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# Solucionario Walpole 8 ED

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```3.81 P (the system does not fail) = P (at least one of the components works)
= 1\u2212P (all components fail) = 1\u2212 (1\u22120.95)(1\u22120.94)(1\u22120.90)(1\u22120.97) = 0.999991.
3.82 Denote by X the number of components (out of 5) work.
Then, P (the system is operational) = P (X \u2265 3) = P (X = 3) + P (X = 4) + P (X =
5) =
(
5
3
)
(0.92)3(1\u2212 0.92)2 + (5
4
)
(0.92)4(1\u2212 0.92) + (5
5
)
(0.92)5 = 0.9955.
Chapter 4
Mathematical Expectation
4.1 E(X) = 1
pia2
\u222b a
\u2212a
\u222b\u221aa2\u2212y2
\u2212
\u221a
a2\u2212y2
x dx dy = 1
pia2
[(
a2\u2212y2
2
)
\u2212
(
a2\u2212y2
2
)]
dy = 0.
4.2 E(X) =
3\u2211
x=0
x f(x) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 3/4.
4.3 µ = E(X) = (20)(1/5) + (25)(3/5) + (30)(1/5) = 25 cents.
4.4 Assigning wrights of 3w and w for a head and tail, respectively. We obtain P (H) = 3/4
and P (T ) = 1/4. The sample space for the experiment is S = {HH,HT, TH, TT}.
Now if X represents the number of tails that occur in two tosses of the coin, we have
P (X = 0) = P (HH) = (3/4)(3/4) = 9/16,
P (X = 1) = P (HT ) + P (TH) = (2)(3/4)(1/4) = 3/8,
P (X = 2) = P (TT ) = (1/4)(1/4) = 1/16.
The probability distribution for X is then
x 0 1 2
f(x) 9/16 3/8 1/16
from which we get µ = E(X) = (0)(9/16) + (1)(3/8) + (2)(1/16) = 1/2.
4.5 µ = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.
4.6 µ = E(X) = (\$7)(1/12)+(\$9)(1/12)+(\$11)(1/4)+(\$13)(1/4)+(\$15)(1/6)+(\$17)(1/6)
= \$12.67.
4.7 Expected gain = E(X) = (4000)(0.3) + (\u22121000)(0.7) = \$500.
4.8 Let X = profit. Then
µ = E(X) = (250)(0.22) + (150)(0.36) + (0)(0.28) + (\u2212150)(0.14) = \$88.
45
46 Chapter 4 Mathematical Expectation
4.9 Let c = amount to play the game and Y = amount won.
y 5\u2212 c 3\u2212 c \u2212c
f(y) 2/13 2/13 9/13
E(Y ) = (5 \u2212 c)(2/13) + (3 \u2212 c)(2/13) + (\u2212c)(9/13) = 0. So, 13c = 16 which implies
c = \$1.23.
4.10 µX =
\u2211
xg(x) = (1)(0.17) + (2)(0.5) + (3)(0.33) = 2.16,
µY =
\u2211
yh(y) = (1)(0.23) + (2)(0.5) + (3)(0.27) = 2.04.
4.11 For the insurance of \$200,000 pilot, the distribution of the claim the insurance company
would have is as follows:
Claim Amount \$200,000 \$100,000 \$50,000 0
f(x) 0.002 0.01 0.1 0.888
So, the expected claim would be
(\$200, 000)(0.002) + (\$100, 000)(0.01) + (\$50, 000)(0.1) + (\$0)(0.888) = \$6, 400.
Hence the insurance company should charge a premium of \$6, 400 + \$500 = \$6, 900.
4.12 E(X) =
\u222b 1
0
2x(1\u2212 x) dx = 1/3. So, (1/3)(\$5, 000) = \$1, 667.67.
4.13 E(X) = 4
pi
\u222b 1
0
x
1+x2
dx = ln 4
pi
.
4.14 E(X) =
\u222b 1
0
2x(x+2)
5
dx = 8
15
.
4.15 E(X) =
\u222b 1
0
x2 dx +
\u222b 2
1
x(2 \u2212 x) dx = 1. Therefore, the average number of hours per
year is (1)(100) = 100 hours.
4.16 P (X1 +X2 = 1) = P (X1 = 1, X2 = 0) + P (X1 = 0, X2 = 1)
=
(9801 )(
20
1 )
(10002 )
+
(9801 )(
20
1 )
(10002 )
= (2)(0.0392) = 0.0784.
4.17 The probability density function is,
x \u22123 6 9
f(x) 1/6 1/2 1/3
g(x) 25 169 361
µg(X) = E[(2X + 1)
2] = (25)(1/6) + (169)(1/2) + (361)(1/3) = 209.
4.18 E(X2) = (0)(27/64) + (1)(27/64) + (4)(9/64) + (9)(1/64) = 9/8.
4.19 Let Y = 1200X \u2212 50X2 be the amount spent.
Solutions for Exercises in Chapter 4 47
x 0 1 2 3
f(x) 1/10 3/10 2/5 1/5
y = g(x) 0 1150 2200 3150
µY = E(1200X \u2212 50X2) = (0)(1/10) + (1150)(3/10) + (2200)(2/5) + (3150)(1/5)
= \$1, 855.
4.20 E[g(X)] = E(e2X/3) =
\u222b\u221e
0
e2x/3e\u2212x dx =
\u222b\u221e
0
e\u2212x/3 dx = 3.
4.21 E(X2) =
\u222b 1
0
2x2(1 \u2212 x) dx = 1
6
. Therefore, the average profit per new automobile is
(1/6)(\$5000.00) = \$833.33.
4.22 E(Y ) = E(X + 4) =
\u222b\u221e
0
32(x+ 4) 1
(x+4)3
dx = 8 days.
4.23 (a) E[g(X, Y )] = E(XY 2) =
\u2211
x
\u2211
y
xy2f(x, y)
= (2)(1)2(0.10) + (2)(3)2(0.20) + (2)(5)2(0.10) + (4)(1)2(0.15) + (4)(3)2(0.30)
+ (4)(5)2(0.15) = 35.2.
(b) µX = E(X) = (2)(0.40) + (4)(0.60) = 3.20,
µY = E(Y ) = (1)(0.25) + (3)(0.50) + (5)(0.25) = 3.00.
4.24 (a) E(X2Y \u2212 2XY ) =
3\u2211
x=0
2\u2211
y=0
(x2y \u2212 2xy)f(x, y) = (1\u2212 2)(18/70) + (4\u2212 4)(18/70) +
· · ·+ (8\u2212 8)(3/70) = \u22123/7.
(b)
x 0 1 2 3
g(x) 5/70 30/70 30/70 5/70
y 0 1 2
h(y) 15/70 40/70 15/70
µX = E(X) = (0)(5/70) + (1)(30/70) + (2)(30/70) + (3)(5/70) = 3/2,
µY = E(Y ) = (0)(15/70) + (1)(40/70) + (2)(15/70) = 1.
4.25 µX+Y = E(X + Y ) =
3\u2211
x=0
3\u2211
y=0
(x+ y)f(x, y) = (0+ 0)(1/55)+ (1+ 0)(6/55)+ · · ·+ (0+
3)(1/55) = 2.
4.26 E(Z) = E(
\u221a
X2 + Y 2) =
\u222b 1
0
\u222b 1
0
4xy
\u221a
x2 + y2 dx dy = 4
3
\u222b 1
0
[y(1 + y2)3/2 \u2212 y4] dy
= 8(23/2 \u2212 1)/15 = 0.9752.
4.27 E(X) = 1
2000
\u222b\u221e
0
x exp(\u2212x/2000) dx = 2000 \u222b\u221e
0
y exp(\u2212y) dy = 2000.
4.28 (a) The density function is shown next.
f(x
)
23.75 26.25
2/5
48 Chapter 4 Mathematical Expectation
(b) E(X) = 2
5
\u222b 26.25
23.75
x dx = 1
5
(26.252 \u2212 23.752) = 25.
(c) The mean is exactly in the middle of the interval. This should not be surprised
due to the symmetry of the density at 25.
4.29 (a) The density function is shown next
f(x
)
1 1.5 2 2.5 3 3.5 40
1
2
3
(b) µ = E(X) =
\u222b\u221e
1
3x\u22123 dx = 3
2
.
4.30 E(Y ) = 1
4
\u222b\u221e
0
ye\u2212y/4 dy = 4.
4.31 (a) µ = E(Y ) = 5
\u222b 1
0
y(1\u2212 y)4 dy = \u2212 \u222b 1
0
y d(1\u2212 y)5 = \u222b\u221e
0
(1\u2212 y)5 dy = 1
6
.
(b) P (Y > 1/6) =
\u222b 1
1/6
5(1\u2212 y)4 dy = \u2212 (1\u2212 y)5|11/6 = (1\u2212 1/6)5 = 0.4019.
4.32 (a) A histogram is shown next.
1 2 3 4 5
xx
f(x
)
f(x
)
0
0.1
0.2
0.3
0.4
(b) µ = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88.
(c) E(X2) = (0)2(0.41) + (1)2(0.37) + (2)2(0.16) + (3)2(0.05) + (4)2(0.01) = 1.62.
(d) V ar(X) = 1.62\u2212 0.882 = 0.8456.
4.33 µ = \$500. So, \u3c32 = E[(X \u2212 µ)2] = \u2211
x
(x \u2212 µ)2f(x) = (\u22121500)2(0.7) + (3500)2(0.3) =
\$5, 250, 000.
Solutions for Exercises in Chapter 4 49
4.34 µ = (\u22122)(0.3) + (3)(0.2) + (5)(0.5) = 2.5 and
E(X2) = (\u22122)2(0.3) + (3)2(0.2) + (5)2(0.5) = 15.5.
So, \u3c32 = E(X2)\u2212 µ2 = 9.25 and \u3c3 = 3.041.
4.35 µ = (2)(0.01) + (3)(0.25) + (4)(0.4) + (5)(0.3) + (6)(0.04) = 4.11,
E(X2) = (2)2(0.01) + (3)2(0.25) + (4)2(0.4) + (5)2(0.3) + (6)2(0.04) = 17.63.
So, \u3c32 = 17.63\u2212 4.112 = 0.74.
4.36 µ = (0)(0.4) + (1)(0.3) + (2)(0.2) + (3)(0.1) = 1.0,
and E(X2) = (0)2(0.4) + (1)2(0.3) + (2)2(0.2) + (3)2(0.1) = 2.0.
So, \u3c32 = 2.0\u2212 1.02 = 1.0.
4.37 It is know µ = 1/3.
So, E(X2) =
\u222b 1
0
2x2(1\u2212 x) dx = 1/6 and \u3c32 = 1/6\u2212 (1/3)2 = 1/18. So, in the actual
profit, the variance is 1
18
(5000)2.
4.38 It is known µ = 8/15.
Since E(X2) =
\u222b 1
0
2
5
x2(x+ 2) dx = 11
30
, then \u3c32 = 11/30\u2212 (8/15)2 = 37/450.
4.39 It is known µ = 1.
Since E(X2) =
\u222b 1
0
x2 dx+
\u222b 2
1
x2(2\u2212 x) dx = 7/6, then \u3c32 = 7/6\u2212 (1)2 = 1/6.
4.40 µg(X) = E[g(X)] =
\u222b 1
0
(3x2 + 4)
(
2x+4
5
)
dx = 1
5
\u222b 1
0
(6x3 + 12x2 + 8x+ 16) dx = 5.1.
So, \u3c32 = E[g(X)\u2212 µ]2 = \u222b 1
0
(3x2 + 4\u2212 5.1)2 (2x+4
5
)
dx
=
\u222b 1
0
(9x4 \u2212 6.6x2 + 1.21) (2x+4
5
)
dx = 0.83.
4.41 It is known µg(X) = E[(2X + 1)
2] = 209. Hence
\u3c32g(X) =
\u2211
x
[(2X + 1)2 \u2212 209]2g(x)
= (25\u2212 209)2(1/6) + (169\u2212 209)2(1/2) + (361\u2212 209)2(1/3) = 14, 144.
So, \u3c3g(X) =
\u221a
14, 144 = 118.9.
4.42 It is known µg(X) = E(X
2) = 1/6. Hence
\u3c32g(X) =
\u222b 1
0
2
(
x2 \u2212 1
6
)2
(1\u2212 x) dx = 7/180.
4.43 µY = E(3X \u2212 2) = 14
\u222b\u221e
0
(3x\u2212 2)e\u2212x/4 dx = 10. So
\u3c32Y = E{[(3X \u2212 2)\u2212 10]2} = 94
\u222b\u221e
0
(x\u2212 4)2e\u2212x/4 dx = 144.
4.44 E(XY ) =
\u2211
x
\u2211
y
xyf(x, y) = (1)(1)(18/70) + (2)(1)(18/70)
+ (3)(1)(2/70) + (1)(2)(9/70) + (2)(2)(3/70) = 9/7;
µX =
\u2211
x
\u2211
y
xf(x, y) = (0)f(0, 1) + (0)f(0, 2) + (1)f(1, 0) + · · ·+ (3)f(3, 1) = 3/2,
and µY = 1.
So, \u3c3XY = E(XY )\u2212 µXµY = 9/7\u2212 (3/2)(1) = \u22123/14.
50 Chapter 4 Mathematical Expectation
4.45 µX =
\u2211
x
xg(x) = 2.45, µY =
\u2211
y
yh(y) = 2.10, and
E(XY ) =
\u2211
x
\u2211
x
xyf(x, y) = (1)(0.05) + (2)(0.05) + (3)(0.10) + (2)(0.05)
+ (4)(0.10) + (6)(0.35) + (3)(0) + (6)(0.20) + (9)(0.10) = 5.15.
So, \u3c3XY = 5.15\u2212 (2.45)(2.10) = 0.005.
4.46 From previous exercise,```