soluciona ejercicios fisica

7/30/2019 soluciona ejercicios fisica

1/11

Chapter 4 Selected Problem Solutions

Section 4-2

4-1. a) 3679.0)()1( 1

11

====<

eedxeXP xx

b) 2858.0)()5.21( 5.215.2

1

5.2

1

==== x

dxx

XP because 0)( =xfX for x > 5.

c) 5625.016

45

168)54(

225

4

25

4

=

=== 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are

mutually exclusive. Then, P(X < 2.25) = 0 and

P(X > 2.75) = 10.0)05.0(228.2

75.2

== dx .

b) If the probability density function is centered at 2.5 meters, then 2)( =xfX for

2.25 < x < 2.75 and all rods will meet specifications.

Section 4-3

4-11. a) P(X


2/11


3/11

Section 4-5

4-33. a) f(x)= 2.0 for 49.75 < x < 50.25.

E(X) = (50.25 + 49.75)/2 = 50.0,

144.0,0208.012

)75.4925.50()(

2

=== xandXV .

b) =x

dxxF75.49

0.2)( for 49.75 < x < 50.25. Therefore,

z) = 0.9, then P(Z < z) = 0.10 and z = 1.28e) P(1.24 < Z < z) = P(Z < z) P(Z < 1.24)

= P(Z < z) 0.10749.Therefore, P(Z < z) = 0.8 + 0.10749 = 0.90749 and z = 1.33


4/11

4-43. a) P(X < 13) = P(Z < (1310)/2)= P(Z < 1.5)

= 0.93319

b) P(X > 9) = 1 P(X < 9)= 1 P(Z < (910)/2)= 1

P(Z 60) =

===60

6

60

1.01.0 0025.01.0 eedxe xx

b) P(X < 10) = === 10

0

110

0

1.01.0 6321.011.0 eedxe xx

4-83. Let X denote the distance between major cracks. Then, X is an exponential random

variable with 2.0)(/1 == XE cracks/mile.

a) P(X > 10) =

===10

2

10

2.02.0 1353.02.0 eedxe xx

b) Let Y denote the number of cracks in 10 miles of highway. Because the distance

between cracks is exponential, Y is a Poisson random variable with 2)2.0(10 ==cracks per 10 miles.

P(Y = 2) = 2707.0!2

222=

e

c) 5/1 == X miles.

4-87. Let X denote the number of calls in 3 hours. Because the time between calls is an

exponential random variable, the number of calls in 3 hours is a Poisson random variable.

Now, the mean time between calls is 0.5 hours and = =1 05 2/ . calls per hour = 6 calls in3 hours.

8488.0!36

!26

!16

!061)3(1)4(

36261606

=

+++==

eeeeXPXP

Section 4-10

4-97. Let Y denote the number of calls in one minute. Then, Y is a Poisson random variable

with = 5 calls per minute.

a) P(Y = 4) = 1755.0!4

545=

e

b) P(Y > 2) = 1 - 8754.0!2

5

!1

5

!0

51)2(

251505

== eee

YP .

Let W denote the number of one minute intervals out of 10 that contain more than 2

calls. Because the calls are a Poisson process, W is a binomial random variable with n =10 and p = 0.8754.

Therefore, P(W = 10) = ( ) 2643.0)8754.01(8754.0 0101010 = .

4-101. Let X denote the number of bits until five errors occur. Then, X has an Erlang distribution

with r = 5 and = 10 5 error per bit.


8/11


9/11

95.03

5)ln())ln(()()( =

=


10/11


11/11

c) If P(X < x) = 0.90, then

>

2.0

5xZP = 0.9. Therefore,

2.0

5x= 1.65 and x = 5.33.

4-139. If P(0.002-x < X < 0.002+x), then P(-x/0.0004 < Z < x/0.0004) = 0.9973. Therefore,

x/0.0004 = 3 and x = 0.0012. The specifications are from 0.0008 to 0.0032.

4-141. If P(X > 10,000) = 0.99, then P(Z > 600000,10

) = 0.99. Therefore, 600000,10

= -2.33 and

398,11= .

4-143 X is an exponential distribution with E(X) = 7000 hours

a.) 5633.017000

1)5800(

5800

0

7000

5800

7000 ===<

edxeXP

x

b.) 9.07000

1

)(7000

==> x

x

dxexXP Therefore, 9.07000

=

x

e

and 5.737)9.0ln(7000 ==x hours

soluciona ejercicios fisica

Documents

Transcript of soluciona ejercicios fisica