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# Solucionario Walpole 8 ED

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```of switches that fail during the first year. Using the normal approximation we find
µ = (100)(0.3935) = 39.35, \u3c3 =
\u221a
(100)(0.3935)(0.6065) = 4.885, and z = (30.5 \u2212
39.35)/4.885 = \u22121.81. Therefore, P (Y \u2264 30) = P (Z < \u22121.81) = 0.0352.
6.47 (a) E(X) =
\u222b\u221e
0
x2e\u2212x
2/2 dx = \u2212xe\u2212x2/2
\u2223\u2223\u2223\u221e
0
+
\u222b\u221e
0
e\u2212x
2/2 dx
= 0 +
\u221a
2\u3c0 · 1\u221a
2pi
\u222b\u221e
0
e\u2212x
2/2 dx =
\u221a
2pi
2
=
\u221a
pi
2
= 1.2533.
(b) P (X > 2) =
\u222b\u221e
2
xe\u2212x
2/2 dx = \u2212e\u2212x2/2
\u2223\u2223\u2223\u221e
2
= e\u22122 = 0.1353.
6.48 µ = E(T ) = \u3b1\u3b2
\u222b\u221e
0
t\u3b2e\u2212\u3b1t
\u3b2
dt. Let y = \u3b1t\u3b2, then dy = \u3b1\u3b2t\u3b2\u22121 dt and t = (y/\u3b1)1/\u3b2.
Then
µ =
\u222b \u221e
0
(y/\u3b1)1/\u3b2e\u2212y dy = \u3b1\u22121/\u3b2
\u222b \u221e
0
y(1+1/\u3b2)\u22121e\u2212y dy = \u3b1\u22121/\u3b2\u393(1 + 1/\u3b2).
78 Chapter 6 Some Continuous Probability Distributions
E(T 2) = \u3b1\u3b2
\u222b \u221e
0
t\u3b2+1e\u2212\u3b1t
\u3b2
dt =
\u222b \u221e
0
(y/\u3b1)2/\u3b2e\u2212y dy = \u3b1\u22122/\u3b2
\u222b \u221e
0
y(1+2/\u3b2)\u22121e\u2212y dy
= \u3b1\u22122/\u3b2\u393(1 + 2/\u3b2).
So, \u3c32 = E(T 2)\u2212 µ2 = \u3b1\u22122/\u3b2{\u393(1 + 2/\u3b2)\u2212 [\u393(1 + 1/\u3b2)]2}.
6.49 R(t) = ce\u2212
R
1/
\u221a
t dt = ce\u22122
\u221a
t. However, R(0) = 1 and hence c = 1. Now
f(t) = Z(t)R(t) = e\u22122
\u221a
t/
\u221a
t, t > 0,
and
P (T > 4) =
\u222b \u221e
4
e\u22122
\u221a
t/
\u221a
t dt = \u2212e\u22122
\u221a
t
\u2223\u2223\u2223\u221e
4
= e\u22124 = 0.0183.
6.50 f(x) = 12x2(1\u2212 x), 0 < x < 1. Therefore,
P (X > 0.8) = 12
\u222b 1
0.8
x2(1\u2212 x) dx = 0.1808.
6.51 \u3b1 = 5; \u3b2 = 10;
(a) \u3b1\u3b2 = 50.
(b) \u3c32 = \u3b1\u3b22 = 500; so \u3c3 =
\u221a
500 = 22.36.
(c) P (X > 30) = 1
\u3b2\u3b1\u393(\u3b1)
\u222b\u221e
30
x\u3b1\u22121e\u2212x/\u3b2 dx. Using the incomplete gamma with y =
x/\u3b2, then
1\u2212 P (X \u2264 30) = 1\u2212 P (Y \u2264 3) = 1\u2212
\u222b 3
0
y4e\u2212y
\u393(5)
dy = 1\u2212 0.185 = 0.815.
6.52 \u3b1\u3b2 = 10; \u3c3 =
\u221a
\u3b1\u3b22
\u221a
50 = 7.07.
(a) Using integration by parts,
P (X \u2264 50) = 1
\u3b2\u3b1\u393(\u3b1)
\u222b 50
0
x\u3b1\u22121e\u2212x/\u3b2 dx =
1
25
\u222b 50
0
xe\u2212x/5 dx = 0.9995.
(b) P (X < 10) = 1
\u3b2\u3b1\u393(\u3b1)
\u222b 10
0
x\u3b1\u22121e\u2212x/\u3b2 dx. Using the incomplete gamma with y =
x/\u3b2, we have
P (X < 10) = P (Y < 2) =
\u222b 2
0
ye\u2212y dy = 0.5940.
6.53 µ = 3 seconds with f(x) = 1
3
e\u2212x/3 for x > 0.
Solutions for Exercises in Chapter 6 79
(a) P (X > 5) =
\u222b\u221e
5
1
3
e\u2212x/3 dx = 1
3
[\u22123e\u2212x/3]\u2223\u2223\u221e
5
= e\u22125/3 = 0.1889.
(b) P (X > 10) = e\u221210/3 = 0.0357.
6.54 P (X > 270) = 1\u2212 \u3a6 ( ln 270\u22124
2
)
= 1\u2212 \u3a6(0.7992) = 0.2119.
6.55 µ = E(X) = e4+4/2 = e6; \u3c32 = e8+4(e4 \u2212 1) = e12(e4 \u2212 1).
6.56 \u3b2 = 1/5 and \u3b1 = 10.
(a) P (X > 10) = 1\u2212 P (X \u2264 10) = 1\u2212 0.9863 = 0.0137.
(b) P (X > 2) before 10 cars arrive.
P (X \u2264 2) =
\u222b 2
0
1
\u3b2\u3b1
x\u3b1\u22121e\u2212x/\u3b2
\u393(\u3b1)
dx.
Given y = x/\u3b2, then
P (X \u2264 2) = P (Y \u2264 10) =
\u222b 10
0
y\u3b1\u22121e\u2212y
\u393(\u3b1)
dy =
\u222b 10
0
y10\u22121e\u2212y
\u393(10)
dy = 0.542,
with P (X > 2) = 1\u2212 P (X \u2264 2) = 1\u2212 0.542 = 0.458.
6.57 (a) P (X > 1) = 1\u2212 P (X \u2264 1) = 1\u2212 10 \u222b 1
0
e\u221210x dx = e\u221210 = 0.000045.
(b) µ = \u3b2 = 1/10 = 0.1.
6.58 Assume that Z(t) = \u3b1\u3b2t\u3b2\u22121, for t > 0. Then we can write f(t) = Z(t)R(t), where
R(t) = ce\u2212
R
Z(t) dt = ce\u2212
R
\u3b1\u3b2t\u3b2\u22121dt = ce\u2212\u3b1t
\u3b2
. From the condition that R(0) = 1, we find
that c = 1. Hence R(t) = e\u3b1t
\u3b2
and f(t) = \u3b1\u3b2t\u3b2\u22121e\u2212\u3b1t
\u3b2
, for t > 0. Since
Z(t) =
f(t)
R(t)
,
where
R(t) = 1\u2212 F (t) = 1\u2212
\u222b t
0
\u3b1\u3b2x\u3b2\u22121e\u2212\u3b1x
\u3b2
dx = 1 +
\u222b t
0
de\u2212\u3b1x
\u3b2
= e\u2212\u3b1t
\u3b2
,
then
Z(t) =
\u3b1\u3b2t\u3b2\u22121e\u2212\u3b1t
\u3b2
e\u2212\u3b1t\u3b2
= \u3b1\u3b2t\u3b2\u22121, t > 0.
6.59 µ = np = (1000)(0.49) = 490, \u3c3 =
\u221a
npq =
\u221a
(1000)(0.49)(0.51) = 15.808.
z1 =
481.5\u2212 490
15.808
= \u22120.54, z2 = 510.5\u2212 490
15.808
= 1.3.
P (481.5 < X < 510.5) = P (\u22120.54 < Z < 1.3) = 0.9032\u2212 0.2946 = 0.6086.
80 Chapter 6 Some Continuous Probability Distributions
6.60 P (X > 1/4) =
\u222b\u221e
1/4
6e\u22126x dx = \u2212e\u22126x|\u221e1/4 = e\u22121.5 = 0.223.
6.61 P (X < 1/2) = 108
\u222b 1/2
0
x2e\u22126x dx. Letting y = 6x and using Table A.24 we have
P (X < 1/2) = P (Y < 3) =
\u222b 3
0
y2e\u2212y dy = 0.577.
6.62 Manufacturer A:
P (X \u2265 10000) = P
(
Z \u2265 100000\u2212 14000
2000
)
= P (Z \u2265 \u22122) = 0.9772.
Manufacturer B:
P (X \u2265 10000) = P
(
Z \u2265 10000\u2212 13000
1000
)
= P (Z \u2265 \u22123) = 0.9987.
Manufacturer B will produce the fewest number of defective rivets.
6.63 Using the normal approximation to the binomial with µ = np = 650 and \u3c3 =
\u221a
npq =
15.0831. So,
P (590 \u2264 X \u2264 625) = P (\u221210.64 < Z < \u22128.92) \u2248 0.
6.64 (a) µ = \u3b2 = 100 hours.
(b) P (X \u2265 200) = 0.01 \u222b\u221e
200
e\u22120.01x dx = e\u22122 = 0.1353.
6.65 (a) µ = 85 and \u3c3 = 4. So, P (X < 80) = P (Z < \u22121.25) = 0.1056.
(b) µ = 79 and \u3c3 = 4. So, P (X \u2265 80) = P (Z > 0.25) = 0.4013.
6.66 1/\u3b2 = 1/5 hours with \u3b1 = 2 failures and \u3b2 = 5 hours.
(a) \u3b1\u3b2 = (2)(5) = 10.
(b) P (X \u2265 12) = \u222b\u221e
12
1
52\u393(2)
xe\u2212x/5 dx = 1
25
\u222b\u221e
12
xe\u2212x/5 dx =
[\u2212x
5
e\u2212x/5 \u2212 e\u2212x/5]\u221e
12
= 0.3084.
6.67 Denote by X the elongation. We have µ = 0.05 and \u3c3 = 0.01.
(a) P (X \u2265 0.1) = P (Z \u2265 0.1\u22120.05
0.01
)
= P (Z \u2265 5) \u2248 0.
(b) P (X \u2264 0.04) = P (Z \u2264 0.04\u22120.05
0.01
)
= P (Z \u2264 \u22121) = 0.1587.
(c) P (0.025 \u2264 X \u2264 0.065) = P (\u22122.5 \u2264 Z \u2264 1.5) = 0.9332\u2212 0.0062 = 0.9270.
6.68 Let X be the error. X \u223c n(x; 0, 4). So,
P (fails) = 1\u2212 P (\u221210 < X < 10) = 1\u2212 P (\u22122.25 < Z < 2.25) = 2(0.0122) = 0.0244.
Solutions for Exercises in Chapter 6 81
6.69 Let X be the time to bombing with µ = 3 and \u3c3 = 0.5. Then
P (1 \u2264 X \u2264 4) = P
(
1\u2212 3
0.5
\u2264 Z \u2264 4\u2212 3
0.5
)
= P (\u22124 \u2264 Z \u2264 2) = 0.9772.
P (of an undesirable product) is 1 \u2212 0.9772 = 0.0228. Hence a product is undesirable
is 2.28% of the time.
6.70 \u3b1 = 2 and \u3b2 = 100. P (X \u2264 200) = 1
\u3b22
\u222b 200
0
xe\u2212x/\u3b2 dx. Using the incomplete gamma
table and let y = x/\u3b2,
\u222b 2
0
ye\u2212y dy = 0.594.
6.71 µ = \u3b1\u3b2 = 200 hours and \u3c32 = \u3b1\u3b22 = 20, 000 hours.
6.72 X follows a lognormal distribution.
P (X \u2265 50, 000) = 1\u2212 \u3a6
(
ln 50, 000\u2212 5
2
)
= 1\u2212 \u3a6(2.9099) = 1\u2212 0.9982 = 0.0018.
6.73 The mean of X, which follows a lognormal distribution is µ = E(X) = eµ+\u3c3
2/2 = e7.
6.74 µ = 10 and \u3c3 =
\u221a
50.
(a) P (X \u2264 50) = P (Z \u2264 5.66) \u2248 1.
(b) P (X \u2264 10) = 0.5.
(c) The results are very similar.
6.75 (a) Since f(y) \u2265 0 and \u222b 1
0
10(1\u2212 y)9 dy = \u2212 (1\u2212 y)10|10 = 1, it is a density function.
(b) P (Y > 0.6) = \u2212 (1\u2212 y)10|10.6 = (0.4)10 = 0.0001.
(c) \u3b1 = 1 and \u3b2 = 10.
(d) µ = \u3b1
\u3b1+\u3b2
= 1
11
= 0.0909.
(e) \u3c32 = \u3b1\u3b2
(\u3b1+\u3b2)2(\u3b1+\u3b2+1)
= (1)(10)
(1+10)2(1+10+1)
= 0.006887.
6.76 (a) µ = 1
10
\u222b\u221e
0
ze\u2212z/10 dz = \u2212 ze\u2212z/10\u2223\u2223\u221e
0
+
\u222b\u221e
0
e\u2212z/10 dz = 10.
(b) Using integral by parts twice, we get
E(Z2) =
1
10
\u222b \u221e
0
z2e\u2212z/10 dz = 200.
So, \u3c32 = E(Z2)\u2212 µ2 = 200\u2212 (10)2 = 100.
(c) P (Z > 10) = \u2212 ez/10\u2223\u2223\u221e
10
= e\u22121 = 0.3679.
6.77 This is an exponential distribution with \u3b2 = 10.
(a) µ = \u3b2 = 10.
82 Chapter 6 Some Continuous Probability Distributions
(b) \u3c32 = \u3b22 = 100.
6.78 µ = 0.5 seconds and \u3c3 = 0.4 seconds.
(a) P (X > 0.3) = P
(
Z > 0.3\u22120.5
0.4
)
= P (Z > \u22120.5) = 0.6915.
(b) P (Z > \u22121.645) = 0.95. So, \u22121.645 = x\u22120.5
0.4
yields x = \u22120.158 seconds. The
negative number in reaction time is not reasonable. So, it means that the normal
model may not be accurate enough.
6.79 (a) For an exponential distribution with parameter \u3b2,
P (X > a+ b | X > a) = P (X > a+ b)
P (X > a)
=
e\u2212a\u2212b
e\u2212a
= e\u2212b = P (X > b).
So, P (it will breakdown in the next 21 days | it just broke down) = P (X > 21) =
e\u221221/15 = e\u22121.4 = 0.2466.
(b) P (X > 30) = e\u221230/15 = e\u22122 = 0.1353.
6.80 \u3b1 = 2 and \u3b2 = 50. So,
P (X \u2264 10) = 100
\u222b 10
0
x49e\u22122x
50
dx.
Let y = 2x50 with x = (y/2)1/50 and dx = 1
21/50(50)
y\u221249/50 dy.
P (X \u2264 10) = 100
21/50(50)
\u222b (2)1050
0
(y
2
)49/50
y\u221249/50e\u2212y dy =
\u222b (2)1050
0
e\u2212y dy \u2248 1.
6.81 The density function of a Weibull distribution is
f(y) = \u3b1\u3b2y\u3b2\u22121e\u2212\u3b1y
\u3b2
, y > 0.
So, for any y \u2265 0,
F (y) =
\u222b y
0
f(t) dt = \u3b1\u3b2
\u222b y
0
t\u3b2\u22121e\u2212\u3b1t
\u3b2
dt.
Let z = t\u3b2 which yields t = z1/\u3b2 and dt = 1
\u3b2
z1/\u3b2\u22121 dz. Hence,
F (y) = \u3b1\u3b2
\u222b y\u3b2
0
z1\u22121/\u3b2
1
\u3b2
z1/\u3b2\u22121e\u2212\u3b1z dz = \u3b1
\u222b y\u3b2
0
e\u2212\u3b1z dz = 1\u2212 e\u2212\u3b1y\u3b2 .
On the other hand, since de\u2212\u3b1y
\u3b2
= \u2212\u3b1\u3b2y\u3b2\u22121e\u2212\u3b1y\u3b2 , the above result follows immedi-
ately.
Solutions for Exercises in Chapter 6 83
6.82```