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68 CHAPTER 2
8. Let X1, ..., X24 be the volumes of the 24 bottles. Let S=X1+···+X24 be the total weight. The
average volume per bottle is X.
9. Let X1and X2denote the lengths of the pieces chosen from the population with mean 30 and standard
deviation 0.1, and let Y1and Y2denote the lengths of the pieces chosen from the population with
mean 45 and standard deviation 0.3.
10. The daily revenue is R= 2.60X1+ 2.75X2+ 2.90X3.
Page 68
11. (a) The number of passenger-miles is 8000(210) = 1,680,000. Let Xbe the number of gallons of fuel used.
Then µX= 1,680,000(0.15) = 252,000 gallons.
12. Let Xdenote the number of mismatches and let Ydenote the number of gaps. Then µX= 5, σX= 2,
µY= 2, and σY= 1. Let Sdenote the Needleman-Wunsch score. Then S=X+ 3Y.
13. (a) µ= 0.0695 + 1.0477
20 +0.8649
20 +0.7356
20 +0.2171
30 +2.8146
60 +0.5913
15 +0.0079
10 + 5(0.0006) = 0.2993
14. (a) µX=µO+2N+2C/3=µO+ 2µN+ (2/3)µC= 0.1668 + 2(0.0255) + (2/3)(0.0247) = 0.2343
15. (a) P(X < 9.98) = Z9.98
9.95
10 dx = 10x
9.98
9.95
= 0.3
5.01
5.1
5.01
Page 69
70 CHAPTER 2
(c) Since Xand Yare independent,
4.9
4.9
16. (a) µX=Z6
43
4x−3x(x−5)2
4dx =−9x2+5x3
2−3x4
16 6
4
= 5
17. (a) Let µ= 40.25 be the mean SiO2content, and let σ= 0.36 be the standard deviation of the SiO2
content, in a randomly chosen rock. Let Xbe the average content in a random sample of 10 rocks.
Page 70
18. (a) Yes. Let Xidenote the number of bytes downloaded in the ith second. The total number of bytes is
X1+X2+X3+X4+X5. The mean of the total number is the sum of the five means, each of which
is equal to 105.
Section 2.6
1. (a) 0.17
(b) P(X≥1 and Y < 2) = P(1,0) + P(1,1) + P(2,0) + P(2,1) = 0.17 + 0.23 + 0.06 + 0.14 = 0.60
2. (a) The marginal probability mass function pX(x) is found by summing along the rows of the joint
probability mass function.
Page 71
72 CHAPTER 2
(b) The marginal probability mass function pY(y) is found by summing down the columns of the joint
probability mass function. So pY(0) = 0.33, pY(1) = 0.48, pY(2) = 0.19, pY(y) = 0 if y6= 0,1,or 2
(c) µX= 0pX(0) + 1pX(1) + 2pX(2) = 0(0.26) + 1(0.48) + 2(0.26) = 1.00
Page 72
SECTION 2.6 73
(b) pX|Y(0 |1) = pX,Y (0,1)
pY(1) =0.11
0.48 = 0.2292
4. (a) The marginal probability mass function pX(x) is found by summing along the rows of the joint
probability mass function.
y
x0 1 2 3 pX(x)
(b) The marginal probability mass function pY(y) is found by summing down the columns of the joint
probability mass function. So pY(0) = 0.34, pY(1) = 0.27, pY(2) = 0.22, pY(3) = 0.17, pY(y) =
Page 73
74 CHAPTER 2
(e) σ2
X= 02pX(0)+12pX(1)+22pX(2)+32pX(3)−µ2
X= 02(0.48)+12(0.25)+22(0.17)+32(0.10)−0.892=
1.0379
(f) Cov(X, Y ) = µXY −µXµY
= 0.97
(g) ρX,Y =Cov(X, Y )
σXσY
=−0.1158
(1.0188)(1.0196) =−0.1041
5. (a) µX+Y=µX+µY= 0.89 + 1.22 = 2.11
Page 74
SECTION 2.6 75
6. (a) pY|X(0 |1) = pX,Y (1,0)
pX(1) =0.09
0.25 = 0.36
(b) pX|Y(0 |2) = pX,Y (0,2)
pY(2) =0.11
0.22 = 1/2
(c) E(Y|X= 1) = 0pY|X(0 |1) + 1pY|X(1 |1) + 2pY|X(2 |1) + 3pY|X(3 |1)
= 0(0.36) + 1(0.28) + 2(0.20) + 3(0.16) = 1.16
7. (a) 2X+ 3Y
Page 75
76 CHAPTER 2
8. (a) P(X= 2 and Y= 2) = P(X= 2)P(Y= 2 |X= 2). Now P(X= 2) = 0.30. Given that X= 2,
(b) P(X= 2 and Y= 6) = P(X= 2)P(Y= 6 |X= 2). Now P(X= 2) = 0.30. Given that X= 2, Let Y1
(c) P(Y= 2) = P(X= 1 and Y= 2) + P(X= 2 and Y= 2).
9. (a) The marginal probability mass function pX(x) is found by summing along the rows of the joint
probability mass function.
y
x01234pX(x)
Page 76
SECTION 2.6 77
(b) The marginal probability mass function pY(y) is found by summing down the columns of the joint
probability mass function. So pY(0) = 0.16, pY(1) = 0.19, pY(2) = 0.26, pY(3) = 0.23, pY(4) = 0.16,
(f) Cov(X, Y ) = µXY −µXµY.
µXY = (0)(0)pX,Y (0,0) + (0)(1)pX,Y (0,1) + (0)(2)pX,Y (0,2) + (0)(3)pX,Y (0,3) + (0)(4)pX,Y (0,4)
78 CHAPTER 2
10. (a) µX+Y=µX+µY= 2.15 + 2.04 = 4.19
11. (a) pY|X(0 |4) = pX,Y (4,0)
pX(4) =0.00
0.15 = 0
(b) pX|Y(0 |3) = pX,Y (0,3)
pY(3) =0.00
0.23 = 0
SECTION 2.6 79
12. (a) The marginal probability mass function pX(x) is found by summing along the rows of the joint
probability mass function.
y
x0 1 2 3 pX(x)
(b) The marginal probability mass function pY(y) is found by summing down the columns of the joint
probability mass function. So pY(0) = 0.28, pY(1) = 0.34, pY(2) = 0.25, pY(3) = 0.13, pY(y) =
0 if y6= 0,1,2,or 3.
(f) σ2
Y= 02pY(0) + 12pY(1) + 22pY(2) −µ2
Y= 02(0.28) + 12(0.34) + 22(0.25) + 32(0.13) −1.232= 0.9971.
σY=√0.9971 = 0.9985
Page 79
80 CHAPTER 2
+ (3)(0)pX,Y (3,0) + (3)(1)pX,Y (3,1) + (3)(2)pX,Y (3,2) + (3)(3)pX,Y (3,3)
(h) ρX,Y =Cov(X, Y )
σXσY
=0.2377
(0.8999)(0.9985) = 0.2645
13. (a) µZ=µX+Y=µX+µY= 1.01 + 1.23 = 2.24
14. (a) T= 50X+ 100Y, so µT=µ50X+100Y= 50µX+ 100µY= 50(1.01) + 100(1.23) = 173.50.
Page 80
SECTION 2.6 81
15. (a) pY|X(0 |3) = pX,Y (3,0)
pX(3) =0.01
0.07 = 0.1429
(b) pX|Y(0 |1) = pX,Y (0,1)
pY(1) =0.10
0.34 = 0.2941
(c) E(Y|X= 3) = 0pY|X(0 |3) + 1pY|X(1 |3) + 2pY|X(2 |3) + 3pY|X(3 |3) = 1.71.
16. (a) P(X > 1 and Y > 1) = Z∞
1Z∞
1
xe−(x+xy)dy dx
∞
0
Page 81
82 CHAPTER 2
If x≤0 then f(x, y) = 0 for all yso fX(x) = 0.
0
∞
17. (a) Cov(X, Y ) = µXY −µXµY
1
fX(x) = 1
6x+9
2for 1 ≤x≤2 (see Example 2.55).
1
SECTION 2.6 83
(b) ρX,Y =Cov(X, Y )
σXσY
.
1
18. (a) P(X > 0.5 and Y > 0.5) = Z1
0.5Z1
0.5
3(x2+y2)
2dy dx
(b) For 0 < x < 1, fX(x) = Z1
0
3(x2+y2)
2dy = 0.5 + 1.5x2. For x≤0 and x≥1, fX(x) = 0.
3(x2+y2)
19. (a) Cov(X, Y ) = µXY −µXµY.
Page 83
84 CHAPTER 2
µX=Z1
0
x1 + 3x2
2dx =x2
4+3x4
81
=5
8.
(b) ρX,Y =Cov(X, Y )
σXσY
.
(c) fY|X(y|0.5) = fX,Y (0.5, y)
fX(0.5) .
0
Page 84
SECTION 2.6 85
20. (a) P(X > 1 and Y > 1) = Z∞
1Z∞
1
4xye−(2x+y)dy dx
∞
(b) For x≤0, fX(x) = 0.
For x > 0, fX(x) = Z∞
0
4xye−(2x+y)dy =−4x(1 + y)e−(2x+y)
∞
0
= 4xe−2x.
4xe−2xx > 0
21. (a) The probability mass function of Yis the same as that of X, so fY(y) = e−yif y > 0 and fY(y) = 0
0
1
Page 85
86 CHAPTER 2
1
∞
(c) µX=Z∞
0
xe−xdx =−xe−x
∞
0−Z∞
0
e−xdx = 0 −(−e−x)
∞
0
= 0 + 1 = 1
22. (a) P(X= 1) = 1/3, P(Y= 1) = 2/3, P(X= 1 and Y= 1) = 1/36=P(X= 1)P(Y= 1).
Page 86
SECTION 2.6 87
23. (a) R= 0.3X+ 0.7Y
(b) µR=µ0.3X+0.7Y= 0.3µX+ 0.7µY= (0.3)(6) + (0.7)(6) = 6.
24. µV=Z21
19 Z6
5
3πr2h(h−20)2(r−5) dr dh
Page 87
