SUPPLEMENTARY EXERCISES FOR CHAPTER 2 99
20. (a) µX= 0.02pX(0.02) + 0.04pX(0.04) + 0.06pX(0.06) + 0.08pX(0.08)
(b) µY= 100pY(100) + 150pY(150) + 200pY(200) = 100(0.14) + 150(0.36) + 200(0.50) = 168
(d) σ2
Y= 1002pY(100) + 1502pY(150) + 2002pY(200) −µ2
Y
= 1002(0.14) + 1502(0.36) + 2002(0.50) −1682
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100 CHAPTER 2
21. (a) pY|X(100 |0.06) = p(0.06,100)
pX(0.06) =0.04
0.29 =4
29 = 0.138
pY|X(150 |0.06) = p(0.06,150)
pX(0.06) =0.08
0.29 =8
29 = 0.276
22. Let Ddenote the event that an item is defective, let S1denote the event that an item is produced
(a) P(S1|D) = P(D|S1)P(S1)
P(D|S1)P(S1) + P(D|S2)P(S2) + P(D|S3)P(S3)
23. (a) Under scenario A:
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102 CHAPTER 2
24. Let Ldenote the loss.
25. (a) p(0,0) = P(X= 0 and Y= 0) = 3
102
9=1
15 = 0.0667
(b) The marginal probability density function of Xis:
SUPPLEMENTARY EXERCISES FOR CHAPTER 2 103
pX(0) = p(0,0) + p(0,1) + p(0,2) = 1
15 +3
15 +1
15 =1
3
(c) The marginal probability density function of Yis:
pY(0) = p(0,0) + p(1,0) + p(2,0) = 1
15 +4
15 +2
15 =7
15
(d) σX=q02pX(0) + 12pX(1) + 22pX(2) −µ2
X
=p02(1/3) + 12(8/15) + 22(2/15) −(12/15)2
(f) Cov(X, Y ) = µXY −µXµY.
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104 CHAPTER 2
σXσY
26. (a) The constant csolves Z1
0Z1
0
c(x+y)2dx dy = 1.
(b) For 0 < x < 1, fX(x) = Z1
0
6
7(x+y)2dy =2(x+y)3
7
1
=6x2+ 6x+ 2
7.
(c) fY|X(y|x) = f(x, y)
fX(x). If x≤0 or x≥1fX(x) = 0, so fY|X(y|x) is undefined.
(d) E(Y|X= 0.4) = Z∞
−∞
yfY|X(y|0.4) dy
3y(0.4 + y)2
(e) No, fY|X(y|x)6=fY(y).
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SUPPLEMENTARY EXERCISES FOR CHAPTER 2 105
27. (a) µX=Z∞
−∞
xfX(x)dx =Z1
0
x6x2+ 6x+ 2
7dx =1
14(3x4+ 4x3+ 2x2)
1
0
=9
14 = 0.6429
µX=9
14, computed in part (a). To compute µY, note that the joint density is symmetric in xand y,
(d) Since the marginal density of Yis the same as that of X,σ2
Y=σ2
X=199
2940.
28. Since the coin is fair, P(H) = P(T) = 1/2. The tosses are independent.
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29. (a) pX(0) = 0.6, pX(1) = 0.4, pX(x) = 0 if x6= 0 or 1.
30. The probability mass function of Xis pX(x) = 1/6 for x= 1,2,3,4,5,or 6, and pX(x) = 0 for other
31. (a) The possible values of the pair (X, Y ) are the ordered pairs (x, y) where each of xand yis equal to 1,
2, or 3. There are nine such ordered pairs, and each is equally likely. Therefore pX,Y (x, y) = 1/9 for
x= 1,2,3 and y= 1,2,3, and pX,Y (x, y) = 0 for other values of (x, y).
(b) Both Xand Yare sampled from the numbers {1,2,3}, with each number being equally likely.
SUPPLEMENTARY EXERCISES FOR CHAPTER 2 107
32. (a) The values of Xand Ymust both be integers between 1 and 3 inclusive, and may not be equal. There
are six possible values for the ordered pair (X, Y ), specifically (1,2),(1,3),(2,1),(2,3),(3,1),(3,2).
Each of these six ordered pairs is equally likely.
Therefore pX,Y (x, y) = 1/6 for (x, y) = (1,2),(1,3),(2,1),(2,3),(3,1),(3,2), and pX,Y (x, y) = 0 for
other
values of (x, y).
33. (a) µX=R∞
−∞ xf(x)dx. Since f(x) = 0 for x≤0, µX=R∞
0xf(x)dx.
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108 CHAPTER 2
34. µA=πµR2. We now find µ2
R.
35. (a) If the pooled test is negative, it is the only test performed, so X= 1. If the pooled test is positive,
then nadditional tests are carried out, one for each individual, so X=n+ 1. The possible values of
Xare therefore 1 and n+ 1.
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SECTION 3.1 109
Chapter 3
Section 3.1
1. (ii). The measurements are close together, and thus precise, but they are far from the true value of
100◦C, and thus not accurate.
3. (a) True
4. (a) The uncertainty is 0.02(100) = 2 g
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