possible value.
12. (a) 0, 1, 2, 3
Page 54
SECTION 2.4 55
(b) P(X= 3) = P(SSS) = (0.8)3= 0.512
13. (a) Z90
80
x−80
800 dx =x2−160x
1600
90
80
= 0.0625
80
800 dx =x3−120x
2400
120
80
Page 55
56 CHAPTER 2
(c) σ2
X=Z120
80
x2x−80
800 dx −(320/3)2=x4
3200 −x3
30
120
80 −(320/3)2= 800/9
σX=p800/9 = 9.428
−∞
80
800 dt +Zx
120
14. (a) Z30
25
x
250 dx =x2
500
30
25
= 0.55
30
28
250 dx =x2
500
28
Page 56
15. (a) µ=Z∞
0
0.1te−0.1tdt
=−te−0.1t
∞
0−Z∞
0−e−0.1tdt
= 0 −10e−0.1t
∞
0
= 10
16. (a) µ=Z10.25
9.75
3x[1 −16(x−10)2]dx =−12x4+ 320x3−2398.5x2
10.25
9.75
= 10
9.75
10.25
9.75
Page 57
58 CHAPTER 2
(c) F(x) = Zx
−∞
f(t)dt.
If x≤9.75, F(x) = Zx
−∞
0dt = 0.
If 9.75 < x < 10.25,
F(x) = Zx
−∞
0dt +Zx
9.75
3[1 −16(t−10)2]dt = 3t−16(t−10)3
x
9.75
= 3x−16(x−10)3−29.5
If x≥10.25, F(x) = 1.
17. With this process, the probability that a ring meets the specification is
9.9
−0.15
−0.15
With the process in Exercise 16, the probability is
9.9
−0.1
−0.1
Therefore this process is better than the one in Exercise 16.
18. (a) P(X > 2) = Z∞
2
64
(x+ 2)5dx =−16
(x+ 2)4
∞
2
= 1/16
64
4
0
(x+ 2)5dx =Z∞
2
u5du = 64 Z∞
2
3u−3+1
2u−4∞
2
Page 58
SECTION 2.4 59
(d) σ2=Z∞
0
x264
(x+ 2)5dx −µ2
(e) F(x) = Zx
f(t)dt.
19. (a) P(X > 3) = Z4
3
(3/64)x2(4 −x)dx =x3
16 −3x4
256 4
3
= 67/256
0
64 −3x5
320 4
0
Page 59
60 CHAPTER 2
(d) σ2=Z4
0
(3/64)x4(4 −x)dx −µ2=3x5
80 −x6
1284
0−2.42= 0.64
0
20. (a) P(X < 0.02) = Z0.02
0
625x dx =625x2
2
0.02
0
= 0.125
0.04
(d) F(x) = Zx
f(t)dt
SECTION 2.4 61
21. (a) P(X < 0.2) = Z0.2
0
12(x2−x3)dx = 4x3−3x4
0
= 0.0272
1
0
0
Page 61
22. (a) P(X > 0.5) = Z1
0.5
2e−2x
1−e−2dx =1
1−e−2(−e−2x)
1
0.5
= 0.2689
(b) µ=Z1
0
2xe−2x
1−e−2dx
= 0.34348
(c) Let Xdenote the concentration. The mean is µX= 0.34348.
2e−2x
(d) The variance is σ2=Z1
0
2x2e−2x
1−e−2dx −µ2
0
Page 62
SECTION 2.4 63
2
(e) Let Xdenote the concentration. The mean is µX= 0.34348. The standard deviation is σ= 0.26265.
2e−2x
(f) F(x) = Zx
−∞
f(t)dt
23. (a) P(X < 2.5) = Z2.5
2
(3/52)x(6 −x)dx = (9x2−x3)/52
2.5
2
= 0.2428
2.5
52
3.5
2.5
Page 63
64 CHAPTER 2
4
(e) Let Xrepresent the thickness. Then Xis within ±σof the mean if 2.4316 < X < 3.5684.
2.4316
52
3.5684
2.4316
(f) F(x) = Zx
−∞
f(t)dt
−∞
2
4
24. (a) csolves the equation Z∞
1
c/x3dx = 1. Therefore −0.5c/x2
∞
1
= 1, so c= 2.
∞
−∞
Page 64
SECTION 2.4 65
−∞
1
t2
1
(d) The median xmsolves F(xm) = 0.5. Therefore 1 −1/x2
m= 0.5, so xm= 1.414.
25. (a) P(X < 2) = Z2
0
xe−xdx = −xe−x
2
0
+Z2
0
e−xdx!= −2e−2−e−x
2
0!= 1 −3e−2= 0.5940
3
3
0
26. (a) P(X < 12.5) = Z12.5
12
6(x−12)(13 −x)dx =−2x3+ 75x2−936x
12.5
12
= 0.5
Page 65
66 CHAPTER 2
13
(d) F(x) = Zx
−∞
f(t)dt
12.3
12.3
Section 2.5
1. (a) µ3X= 3µX= 3(9.5) = 28.5
Page 66
SECTION 2.5 67
3. Let X1, ..., X4be the lifetimes of the four transistors. Let S=X1+··· +X4be the total lifetime.
4. (a) µV=µV1+µV2= 12 + 6 = 18
5. Let X1, ..., X5be the thicknesses of the five layers. Let S=X1+···+X5be the total thickness.
Page 67
