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# Solucionario Walpole 8 ED

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```\u2202eb0+b1k1+b2k2
\u2202bi
)2\u2223\u2223\u2223\u2223\u2223
bi=\u3b2i, 0\u2264i\u22642
\u3c32bi = e
2(\u3b20+k1\u3b21+k2\u3b22)(\u3c320 + k
2
1\u3c3
2
1 + k
2
2\u3c3
2
2).
4.103 (a) E(Y ) = 10
\u222b 1
0
y(1\u2212 y)9 dy = \u2212 y(1\u2212 y)10|10 +
\u222b 1
0
(1\u2212 y)10 dy = 1
11
.
(b) E(1\u2212 Y ) = 1\u2212 E(Y ) = 10
11
.
(c) V ar(Z) = V ar(1\u2212 Y ) = V ar(Y ) = E(Y 2)\u2212 [E(Y )]2 = 10
112×12 = 0.006887.
Chapter 5
Some Discrete Probability
Distributions
5.1 This is a uniform distribution: f(x) = 1
10
, for x = 1, 2, . . . , 10.
Therefore P (X < 4) =
3\u2211
x=1
f(x) = 3
10
.
5.2 Binomial distribution with n = 12 and p = 0.5. Hence
P (X = 3) = P (X \u2264 3)\u2212 P (X \u2264 2) = 0.0730\u2212 0.0193 = 0.0537.
5.3 µ =
10\u2211
x=1
x
10
= 5.5, and \u3c32 =
10\u2211
x=1
(x\u22125.5)2
10
= 8.25.
5.4 For n = 5 and p = 3/4, we have
(a) P (X = 2) =
(
5
2
)
(3/4)2(1/4)3 = 0.0879,
(b) P (X \u2264 3) =
3\u2211
x=0
b(x; 5, 3/4) = 1\u2212 P (X = 4)\u2212 P (X = 5)
= 1\u2212 (5
4
)
(3/4)4(1/4)1 \u2212 (5
5
)
(3/4)5(1/4)0 = 0.3672.
5.5 We are considering a b(x; 20, 0.3).
(a) P (X \u2265 10) = 1\u2212 P (X \u2264 9) = 1\u2212 0.9520 = 0.0480.
(b) P (X \u2264 4) = 0.2375.
(c) P (X = 5) = 0.1789. This probability is not very small so this is not a rare event.
Therefore, P = 0.30 is reasonable.
5.6 For n = 6 and p = 1/2.
(a) P (2 \u2264 X \u2264 5) = P (X \u2264 5)\u2212 P (X \u2264 1) = 0.9844\u2212 0.1094.
(b) P (X < 3) = P (X \u2264 2) = 0.3438.
5.7 p = 0.7.
59
60 Chapter 5 Some Discrete Probability Distributions
(a) For n = 10, P (X < 5) = P (X \u2264 4) = 0.0474.
(b) For n = 20, P (X < 10) = P (X \u2264 9) = 0.0171.
5.8 For n = 8 and p = 0.6, we have
(a) P (X = 3) = b(3; 8, 0.6) = P (X \u2264 3)\u2212 P (X \u2264 2) = 0.1737\u2212 0.0498 = 0.1239.
(b) P (X \u2265 5) = 1\u2212 P (X \u2264 4) = 1\u2212 0.4059 = 0.5941.
5.9 For n = 15 and p = 0.25, we have
(a) P (3 \u2264 X \u2264 6) = P (X \u2264 6)\u2212 P (X \u2264 2) = 0.9434\u2212 0.2361 = 0.7073.
(b) P (X < 4) = P (X \u2264 3) = 0.4613.
(c) P (X > 5) = 1\u2212 P (X \u2264 5) = 1\u2212 0.8516 = 0.1484.
5.10 From Table A.1 with n = 12 and p = 0.7, we have
(a) P (7 \u2264 X \u2264 9) = P (X \u2264 9)\u2212 P (X \u2264 6) = 0.7472\u2212 0.1178 = 0.6294.
(b) P (X \u2264 5) = 0.0386.
(c) P (X \u2265 8) = 1\u2212 P (X \u2264 7) = 1\u2212 0.2763 = 0.7237.
5.11 From Table A.1 with n = 7 and p = 0.9, we have
P (X = 5) = P (X \u2264 5)\u2212 P (X \u2264 4) = 0.1497\u2212 0.0257 = 0.1240.
5.12 From Table A.1 with n = 9 and p = 0.25, we have P (X < 4) = 0.8343.
5.13 From Table A.1 with n = 5 and p = 0.7, we have
P (X \u2265 3) = 1\u2212 P (X \u2264 2) = 1\u2212 0.1631 = 0.8369.
5.14 (a) n = 4, P (X = 4) = 1\u2212 0.3439 = 0.6561.
(b) Assuming the series went to the seventh game, the probability that the Bulls won
3 of the first 6 games and then the seventh game is given by[(
6
3
)
(0.9)3(0.1)3
]
(0.9) = 0.0131.
(c) The probability that the Bulls win is always 0.9.
5.15 p = 0.4 and n = 5.
(a) P (X = 0) = 0.0778.
(b) P (X < 2) = P (X \u2264 1) = 0.3370.
(c) P (X > 3) = 1\u2212 P (X \u2264 3) = 1\u2212 0.9130 = 0.0870.
Solutions for Exercises in Chapter 5 61
5.16 Probability of 2 or more of 4 engines operating when p = 0.6 is
P (X \u2265 2) = 1\u2212 P (X \u2264 1) = 0.8208,
and the probability of 1 or more of 2 engines operating when p = 0.6 is
P (X \u2265 1) = 1\u2212 P (X = 0) = 0.8400.
The 2-engine plane has a slightly higher probability for a successful flight when p = 0.6.
5.17 Since µ = np = (5)(0.7) = 3.5 and \u3c32 = npq = (5)(0.7)(0.3) = 1.05 with \u3c3 = 1.025.
Then µ± 2\u3c3 = 3.5± (2)(1.025) = 3.5± 2.050 or from 1.45 to 5.55. Therefore, at least
3/4 of the time when 5 people are selected at random, anywhere from 2 to 5 are of the
opinion that tranquilizers do not cure but only cover up the real problem.
5.18 (a) µ = np = (15)(0.25) = 3.75.
(b) With k = 2 and \u3c3 =
\u221a
npq =
\u221a
(15)(0.25)(0.75) = 1.677, µ ± 2\u3c3 = 3.75± 3.354
or from 0.396 to 7.104.
5.19 Let X1 = number of times encountered green light with P (Green) = 0.35,
X2 = number of times encountered yellow light with P (Yellow) = 0.05, and
X3 = number of times encountered red light with P (Red) = 0.60. Then
f(x1, x2, x3) =
(
n
x1, x2, x3
)
(0.35)x1(0.05)x2(0.60)x3.
5.20 (a)
(
10
2,5,3
)
(0.225)2(0.544)5(0.231)3 = 0.0749.
(b)
(
10
10
)
(0.544)10(0.456)0 = 0.0023.
(c)
(
10
0
)
(0.225)0(0.775)10 = 0.0782.
5.21 Using the multinomial distribution with required probability is(
7
0, 0, 1, 4, 2
)
(0.02)(0.82)4(0.1)2 = 0.0095.
5.22 Using the multinomial distribution, we have
(
8
5,2,1
)
(1/2)5(1/4)2(1/4) = 21/256.
5.23 Using the multinomial distribution, we have(
9
3, 3, 1, 2
)
(0.4)3(0.2)3(0.3)(0.2)2 = 0.0077.
5.24 p = 0.40 and n = 6, so P (X = 4) = P (X \u2264 4)\u2212P (X \u2264 3) = 0.9590\u22120.8208 = 0.1382.
5.25 n = 20 and the probability of a defective is p = 0.10. So, P (X \u2264 3) = 0.8670.
62 Chapter 5 Some Discrete Probability Distributions
5.26 n = 8 and p = 0.60;
(a) P (X = 6) =
(
8
6
)
(0.6)6(0.4)2 = 0.2090.
(b) P (X = 6) = P (X \u2264 6)\u2212 P (X \u2264 5) = 0.8936\u2212 0.6846 = 0.2090.
5.27 n = 20 and p = 0.90;
(a) P (X = 18) = P (X \u2264 18)\u2212 P (X \u2264 17) = 0.6083\u2212 0.3231 = 0.2852.
(b) P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212 0.0113 = 0.9887.
(c) P (X \u2264 18) = 0.6083.
5.28 n = 20;
(a) p = 0.20, P (X \u2265 x) \u2264 0.5 and P (X < x) > 0.5 yields x = 4.
(b) p = 0.80, P (Y \u2265 y) \u2265 0.8 and P (Y < y) < 0.2 yields y = 14.
5.29 Using the hypergeometric distribution, we get
(a)
(122 )(
40
5 )
(527 )
= 0.3246.
(b) 1\u2212 (
48
7 )
(527 )
= 0.4496.
5.30 P (X \u2265 1) = 1\u2212 P (X = 0) = 1\u2212 h(0; 15, 3, 6) = 1\u2212 (
6
0)(
9
3)
(153 )
= 53
65
.
5.31 Using the hypergeometric distribution, we get h(2; 9, 6, 4) =
(42)(
5
4)
(96)
= 5
14
.
5.32 (a) Probability that all 4 fire = h(4; 10, 4, 7) = 1
6
.
(b) Probability that at most 2 will not fire =
2\u2211
x=0
h(x; 10, 4, 3) = 29
30
.
5.33 h(x; 6, 3, 4) =
(4x)(
2
3\u2212x)
(63)
, for x = 1, 2, 3.
P (2 \u2264 X \u2264 3) = h(2; 6, 3, 4) + h(3; 6, 3, 4) = 4
5
.
5.34 h(2; 9, 5, 4) =
(42)(
5
3)
(95)
= 10
21
.
5.35 P (X \u2264 2) =
2\u2211
x=0
h(x; 50, 5, 10) = 0.9517.
5.36 (a) P (X = 0) = h(0; 25, 3, 3) = 77
115
.
(b) P (X = 1) = h(1; 25, 3, 1) = 3
25
.
5.37 (a) P (X = 0) = b(0; 3, 3/25) = 0.6815.
Solutions for Exercises in Chapter 5 63
(b) P (1 \u2264 X \u2264 3) =
3\u2211
x=1
b(x; 3, 1/25) = 0.1153.
5.38 Since µ = (4)(3/10) = 1.2 and \u3c32 = (4)(3/10)(7/10)(6/9) = 504/900 with \u3c3 = 0.7483,
at least 3/4 of the time the number of defectives will fall in the interval
µ± 2\u3c3 = 1.2± (2)(0.7483), or from \u2212 0.297 to 2.697,
and at least 8/9 of the time the number of defectives will fall in the interval
µ± 3\u3c3 = 1.2± (3)(0.7483) or from \u2212 1.045 to 3.445.
5.39 Since µ = (13)(13/52) = 3.25 and \u3c32 = (13)(1/4)(3/4)(39/51) = 1.864 with \u3c3 = 1.365,
at least 75% of the time the number of hearts lay between
µ± 2\u3c3 = 3.25± (2)(1.365) or from 0.52 to 5.98.
5.40 The binomial approximation of the hypergeometric with p = 1 \u2212 4000/10000 = 0.6
gives a probability of
7\u2211
x=0
b(x; 15, 0.6) = 0.2131.
5.41 Using the binomial approximation of the hypergeometric with p = 0.5, the probability
is 1\u2212
2\u2211
x=0
b(x; 10, 0.5) = 0.9453.
5.42 Using the binomial approximation of the hypergeometric distribution with p = 30/150 =
0.2, the probability is 1\u2212
2\u2211
x=0
b(x; 10, 0.2) = 0.3222.
5.43 Using the binomial approximation of the hypergeometric distribution with 0.7, the
probability is 1\u2212
13\u2211
x=10
b(x; 18, 0.7) = 0.6077.
5.44 Using the extension of the hypergeometric distribution the probability is(
13
5
)(
13
2
)(
13
3
)(
13
3
)(
52
13
) = 0.0129.
5.45 (a) The extension of the hypergeometric distribution gives a probability(
2
1
)(
3
1
)(
5
1
)(
2
1
)(
12
4
) = 4
33
.
(b) Using the extension of the hypergeometric distribution, we have(
2
1
)(
3
1
)(
2
2
)(
12
4
) + (22)(31)(21)(12
4
) + (21)(32)(21)(12
4
) = 8
165
.
64 Chapter 5 Some Discrete Probability Distributions
5.46 Using the extension of the hypergeometric distribution```