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# Solucionario Walpole 8 ED

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```k =
(
3
392
)
10\u22124, and g(x) = k
(
20x2 + 98000
3
)
, with
µX = E(X) =
\u222b 50
30
xg(x) dx = k
\u222b 50
30
(
20x3 + 98000
3
x
)
dx = 40.8163.
Similarly, µY = 40.8163. On the other hand,
E(XY ) = k
\u222b 50
30
\u222b 50
30
xy(x2 + y2) dy dx = 1665.3061.
Hence, \u3c3XY = E(XY )\u2212 µXµY = 1665.3061\u2212 (40.8163)2 = \u22120.6642.
4.47 g(x) = 2
3
\u222b 1
0
(x+ 2y) dy = 2
3
(x+ 1, for 0 < x < 1, so µX =
2
3
\u222b 1
0
x(x+ 1) dx = 5
9
;
h(y) = 2
3
\u222b 1
0
(x+ 2y) dx = 2
3
(
1
2
+ 2y
)
, so µY =
2
3
\u222b 1
0
y
(
1
2
+ 2y
)
dy = 11
18
; and
E(XY ) = 2
3
\u222b 1
0
\u222b 1
0
xy(x+ 2y) dy dx = 1
3
.
So, \u3c3XY = E(XY )\u2212 µXµY = 13 \u2212
(
5
9
) (
11
18
)
= \u22120.0062.
4.48 Since \u3c3XY = Cov(a+ bX,X) = b\u3c3
2
X and \u3c3
2
Y = b
2\u3c32X , then
\u3c1 = \u3c3XY
\u3c3X
\u3c3Y =
b\u3c32X\u221a
\u3c32Xb
2\u3c32X
= b|b| = sign of b.
Hence \u3c1 = 1 if b > 0 and \u3c1 = \u22121 if b < 0.
4.49 E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88
and E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.
So, V ar(X) = 1.62\u2212 0.882 = 0.8456 and \u3c3 = \u221a0.8456 = 0.9196.
4.50 E(X) = 2
\u222b 1
0
x(1\u2212 x) dx = 2
(
x2
2
\u2212 x3
3
)\u2223\u2223\u22231
0
= 1
3
and
E(X2) = 2
\u222b 1
0
x2(1\u2212 x) dx = 2
(
x3
3
\u2212 x4
4
)\u2223\u2223\u22231
0
= 1
6
. Hence,
V ar(X) = 1
6
\u2212 (1
3
)2
= 1
18
, and \u3c3 =
\u221a
1/18 = 0.2357.
4.51 Previously we found µ = 4.11 and \u3c32 = 0.74, Therefore,
µg(X) = E(3X \u2212 2) = 3µ\u2212 2 = (3)(4.11)\u2212 2 = 10.33 and \u3c3g(X) = 9\u3c32 = 6.66.
4.52 Previously we found µ = 1 and \u3c32 = 1. Therefore,
µg(X) = E(5X + 3) = 5µ+ 3 = (5)(1) + 3 = 8 and \u3c3g(X) = 25\u3c3
2 = 25.
4.53 Let X = number of cartons sold and Y = profit.
We can write Y = 1.65X + (0.90)(5\u2212X)\u2212 6 = 0.75X \u2212 1.50. Now
E(X) = (0)(1/15)+(1)(2/15)+(2)(2/15)+(3)(3/15)+(4)(4/15)+(5)(3/15) = 46/15,
and E(Y ) = (0.75)E(X)\u2212 1.50 = (0.75)(46/15)\u2212 1.50 = \$0.80.
4.54 µX = E(X) =
1
4
\u222b\u221e
0
xe\u2212x/4 dx = 4.
Therefore, µY = E(3X \u2212 2) = 3E(X)\u2212 2 = (3)(4)\u2212 2 = 10.
Since E(X2) = 1
4
\u222b\u221e
0
x2e\u2212x/4 dx = 32, therefore, \u3c32X = E(X
2)\u2212 µ2X = 32\u2212 16 = 16.
Hence \u3c32Y = 9\u3c3
2
X = (9)(16) = 144.
Solutions for Exercises in Chapter 4 51
4.55 E(X) = (\u22123)(1/6) + (6)(1/2) + (9)(1/3) = 11/2,
E(X2) = (\u22123)2(1/6) + (6)2(1/2) + (9)2(1/3) = 93/2. So,
E[(2X + 1)2] = 4E(X2) + 4E(X) + 1 = (4)(93/2) + (4)(11/2) + 1 = 209.
4.56 Since E(X) =
\u222b 1
0
x2 dx+
\u222b 2
1
x(2\u2212 x) dx = 1, and
E(X2) =
\u222b 1
0
x32 dx+
\u222b 2
1
x2(2\u2212 x) dx = 7/6,then
E(Y ) = 60E(X2) + 39E(X) = (60)(7/6) + (39)(1) = 109 kilowatt hours.
4.57 The equations E[(X \u2212 1)2] = 10 and E[(X \u2212 2)2] = 6 may be written in the form:
E(X2)\u2212 2E(X) = 9, E(X2)\u2212 4E(X) = 2.
Solving these two equations simultaneously we obtain
E(X) = 7/2, and E(X2) = 16.
Hence µ = 7/2 and \u3c32 = 16\u2212 (7/2)2 = 15/4.
4.58 E(X) = (2)(0.40) + (4)(0.60) = 3.20, and
E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3. So,
(a) E(2X \u2212 3Y ) = 2E(X)\u2212 3E(Y ) = (2)(3.20)\u2212 (3)(3.00) = \u22122.60.
(b) E(XY ) = E(X)E(Y ) = (3.20)(3.00) = 9.60.
4.59 E(2XY 2 \u2212X2Y ) = 2E(XY 2)\u2212 E(X2Y ). Now,
E(XY 2) =
2\u2211
x=0
2\u2211
y=0
xy2f(x, y) = (1)(1)2(3/14) = 3.14, and
E(X2Y ) =
2\u2211
x=0
2\u2211
y=0
x2yf(x, y) = (1)2(1)(3/14) = 3.14.
Therefore, E(2XY 2 \u2212X2Y ) = (2)(3/14)\u2212 (3/14) = 3/14.
4.60 Using µ = 60 and \u3c3 = 6 and Chebyshev\u2019s theorem
P (µ\u2212 k\u3c3 < X < µ+ k\u3c3) \u2265 1\u2212 1
k2
,
since from µ+ k\u3c3 = 84 we obtain k = 4.
So, P (X < 84) \u2265 P (36 < X < 84) \u2265 1\u2212 1
42
= 0.9375. Therefore,
P (X \u2265 84) \u2264 1\u2212 0.9375 = 0.0625.
Since 1000(0.0625) = 62.5, we claim that at most 63 applicants would have a score as
84 or higher. Since there will be 70 positions, the applicant will have the job.
4.61 µ = 900 hours and \u3c3 = 50 hours. Solving µ\u2212 k\u3c3 = 700 we obtain k = 4.
So, using Chebyshev\u2019s theorem with P (µ \u2212 4\u3c3 < X < µ + 4\u3c3) \u2265 1 \u2212 1/42 = 0.9375,
we obtain P (700 < X < 1100) \u2265 0.9375. Therefore, P (X \u2264 700) \u2264 0.03125.
52 Chapter 4 Mathematical Expectation
4.62 µ = 52 and \u3c3 = 6.5. Solving µ+ k\u3c3 = 71.5 we obtain k = 3. So,
P (µ\u2212 3\u3c3 < X < µ+ 3\u3c3) \u2265 1\u2212 1
32
= 0.8889,
which is
P (32.5 < X < 71.5) \u2265 0.8889.
we obtain P (X > 71.5) < 1\u22120.8889
2
= 0.0556 using the symmetry.
4.63 n = 500, µ = 4.5 and \u3c3 = 2.8733. Solving µ+ k(\u3c3/
\u221a
500) = 5 we obtain
k =
5\u2212 4.5
2.87333/
\u221a
500
=
0.5
0.1284
= 3.8924.
So, P (4 \u2264 X¯ \u2264 5) \u2265 1\u2212 1
k2
= 0.9340.
4.64 \u3c32Z = \u3c3
2
\u22122X+4Y \u22123 = 4\u3c3
2
X + 16\u3c3
2
Y = (4)(5) + (16)(3) = 68.
4.65 \u3c32Z = \u3c3
2
\u22122X+4Y \u22123 = 4\u3c3
2
X + 16\u3c3
2
Y \u2212 16\u3c3XY = (4)(5) + (16)(3)\u2212 (16)(1) = 52.
4.66 (a) P (6 < X < 18) = P [12\u2212 (2)(3) < X < 12 + (2)(3)] \u2265 1\u2212 1
22
= 3
4
.
(b) P (3 < X < 21) = P [12\u2212 (3)(3) < X < 12 + (3)(3)] \u2265 1\u2212 1
32
= 8
9
.
4.67 (a) P (|X \u2212 10| \u2265 3) = 1\u2212 P (|X \u2212 10| < 3)
= 1\u2212 P [10\u2212 (3/2)(2) < X < 10 + (3/2)(2)] \u2264 1\u2212
[
1\u2212 1
(3/2)2
]
= 4
9
.
(b) P (|X \u2212 10| < 3) = 1\u2212 P (|X \u2212 10| \u2265 3) \u2265 1\u2212 4
9
= 5
9
.
(c) P (5 < X < 15) = P [10\u2212 (5/2)(2) < X < 10 + (5/2)(2)] \u2265 1\u2212 1
(5/2)2
= 21
25
.
(d) P (|X \u2212 10| \u2265 c) \u2264 0.04 implies that P (|X \u2212 10| < c) \u2265 1\u2212 0.04 = 0.96.
Solving 0.96 = 1\u2212 1
k2
we obtain k = 5. So, c = k\u3c3 = (5)(2) = 10.
4.68 µ = E(X) = 6
\u222b 1
0
x2(1 \u2212 x) dx = 0.5, E(X2) = 6 \u222b 1
0
x3(1 \u2212 x) dx = 0.3, which imply
\u3c32 = 0.3\u2212 (0.5)2 = 0.05 and \u3c3 = 0.2236. Hence,
P (µ\u2212 2\u3c3 < X < µ+ 2\u3c3) = P (0.5\u2212 0.4472 < X < 0.5 + 0.4472)
= P (0.0528 < X < 0.9472) = 6
\u222b 0.9472
0.0528
x(1\u2212 x) dx = 0.9839,
compared to a probability of at least 0.75 given by Chebyshev\u2019s theorem.
4.69 It is easy to see that the expectations of X and Y are both 3.5. So,
(a) E(X + Y ) = E(X) + E(Y ) = 3.5 + 3.5 = 7.0.
(b) E(X \u2212 Y ) = E(X)\u2212 E(Y ) = 0.
Solutions for Exercises in Chapter 4 53
(c) E(XY ) = E(X)E(Y ) = (3.5)(3.5) = 12.25.
4.70 E(Z) = E(XY ) = E(X)E(Y ) =
\u222b 1
0
\u222b\u221e
2
16xy(y/x3) dx dy = 8/3.
4.71 E[g(X, Y )] = E(X/Y 3 +X2Y ) = E(X/Y 3) + E(X2Y ).
E(X/Y 3) =
\u222b 2
1
\u222b 1
0
2x(x+2y)
7y3
dx dy = 2
7
\u222b 2
1
(
1
3y3
+ 1
y2
)
dy = 15
84
;
E(X2Y ) =
\u222b 2
1
\u222b 1
0
2x2y(x+2y)
7
dx dy = 2
7
\u222b 2
1
y
(
1
4
+ 2y
3
)
dy = 139
252
.
Hence, E[g(X, Y )] = 15
84
+ 139
252
= 46
63
.
4.72 µX = µY = 3.5. \u3c3
2
X = \u3c3
2
Y = [(1)
2 + (2)2 + · · ·+ (6)2](1/6)\u2212 (3.5)2 = 35
12
.
(a) \u3c32X\u2212Y = 4\u3c32X + \u3c3
2
Y =
175
12
;
(b) \u3c3X+3Y \u22125 = \u3c32X + 9\u3c3
2
Y =
175
6
.
4.73 (a) µ = 1
5
\u222b 5
0
x dx = 2.5, \u3c32 = E(X2)\u2212 µ2 = 1
5
x2
\u222b 5
0
x2 dx\u2212 2.52 = 2.08.
So, \u3c3 =
\u221a
\u3c32 = 1.44.
(b) By Chebyshev\u2019s theorem,
P [2.5\u2212 (2)(1.44) < X < 2.5 + (2)(1.44)] = P (\u22120.38 < X < 5.38) \u2265 0.75.
Using integration, P (\u22120.38 < X < 5.38) = 1 \u2265 0.75;
P [2.5\u2212 (3)(1.44) < X < 2.5 + (3)(1.44)] = P (\u22121.82 < X < 6.82) \u2265 0.89.
Using integration, P (\u22121.82 < X < 6.82) = 1 \u2265 0.89.
4.74 P = I2R with R = 50, µI = E(I) = 15 and \u3c3
2
I = V ar(I) = 0.03.
E(P ) = E(I2R) = 50E(I2) = 50[V ar(I) + µ2I ] = 50(0.03 + 15
2) = 11251.5. If we use
the approximation formula, with g(I) = I2, g\u2032(I) = 2I and g\u2032\u2032(I) = 2, we obtain,
E(P ) \u2248 50
[
g(µI) + 2
\u3c32I
2
]
= 50(152 + 0.03) = 11251.5.
Since V ar[g(I)] \u2248
[
\u2202g(i)
\u2202i
]2
i=µI
\u3c32I , we obtain
V ar(P ) = 502V ar(I2) = 502(2µI)
2\u3c32I = 50
2(30)2(0.03) = 67500.
4.75 For 0 < a < 1, since g(a) =
\u221e\u2211
x=0
ax = 1
1\u2212a , g
\u2032(a) =
\u221e\u2211
x=1
xax\u22121 = 1
(1\u2212a)2 and
g\u2032\u2032(a) =
\u221e\u2211
x=2
x(x\u2212 1)ax\u22122 = 2
(1\u2212a)3 .
54 Chapter 4 Mathematical Expectation
(a) E(X) = (3/4)
\u221e\u2211
x=1
x(1/4)x = (3/4)(1/4)
\u221e\u2211
x=1
x(1/4)x\u22121 = (3/16)[1/(1\u2212 1/4)2]
= 1/3, and E(Y ) = E(X) = 1/3.
E(X2)\u2212E(X) = E[X(X \u2212 1)] = (3/4)
\u221e\u2211
x=2
x(x\u2212 1)(1/4)x
= (3/4)(1/4)2
\u221e\u2211
x=2
x(x\u2212 1)(1/4)x\u22122 = (3/43)[2/(1\u2212 1/4)3] = 2/9.
So, V ar(X) = E(X2)\u2212 [E(X)]2 = [E(X2)\u2212 E(X)] + E(X)\u2212 [E(X)]2
2/9 + 1/3\u2212 (1/3)2 = 4/9, and V ar(Y ) = 4/9.
(b) E(Z) = E(X) + E(Y ) = (1/3) + (1/3) = 2/3, and
V ar(Z)```