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88 CHAPTER 2
(b) µM1M2=µR2+E1R+E2R+E1E2=µR2+µE1µR+µE2µR+µE1µE2=µR2
26. Cov(X, X) = µX·X−µXµX=µX2−(µX)2=σ2
X.
28. Cov(X+Y, Z) = µ(X+Y)Z−µX+YµZ
SECTION 2.6 89
(b) V(X−(σX/σY)Y)≥0
2σ2
X−2(σX/σY)Cov(X, Y )≥0
30. (a) µX=µ1.12C+2.69N+O−0.21F e
= 1.12µC+ 2.69µN+µO−0.21µF e
= 1.12(0.0247) + 2.69(0.0255) + 0.1668 −0.21(0.0597)
Page 89
90 CHAPTER 2
31. µY=µ7.84C+11.44N+O−1.58F e
= 7.842σ2
C+ 11.442σ2
N+σ2
O+ 1.582σ2
F e + 2(7.84)(11.44)Cov(C, N) + 2(7.84)Cov(C, O)−2(7.84)(1.58)Cov(C,
32. (a) Let c=R∞
−∞ h(y)dy.
Now fX(x) = R∞
33. (a) R∞
−∞ R∞
−∞ f(x, y)dx dy =Rd
cRb
ak dx dy =kRd
cRb
adx dy =k(d−c)(b−a) = 1.
Page 90
SUPPLEMENTARY EXERCISES FOR CHAPTER 2 91
Supplementary Exercises for Chapter 2
1. Let Abe the event that component A functions, let Bbe the event that component B functions, let C
2. P(more than 3 tosses necessary) = Pfirst 3 tosses are tails) = 1
23
=1
8.
3. Let Adenote the event that the resistance is above specitication, and let Bdenote the event that the
Page 91
92 CHAPTER 2
(a) P(L1) = 2
3
5. Let Rbe the event that the shipment is returned. Let B1be the event that the tirst brick chosen
6. (a) (0.99)10 = 0.904
7. Let Abe the event that the bit is reversed at the tirst relay, and let Bbe the event that the bit is
Page 92
8. (a) Z1
−1
k(1 −x2)dx =kZ1
−1
(1 −x2)dx = 1. Since Z1
−1
(1 −x2)dx =x−x3
31
−1
=4
3,k=3
4= 0.75.
9. Let Abe the event that two different numbers come up, and let Bbe the event that one of the
10. Let Abe the event that the tirst component is defective and let Bbe the event that the second
component is defective.
Page 93
94 CHAPTER 2
(a) P(X= 0) = P(Ac∩Bc) = P(Ac)P(Bc|Ac) = 8
107
9= 0.6222
11. (a) P(X≤2 and Y≤3) = Z2
0Z3
0
1
6e−x/2−y/3dy dx
1
3
Page 94
SUPPLEMENTARY EXERCISES FOR CHAPTER 2 95
∞
(c) If x≤0, f(x, y) = 0 for all yso fX(x) = 0.
If x > 0, fX(x) = Z∞
0
1
6e−x/2−y/3dy =1
2e−x/2 −e−y/3
∞
2e−x/2.
12. (a) Aand Bare mutually exclusive if P(A∩B) = 0, or equivalently, if P(A∪B) = P(A) + P(B).
So if P(B) = P(A∪B)−P(A) = 0.7−0.3 = 0.4, then Aand Bare mutually exclusive.
96 CHAPTER 2
(a) P(D) = P(D|E)P(E) + P(D|W)P(W) + P(D|C)P(C)
(b) P(D∩C) = P(D|C)P(C) = 8
28 4
100=32
2800 = 0.0114
14. (a) Discrete. The possible values are 10, 60, and 80.
15. The total number of pairs of cubicles is 6
Page 96
16. The total number of combinations of four shoes that can be selected from eight is 8
4=8!
4!4! = 70. The
17. (a) µ3X= 3µX= 3(2) = 6, σ2
3X= 32σ2
X= (32)(12) = 9
18. Cov(X, Y ) = ρX,Y σXσY= (0.5)(2)(1) = 1
19. The marginal probability mass function pX(x) is found by summing along the rows of the joint
probability mass function.
98 CHAPTER 2
y
x100 150 200 pX(x)
(a) For additive concentration (X): pX(0.02) = 0.22, pX(0.04) = 0.19, pX(0.06) = 0.29, pX(0.08) = 0.30,
(d) P(Y > 125 |X= 0.08) = P(Y > 125 and X= 0.08)
P(X= 0.08)
Page 98
