ch03a

Chapter 3

Section 3-2

3-1. Continuous3-2. Discrete3-3. Continuous3-4. Discrete3-5. Discrete3-6. Continuous3-7. Discrete3-8. Continuous3-9. Continuous

Section 3-3

3-10. a) Engineers with at least 36 months of full-time employment.b) Samples of cement blocks with compressive strength of at least 6000 kg per square centimeter.c) Measurements of the diameter of forged pistons that conform to engineering specifications.d) Cholesterol levels at most 180 or at least 220.

3-11. The intersection of A and B is empty, therefore P(XAB) = 0.a) Yes, since P(XAB) = 0.b) P(XA) = 1 - P(XA) = 1 – 0.4 = 0.6c) P(XB) = 1 - P(XB) = 1 – 0.6 = 0.4d) P(XAB) = P(XA) + P(XB) - P(XAB) = 0.4 + 0.6 – 0 = 1

3-12. a) P(XA) = 1 - P(XA) = 1 – 0.3 = 0.7b) P(XB) = 1 - P(XB) = 1 – 0.25 = 0.75c) P(XC) = 1 - P(XC) = 1 – 0.6 = 0.4d) A and B are mutually exclusive if P(XAB) = 0. To determine if A and B are mutually exclusive, solve the following for P(XAB): P(XAB) = P(XA) + P(XB) - P(XAB) 0.55 = 0.3 + 0.25 - P(XAB) 0.55 = 0.55 - P(XAB) and P(XAB) = 0. Therefore, A and B are mutually exclusive.e) B and C are mutually exclusive if P(XBC) = 0. To determine if B and C are mutually exclusive, solve the following for P(XBC): P(XBC) = P(XB) + P(XC) - P(XBC) 0.70 = 0.25 + 0.60 - P(XBC) 0.70 = 0.85 - P(XBC) and P(XBC) = 0.15. Therefore, B and C are not mutually exclusive.

3-13. a) P(X > 15) = 1 – P(X 15) = 1 – 0.3 = 0.7b) P(X 24) = P(X 15) + P(15 < X 24) = 0.3 + 0.6 = 0.9c) P(15 < X 20) = P(X 20) – P(X 15) = 0.5 – 0.3 = 0.2d) P(X 18) = P(15 < X 18) + P(X 15) where P(15 < X 18) = P(15 < X 24) – P(18 < X 24) = 0.6 – 0.4 = 0.2 Therefore, P(X 18) = P(15 < X 18) + P(X 15) = 0.2 + 0.3 = 0.5 Alternatively, P(X 18) = P(X 24) - P(18 < X 24) = 0.9 – 0.4 = 0.5.

3-14. A - Overfilled, B - Medium filled, C - Underfilleda) P(XC’) = 1 - P(XC) = 1 – 0.15 = 0.85

b) P(XAC) = P(XA) + P(XC) – P(XAC) = 0.40 + 0.15 – 0 = 0.55 (P(XAC) = 0 since A and C are mutually exclusive)

3-15. a) P(X 7000) = 1 – P(X > 7000) = 1 – 0.45 = 0.55b) P(X > 5000) = 1 – P(X 5000) = 1 – 0.05 = 0.95c) P(5000 < X 7000) = P(X 7000) – P(X 5000) = 0.55 – 0.05 = 0.50

3-16. a) Probability that a component does not fail: )E(P 1 = 1 – P(E1) = 1 – 0.15 = 0.85

b) P(E1 or E2) = P(E1) + P(E2) = 0.15 + 0.30 = 0.45c) P(E1 or E2) = 1 – 0.45 = 0.55

3-17. a) P(X = 2 or X = 3) = 0.25 + 0.25 = 0.50b) P(X < 2) = P(X = 0) + P(X = 1) = 0.10 +0.15 = 0.25c) P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.10 = 0.25d) P(X > 0) = 1 – P(X = 0) = 1 – 0.1 = 0.9

3-18. a) P(X < 1) = P(X = 0) = 0.7b) P(X > 3) = 1 – P(X ≤ 3) = 1 – 1 = 0c) P(X > 0) = 1 – P(X = 0) = 1 – 0.7 = 0.3d) P(X = 0) = 0.7

Section 3-4

3-19. a) kx dx k kx2

0

4

3 0

43 643 . Therefore, k = 3/64.

E X x dx x( ) 3

643

64 433

4

0

4

0

4

V X x x dx x x x dx

x x x

( ) ( ) ( )

.

3

643 3

646 9

364

0 6

2 2

0

44 3 2

0

4

56

49

3 0

45 4 3

b) k x dx k x x k( ) ( )1 2 60

22

0

2 . Therefore, k = 1/6.

E X x x dx x x dx x x( ) ( ) ( ) ( ) / 16

1 2 16

2 16 2

23

11 92

0

2

0

2 2 3

0

2

284.06704.12

)2(

))(21()(

2

0281242

3944

481121

2922

361

2

081

24229443

81121

9222

61

2

0

2911

61

23423 ==+−++−=

+−++−=

−+=

∫

∫

xxxxx x

dxxxxxx

dxxxXV

c) ke dx k e kx x

0 0

( ) . Therefore, k = 1.

E X xe dxx( )

0

, using integration by parts E X xe e dx ex x x( )

0 0 0

1

V X x e dx x e xe e dxx x x x( ) ( ) ( )

1 22

0

2

0. Now, using integration by parts

x e dx xex x2

0 02

. Therefore, V X e dxx( )

0

1 , because e x is a probability density

function.

d) 2100100

100

100kkxkdx kk

, k2 = 1, so k =1 (where k > 0)

E(X) = 5.1001101

100 xdx

V(X) = 08333.01)5.100(101

100

2 − dxx

3-20. For 3-19a:

a) 40,64

3)(2

<< xxxf

x

f(x)

43210

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

b) F(X) = , 0 < x < 4

mean

c)

x 43210

1.0

0.5

0.0

x

F(X)

3-19b:

a)

210

0.85

0.75

0.65

0.55

0.45

0.35

0.25

0.15

x

f(x)

b) F(X) = 20,61

6)21(

0

2 <<∫ +=+

xxxdttx

c)

43210

1.0

0.5

0.0

x

F(x)

3-19c: xexf − , x >0a)

0

1.0

0.5

0.0

x

f(x)

b) F(x) = 1 – e-x, x > 0

c)

0

1.0

0.5

0.0

x

F(x)

3-19 d: 1)( xf , 100 < x < 101a)

101.0100.5100.0

1.05

1.00

0.95

x

f(x)

b) F(X) = x – 100, 100 < x < 101

c)

101.0100.5100.0

1.0

0.5

0.0

x

F(x)

3-21. a) P(X > 6) = 1dxe6

)6x( ∞

−−

b) P(6 X < 8) = 8647.01353.0edxe 08

6

)6x( − −−

c) P(X < 8) = 8647.0dxe8

6

)6x( −−

d) P(X > 8) = 1 – P(X 8) = 1 – P(X < 8) = 1 – 0.8647 = 0.1353

e) P(X < x) = 95.0dxex

6

)6x( −−

9x6)05.0ln(x

95.0e1

e1dxe

)6x(

)6x(x

6

)6x(

−

−

−

−−

−−−−

3-22. a) 5.0)X0(P < , by symmetry.

b) 4375.00625.05.0x5.0dxx5.1)X5.0(P1

5.03

1

5.0

2 −<

c) 125.0x5.0dxx5.1)5.0X5.0(P5.0

5.03

5.0

5.0

2 ≤≤−−−

d) P(X < 2) = 0

e) P(X < 0 or X > 0.5) = 1

f) 05.0x5.05.0x5.0dxx5.1)Xx(P 31

x3

1

x

2 −< Then, x = 0.9655

3-23. a) P(X > 1000) = 7165.0dx3000

e

1000

3000x

∞ −

b) P(1000 < X < 2000) = 2031.0dx3000

e2000

1000

3000x

−

c) P(X < 3000) = 6321.0dx3000

e3000

0

3000x

−

d) P(X < x) = 10.0dx3000

ex

0

3000x

−

2.316x9.0e

1.0e1edx3000

e

3000x

3000xx

0

3000xx

0

3000x

−−

−

−−−

e) E(X) = ∞ −

0

3000x

dx3000

ex 3000

V(X) = (3000)2 = 9,000,000

3-24. a) P(19.75 < X < 20) = 5.00.220

75.19 dx

b) P(19.9 < X < 20.1) = 4.00.21.20

9.19 dx

c) E(X) =

V(X) =

d) F(x) = , 19.75 < x < 20.25

3-25. a) P(X < 805) = F(805) = 0.1(805) – 80 = 0.5 b) P(800 < X < 805) = F(805) – F(800) = 0.5 – 0 = 0.5c) P(X > 808) = 1 – P(X 808) = 1 – [0.1(808) – 80] = 0.2d) In specification = P(802 < X < 808) =F(808) – F(802) = 0.8 – 0.2 = 0.6. Therefore, out of specification is 1 – 0.6 = 0.4.

3-26. a) 6.0401)0080.2(200)0080.2()0080.2( −≤ FXP b) 9.01.01)0055.2(1)0055.2(1)0055.2( −−≤−> FXPXPc) 2.06.08.0)0080.2()0090.2()0090.20080.2( −−<< FFXP

3-27. a) Show that 121

3 ∞ − dxx :

1)1(0xdxx2 21

2

1

3 −−− −∞−∞

−

b) F(X) = 21

2

1

3 12 −−− −− xxdxxxx

c) E(X) = 0.2dxx21

2 ∞

−

d) P(X < 5) = 96.0dxx25

1

3 −

e) P(X > 7) = 0204.0dxx27

3 ∞

−

3-28. a) E X x e dxx( ) ( )

10 10 5

5.

Using integration by parts with u = x and dv e dxx 10 10 5( ) , we obtain

Now, . Using the integration by parts with u x ( . )51 2 and

dv e x 10 10 5( ) , we obtain V X x e x e dxx x( ) ( . ) ( . )( ) ( )

51 2 512 10 55

10 5

5.

From the definition of E(X), the integral above is recognized to equal 0. Therefore, V X( ) ( . ) . 5 51 0 012 .

b)

3-29. a) P(X < 1208) = F(1208) = 0.1(1208) – 120 = 0.8

b) P(1195 < X < 1205) = F(1205) – F(1195) = [0.1(1205)-120] – 0 = 0.5

3-30. a) E(X) =

V(X) =

b) Average Cost = Coating Cost Mean of the Coating Thickness= 0.50 109.39 = $54.70

3-31. a) P(X < 209) = 0.1(209) – 20 = 0.9b) P(200 < X < 208) = F(208) – F(200) = 0.8 – 0 = 0.8c) P(X > 209) = 1 – F(209) = 1 – 0.9 = 0.1d) f(x) = 0.1 for 200 x 210e)

f) E(X) = 205, V(X) = 8.3333

3-32. Now, for 0 < x and

for 0 < x.

Then, a) P(X<60) = F(60) =1 - e-6 = 1 - 0.002479 = 0.9975

b)

c)

d) P(15<X<30) = F(30)-F(15) = e-1.5- e-3=0.173343

e)

3-33. a)

b)

706970)()(

70

1

70

1

22 dxdxxfxXE

Var(X) = 51.423018.5770 70)]([)( 22 −− XEXEc) 3*4.3101 = 12.93

Section 3-5

3-34. a) P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.841345 – 0.158655 = 0.68269b) P(-2 < Z < 2) = P(Z < 2) – P(Z < -2) = 0.97725 – 0.02275 = 0.9545c) P(-3 < Z < 3) = P(Z < 3) – P(Z < -3) = 0.998650 – 0.00135 = 0.997300d) P(Z < -3) = 0.00135e) P(0 < Z 3) = P(Z < 3) – P(Z < 0) = 0.998650 – 0. 5 = 0.498650

3-35. a) P(Z < z) = 0.500000z = 0

b) P(Z < z) = 0.001001z = -3.09

c) P(Z > z) = 0.881000 Can be rewritten as: 1 – P(Z < z) = 0.881000

P(Z < z) = 0.119000 z = -1.18

d) P(Z > z) = 0.866500 Can be rewritten as: 1 – P(Z < z) = 0.866500

P(Z < z) = 0.133500 z = -1.11

e) P(-1.3 < Z < z) = 0.863140 P(Z < z) – P(Z < -1.3) = 0.863140 P(Z < z) – 0.096801 = 0.863140 P(Z < z) = 0.959941, Z = 1.75

3-36. a) P(-z < Z < z) = P(Z < z) – P(Z < -z) = 1 – 2P(Z<-z) = 0.95 So P(Z<-z) = 0.5(1 – 0.95) = 0.025

z = 1.96b) P(-z < Z < z) = P(Z < z) – P(Z < -z) = 1 – 2P(Z<-z) = 0.99 So P(Z<-z) = 0.5(1 – 0.99) = 0.005

z = 2.58c) P(-z < Z < z) = P(Z < z) – P(Z < -z) = 1 – 2P(Z<-z) = 0.68 So P(Z<-z) = 0.5(1 – 0.68) = 0.16

z = 1d) P(-z < Z < z) = P(Z < z) – P(Z < -z) = 1 – 2P(Z<-z) = 0.9973 So P(Z<-z) = 0.5(1 – 0.9973) = 0.00135

z = 3

3-37. a) P(X < 24) = P(Z < 2) = 0.97725b) P(X > 18) = P(Z > -1) = 0.841345c) P(18 < X < 22) = P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) = 0.841345 – 0.158655 = 0.68269d) P(14 < X < 26) = P(-3 < Z < 3) = 0.99865 – 0.00135 = 0.9973e) P(16 < X < 20) = P(-2 < Z < 0) = P(Z < 0) – P(Z < -2) = 0.5 – 0.02275 = 0.47725f) P(20 < X < 26) = P(0 < Z < 3) = P(Z < 3) – P(Z < 0) = 0.99865 – 0.50 = 0.49865

3-38. a) P(X > x) = P(Z > z) = 0.5 where z = 220−x

= 0. Thus x =20

b) P(X > x) = P(Z > z) = 0.95 where z = 220−x

= -1.645. Thus x =16.71

c) P(x < X < 20) = P(Z < 0) – P(Z < z)= 0.5 – P(Z < z) = 0.2

So P(Z < z) = 0.3 where z = 220−x

= -0.53. Thus x =18.94

3-39. a) P(X < 31) = P(Z < -3) = 0.001350b) P(X > 30) = P(Z > -3.5) = 0.999767c) P(33 < X < 37) = P(-2 < Z < 0) = P(Z < 0) – P(Z < -2) = 0.5 – 0.02275 = 0.47725d) P(32 < X < 39) = P(-2.5 < Z < 1) = 0.841345 – 0.00621 = 0.835135e) P(30 < X < 38) = P(-3.5 < Z < 0.5) = 0.691462 – 0.000233 = 0.691229

3-40. a) P(X > x) = 1 – P(X < x)

= 1 - ⎟⎠⎞⎜

⎝⎛ −

<3

6xZP = 0.50.

And, ⎟⎠⎞⎜

⎝⎛ −

<3

6xZP = 0.50

Therefore, 3

6x −= 0 and x = 6.

b) P(X > x) = 1 – P(X < x)

= 1 - ⎟⎠⎞⎜

⎝⎛ −

<3

6xZP = 0.95.

And, ⎟⎠⎞⎜

⎝⎛ −

<3

6xZP = 0.05

Therefore, 3

6x −= -1.645 and x = 1.065.

c) P(x < X < 9) = 0.20

641345.03

6xZP

20.0841345.03

6xZP

20.03

6xZP841345.0

20.03

6xZP)1Z(P

1Z3

6xP3

69Z3

6xP

⎟⎠⎞⎜

⎝⎛ −

<

−⎟⎠⎞⎜

⎝⎛ −

<

⎟⎠⎞⎜

⎝⎛ −

<−

⎟⎠⎞⎜

⎝⎛ −

<−<

⎟⎠⎞⎜

⎝⎛ <<

−⎟

⎠⎞⎜

⎝⎛ −

<<−

Therefore, 3

6x −=0.36, and x = 7.08.

d) P(3 < X < x) = 0.80

P Zx

P Zx

P Zx

P Z

P Zx

P Zx

3 63

63

16

36

31 0 80

63

0158655 0 80

63

0 958655

−< <

−⎛⎝⎜

⎞⎠⎟ − < <

−⎛⎝⎜

⎞⎠⎟

<−⎛

⎝⎜⎞⎠⎟− < −

<−⎛

⎝⎜⎞⎠⎟−

<−⎛

⎝⎜⎞⎠⎟

.

. .

.

Therefore, x−63

1.74, and x = 11.22.

3-41. a) P(X < 6250) =

= P(Z < 2.5) = 0.9938

b) P(5800 < X < 5900) = P(2 < Z < 1) = P(Z < 1) P(Z < 2) = 0.1359

c) P(X > x) = = 0.95.

Therefore, = 1.64 and x = 5835.51.

3-42. a) P(X < 39) = P(Z < 2) = 0.97725b) P(X < 29) = P(Z < -3) = 0.00135

3-43. a) P(X > 0.62) =

= P(Z > 2.4) = 1 P(Z < 2.4) = 0.0082

b) P(0.47 < X < 0.63) = P(0.6 < Z < 2.6) = P(Z < 2.6) P(Z < 0.6) = 0.99534 0.27425 = 0.7211

c) P(X < x) = = 0.90.

Therefore, = 1.28 and x = 0.5641.

3-44. a) P(X < 12) = P(Z < ) = P(Z < 4)

b) P(X < 12.1) =

= P(Z < 3) = 0.00135

and

P(X > 12.6) = P Z >

12 6 12 401

. ..

= P(Z > 2) = 0.02275

Therefore, the proportion of cans scrapped is 0.00135 + 0.02275 = 0.0241.

c) P(12.4 x < X < 12.4 + x) = 0.99.

Therefore, Px

Zx

< <

01 01. .

= 0.99

Consequently, P Z x<

01.

= 0.995 and x = 0.1(2.58) = 0.258.

The specifications are (12.142, 12.658).

3-45. d) If P(X > 12) = 0.999, then = 0.999.

Therefore, = 3.09 and = 12.309.

e) If P(X > 12) = 0.999, then = 0.999.

Therefore, = -3.09 and = 12.155.

3-46. a) P(X > 0.5) = P Z >

0 5 0 40 05

. ..

= P(Z > 2) = 1 0.97725 = 0.02275

b) P(0.4 < X < 0.5) = P(0 < Z < 2) = P(Z < 2) P(Z < 0) = 0.47725

c) P(X > x) = 0.90, then P Z x>

0 40 05

..

= 0.90.

Therefore, = 1.28 and x = 0.336.

3-47. a) P(90.3 < X) =

= P(1 < Z) = P(Z > 1) = 1 P(Z < 1) =1 0.841345 = 0.158655

P(X < 89.7) =

= P(Z < 5) 0 .

Therefore, the answer is 0.1587.b) The process mean should be set at the center of the specifications; that is, at = 90.0.

c) P(89.7 < X < 90.3) =

= P(3 < Z < 3) = 0.9973.

3-48. a) P(X > 1140) = P(Z > ) = P(Z > 2.33) = P(Z < -2.33) = 0.009903

b) P(X < 900) = P(Z > ) = P(Z > -0.17) = 0.432505

3-49. a) P(X > 9) = P(Z > 1.34) = 0.09012b) P(5.5 < X < 8.5) = P(-0.35 < Z < 1.10) = 0.864334 - 0.363169 = 0.501165c) Threshold = + 3.75 = 6.23 + 3.75(2.064) = 13.97

3-50. a) P(X < 5000) =

= P(Z < 3.33) = 0.0004.

b) P(X > x) = 0.95. Therefore, P Zx

>

7000600

= 0.95 and = 1.64. Consequently, x = 6016.

3-51. a) P(X > 0.0026) =

= P(Z > 1.5) = 0.0668.

b) P(0.0014 < X < 0.0026) = P(1.5 < Z < 1.5) = 0.8664.

c) P(0.0014 < X < 0.0026) =

= P Z< <

0 0006 0 0006. .

= 0.995

Therefore, P Z <

0 0006.

= 0.9975. Consequently, = 2.81 and = 0.000214.

3-52. a) P(X > 2.10) = P(Z > 2) = 1 – P(Z < 2) = 1 – 0.977250 = 0.02275

b) P(X < 2.10) = 0.999

c) P(X < 2.10) = P(Z < z) = 0.999 where z = = 3.09. Thus = 1.9455

3-53. X is a lognormal distribution with =5 and 2=9

a)

b) Find the value for which P(Xx)=0.95

c)

3-54. a) X is a lognormal distribution with =2 and 2=4

b)

c)

d) The product has significantly degraded over the first 500 hours. The degradation is less significant after 500 hours.

3-55. X is a lognormal distribution with =0.5 and 2=1a)

b)

seconds

c)

3-56. a) Find the values of and 2 given that E(X) = 10000 and = 20,000

Let and then (1) and (2)

Square (1): and substitute into (2)

Substitute y into (1) and solve for x:

and

b)

c)

hours

3-57. = 0.2 and d = 100 hours

3-58. a)

b)

3-59. Let X denote lifetime of a bearing. =2 and d=10000 hours

a)

b)

= 8862.3 hoursc) Let Y denote the number of bearings out of 10 that last at least 8000 hours. Then, Y is a binomial random variable with n = 10 and p = 0.5273.

.

3-60. a) hoursb)

c)

3-61.

So

Stdev(X)= 1.3068

3-62. . Use integration by parts with u and dv =e-x. Then,

.

3-63. a)

b)

c)

3-64. E(X) = 6/3 = 2; V(X) = 6/32 = 0.6667 3-65. E(X) = 3.2/2.5 = 1.28; V(X) = 3.2/(2.5)2 = 0.5123-66. r/ = 3, so r = 3; Also, r/2 = 1.5; Using substitution, 3/2 = 1.5, and = 2;

Giving r = 6.3-67. r/ = 4.5, so r = 4.5; Also, r/2 = 6.25; Using substitution, 4.5/2 = 6.25, and = 0.72;

Giving r = 3.24.3-68. r/ = 4, so r = 4; Also, r/2 = 2; Using substitution, 4/2 = 2, and = 2;

Giving r = 8.

Section 3-6

3-69.

P-Value: 0.207A-Squared: 0.475

Anderson-Darling Normality Test

N: 16StDev: 11.5717Average: 206.045

228218208198188

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

Strength

Normal Probability Plot

Because the points in normal probability plot pass the fat pencil test, we conclude that the normal distribution is an appropriate model.

3-70.



N: 20StDev: 2.27806Average: 100.27

105104103102101100999897

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

MC1




N: 20StDev: 7.58062Average: 100.105

1161069686

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

MC2


The dimension from both machine one and two seem to have a normal distribution with approximately the same mean. Dimension from machine two tends to have higher variability than those from machine one, note the difference in the horizontal scale.

3-71.



N: 20StDev: 2.08136Average: 105.395

108.5107.5106.5105.5104.5103.5102.5101.5

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

MC2


The normal distribution is a reasonable model for the data. It also appears that the variance has been reduced.

3-72.



N: 22StDev: 240.653Average: 867.364

120011001000900800700600500

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

babi

lity

Thickness


It is reasonable to model these data using a normal probability distribution.

3-73.

840

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

1.478AD*

Goodness of Fit

Normal Probability Plot for FlatnessML Estimates - 95% CI

Mean

StDev

3.18533

1.78176

ML Estimates

101

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

0.743AD*

Goodness of Fit

Lognormal base e Probability Plot for FlatnessML Estimates - 95% CI

Location

Scale

1.00490

0.558967

ML Estimates

10.01.00.1

99

9590807060504030

20

10

5

3

2

1

Data

Per

cent

0.983AD*

Goodness of Fit

Weibull Probability Plot for FlatnessML Estimates - 95% CI

Shape

Scale

1.92093

3.61414

ML Estimates

Based on three probability plots, the log normal (base e) appears to be the most appropriate choice as a model for the flatness data.

3-74.

151050-5

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

1.052AD*

Goodness of Fit

Normal Probability Plot for TimeML Estimates - 95% CI

Mean

StDev

5.3036

4.22784

ML Estimates

100.010.01.00.1

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

1.12AD*

Goodness of Fit

Lognormal base e Probability Plot for TimeML Estimates - 95% CI

Location

Scale

1.21821

1.12824

ML Estimates

10.01.00.1

99

9590807060504030

20

10

5

3

2

1

Data

Per

cent

0.68AD*

Goodness of Fit

Weibull Probability Plot for TimeML Estimates - 95% CI

Shape

Scale

1.18814

5.60911

ML Estimates

Weibull probability density model appears to provide the most suitable fit to the data. 3-75.

1050

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

1.011AD*

Goodness of Fit

Normal Probability Plot for DurationML Estimates - 95% CI

Mean

StDev

4.43

2.45595

ML Estimates

100.010.01.00.1

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

2.205AD*

Goodness of Fit

Lognormal base e Probability Plot for DurationML Estimates - 95% CI

Location

Scale

1.21103

0.908494

ML Estimates

10.01.00.1

99

9590807060504030

20

10

5

3

2

1

Data

Perc

ent 1.437AD*

Goodness of Fit

Weibull Probability Plot for DurationML Estimates - 95% CI

Shape

Scale

1.72033

4.91235

ML Estimates

The normal distribution appears to be the most appropriate choice as a model for the data.

3-76.

20100

99

95

90

80

7060504030

20

10

5

1

Data

Per

cent

2.192AD*

Goodness of Fit

Normal Probability Plot for TimetoFailurML Estimates - 95% CI

Mean

StDev

4.38

4.24880

ML Estimates

100.010.01.00.1

99

95

90

80

7060504030

20

10

5

1

Data

Perc

ent 0.461AD*

Goodness of Fit

Lognormal base e Probability Plot for TimetoFailurML Estimates - 95% CI

Location

Scale

1.07042

0.930132

ML Estimates

10.001.000.100.01

99

9590807060504030

20

10

5

3

2

1

Data

Per

cent

0.643AD*

Goodness of Fit

Weibull Probability Plot for TimetoFailurML Estimates - 95% CI

Shape

Scale

1.14500

4.61938

ML Estimates

Based on three probability plots, the log normal (base e) appears to be the most appropriate choice as a model for the time-to-failure data.

3-77. The normal distribution does not provide a reasonable model based on the normal probability plot. The data did not lie along the line and did not pass the fat pencil test.

3-78. Based on three probability plots, the normal appears to be the most appropriate choice as a model for the temperatures of effluent from a sewage treatment facility.

Section 3-7

3-79. The function is a probability mass function. All probabilities are nonnegative and sum to 1.

a) P(X 3) = P(X = 1) + P(X = 2) + (PX = 3) = 0.326 + 0.088 + 0.019 = 0.433

b) P(3 < X < 5.1) = P(X = 4) + P(X = 5) = 0.251 + 0.158 = 0.409

c) P(X > 4.5) = P(X = 5) + P(X = 6) + P(X = 7) = 0.158 + 0.140 + 0.018 = 0.316

d) E(X) = 1(0.326) + 2(0.088) + 3(0.019) + 4(0.251) + 5(0.158) + 6(0.140) + 7(0.018) = 3.319 V(X) = 12 (0.326) + 22 (0.088) + 32 (0.019) + 42 (0.251) + 52 (0.158) + 62 (0.140)

+ 72 (0.018) – (3.319)2

= 3.7212

e) Graph of F(x)

7654321

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

x

F(x)


a) P(X 1) = P(X=0) + P(X=1) = 0.025 + 0.041 = 0.066

b) P(2<X<7.2) = P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 0.762

c) P(X6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) = 0.508

d) E(X) = 0(0.025) + 1(0.041) + 2(0.049) + 3(0.074) + 4(0.098) + 5(0.205) +6(0.262) + 7(0.123) + 8(0.074) + 9(0.049)

= 5.244 V(X) = 02 (0.025) + 12 (0.041) + 22 (0.049) + 32 (0.074) + 42 (0.098) + 52 (0.205)

+ 62 (0.262) + 72 (0.123) + 82 0.074) + 92 (0.049) – (5.244)2

= 4.260

e) Graph of F(x)

9876543210

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

F(x)


a) P(X 1) = (8/7)(1/2) = 4/7

b) P(X > 1) = 1 P(X 1) = 1 4/7 = 3/7

c) E(X) =

€

1 87

× 12

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+ 2 87

× 14

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+ 3 87

× 18

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

= 117

V(X) =

€

12 87

× 12

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+ 22 87

× 14

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+ 32 87

× 18

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

− 117 ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

= 2649

d) Graph of F(x)

321

1.0

0.9

0.8

0.7

0.6

x

F(x)


a) P(X = 2) = (1/2)(2/5) = 1/5

b) P(X 3) = P(X=1) + P(X=2) + P(X=3) = 1/10 + 1/5 + 3/10 = 6/10 = 3/5

c) P(X > 2.5) = 1 – P(X 2.5) = 1 – P(X 2) = 1 – (3/10) = 7/10

d) P(X 1) = 1

e) E(X) =

V(X) =

f) Graph of F(x)


a) P(X < 9) = P(X = 7) + P(X = 8) = 0.170

b) P(X > 11) = P(X = 12) + P(X = 13) = 0.05 + 0.05 = 0.10

c) P(8 X 12) = P(X = 8) + P( X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.91

d) E(X) = 7(0.040) + 8(0.130) + 9(0.190) + 10(0.240) + 11(0.300) + 12(0.050) + 13(0.050) = 9.98

V(X) = 72(0.040) + 82(0.130) + 92(0.190) + 102(0.240) + 112(0.300) + 122(0.050) + 132(0.050) (9.98)2

= 2.02

3-84. a) P(X = 10 million) = 0.4, P(X = 5 million) = 0.5, P(X = 1 million) = 0.1

b) E(X) = 10(0.4) + 5(0.5) + 1(0.1) = 6.6 million V(X) = 102(0.4) + 52(0.5) + 12(0.1) –(6.6)2 = 9.04 (million)2

Standard Deviation is then 006.304.9 million

c)

109876543210

0.5

0.4

0.3

0.2

0.1

x

p(x)

E(X) = 6.6

d)

109876543210

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

F(x)

3-85. a) F(x)x 0 1 2 3 4 5

F(x) 0.1 0.25 0.50 0.75 0.90 1.00

b) Mean: 5

0( ) ( )

0(0.1) 1(0.15) 2(0.25) 3(0.25) 4(0.15) 5(0.1) 2.5x

E x xf x

∑

Variance:

5 2 2

02 2 2 2 2 2 2

( ) ( )

0 (0.1) 1 (0.15) 2 (0.25) 3 (0.25) 4 (0.15) 5 (0.1) 2.5 2.05x

V x x f x

−∑

−

c) ( 2) ( 0) ( 1) ( 2) 0.1 0.15 0.25 0.50P x P x P x P x≤

d) ( 3.5) ( 0) ( 1) ( 2) ( 3) 0.1 0.15 0.25 0.25 0.75P x P x P x P x P x≤

3-86. a) F(x)

x 0 1 2 3F(x) 0.7 0.85 0.95 1.00

b) Mean: 3

0( ) ( )

0(0.7) 1(0.15) 2(0.1) 3(0.05) 0.5x

E x xf x

∑

Variance:

5 2 2

02 2 2 2 2

( ) ( )

0 (0.7) 1 (0.15) 2 (0.1) 3 (0.05) 0.5 0.75x

V x x f x

−∑

−

c) ( 1.5) ( 2) ( 3) 0.1 0.05 0.15P x P x P x> d) ( 2) ( 0) ( 1) ( 2) 0.7 0.15 0.1 0.95P x P x P x P x≤

Section 3-8

3-87. A binomial distribution is based on a fixed number of independent trials with two outcomes and a constant probability of success on each trial.a) reasonable.b) independence assumption not reasonable.c) The probability that the second component fails depends on the failure time of the first component. The binomial distribution is not reasonable.d) not independent trials with constant probability.e) probability of a correct answer not constant.f) reasonable.g) probability of finding a hole on the chip is not constant.h) If the fills are independent with a constant probability of an underfill, then the binomial distribution for the number

packages underfilled is reasonable.i) Because of the bursts, each trial (that consists of sending a bit) is not independent.j) not independent trials with constant probability

3-88. a)

b)

c) The value X = 5 appears to be most likely.d) The least likely values are X = 0, 1, 9, and 10.

3-89. a) 0.0148

b) = 0.8684

c) 0

d) 0.1109

e)

3-90. a) The probability mass function is

1050

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

x

f(x)

b) The cumulative distribution function is

c) The value of X that appears to be most likely is 0.

d) The value of X that appears to be least likely is 10, although the probabilities for values of x greater than 1 are very small.

3-91. a) 0015.0)90.0(1.05

10)5X(P 55 ⎟⎟⎠

⎞⎜⎜⎝⎛

b) 8291100 90.010.02

1090.010.0

110

90.010.00

10)2X(P ⎟⎟⎠

⎞⎜⎜⎝⎛

⎟⎟⎠⎞

⎜⎜⎝⎛

⎟⎟⎠⎞

⎜⎜⎝⎛

≤

0.9298

c) ⎟⎟⎠⎞

⎜⎜⎝⎛

⎟⎟⎠⎞

⎜⎜⎝⎛

≥ 01019 )90.0(10.01010

)90.0(10.09

10)9X(P 0

d) ⎟⎟⎠⎞

⎜⎜⎝⎛

⎟⎟⎠⎞

⎜⎜⎝⎛

<≤ 6473 90.010.04

1090.010.0

310

)5X3(P 0.0686

3-92. Let X denote the number of defective circuits. Then, X has a binomial distribution with

n = 40 and p = 0.01. Then, P(X = 0) = 400

0 01 0 99 0 66900 40

. . . .

3-93. Let X denote the number of parts that will successfully pass the test. Then X has a

binomial distribution with n = 7 and p = 0.80. P(X=2) =72

0 8 0 2 0 00432 5⎛⎝⎜⎞⎠⎟ . . .

3-94. Let X denote the number of times the line is occupied. Then, X has a binomial distribution with n = 8 and p = 0.45

a) ⎟⎟⎠⎞

⎜⎜⎝⎛

62 )55.0(45.028

)2(XP 0.1569

b) ⎟⎟⎠⎞

⎜⎜⎝⎛

−−≥ 80 55.045.008

1)0(1)1( XPXP 0.9916

c) E(X) = 8(0.45) = 3.6

3-95. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np.

b) P X( ) . . . . . . .≤ =⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ =2

500

01 0 9501

01 0 9502

01 0 9 011170 50 1 49 2 48

c) 09.01.05050

9.01.04950

)49X(P 050149 =⎟⎟⎠⎞

⎜⎜⎝⎛

+⎟⎟⎠⎞

⎜⎜⎝⎛

=≥

3-96. E(X) = 20 (0.01) = 0.2

V(X) = 20 (0.01) (0.99) = 0.198

X X 3 0 2 3 0198 153. . .

a) P X P X P X( . ) ( ) ( )> 153 2 1 1

1 0 01 0 99 0 01 0 99

0 0169

020 0 20

120 1 19. . . .

.

b) X is binomial with n = 20 and p = 0.04

P X P X( ) ( )

. . . .

.

>

1 1 1

1200

0 04 0 96201

0 04 0 96

01897

0 20 1 19

c) Let Y denote the number of times X exceeds 1 in the next five samples. Then, Y is binomial with n = 5 and p = 0.1897 from part b.

6507.08103.01897.005

1)0Y(P1)1Y(P 50 =⎥⎥⎦

⎤⎢⎢⎣

⎡⎟⎟⎠⎞

⎜⎜⎝⎛

−==−=≥

The probability is 0.6507 that at least one sample from the next five will contain more than one defective.

3-97. Let X denote the passengers with tickets that do not show up for the flight. Then, X is Binomial with n = 125 and p = 0.1.

989.0)5(1)6()9961.0

9.01.04

1259.01.0

3125

9.01.02

1259.01.0

1125

9.01.00

125

1

)4(1)5()

12141223

123212411250

=≤−=≥=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠⎞

⎜⎜⎝⎛

+⎟⎟⎠⎞

⎜⎜⎝⎛

+

⎟⎟⎠⎞

⎜⎜⎝⎛

+⎟⎟⎠⎞

⎜⎜⎝⎛

+⎟⎟⎠⎞

⎜⎜⎝⎛

−=

≤−=≥

XPXPb

XPXPa

c) E(X) = np = 125(0.9) = 112.5 X = np p( ) ( . )( . ) .1 125 0 9 01 3 354−

3-98. Let X denote the number of defective components among those stocked.

a P X

b P X

c P X

) ( ) . . .

) ( ) . . . . . . .

) ( ) .

0100

00 02 0 98 0133

2102

00 02 0 98

1021

0 02 0 98102

20 02 0 98 0 666

5 0 981

0 100

0 102 1 101 2 100

3-99. Let X denote the number of students who successfully land the plane.

a) ⎥⎥⎦⎤

⎢⎢⎣⎡

⎟⎟⎠⎞

⎜⎜⎝⎛

09 20.0)80.0(99

)9(XP 0.13422

b) ⎥⎥⎦⎤

⎢⎢⎣⎡

⎟⎟⎠⎞

⎜⎜⎝⎛

90 20.0)80.0(09

)0(XP 0.000001

c) ⎥⎥⎦⎤

⎢⎢⎣⎡

⎟⎟⎠⎞

⎜⎜⎝⎛

18 20.0)80.0(89

)8(XP 0.301990

3-100. Let X denote the number of bulbs that fail.

a) P(X<2) = ⎥⎥⎦⎤

⎢⎢⎣⎡

⎟⎟⎠⎞

⎜⎜⎝⎛

⎟⎟⎠⎞

⎜⎜⎝⎛

91100 75.0)25.0(1

1075.0)25.0(

010

)1()0( XPXP 0.2440

b) P(X = 0) = 0.0563

c) P(X>4) = 1 – P(X 4) = 1 – 0.921873 = 0.078127

3-101. (a) n=20, p=0.6122, P(X≥1) = 1-P(X=0) = 1

(b) P(X≥3) = 1- P(X<3)= 0.999997

(c) µ=E(X)= np=20*0.6122=12.244

σ2 = V(X)=np(1-p) = 4.748

σ = )(XV =2.179

3-102. n=20,p=0.13

(a) P(X=3) = 173 )1(320

pp −⎟⎟⎠⎞

⎜⎜⎝⎛

=0.235

(b) P(X≥3) = 1-P(X<3)=0.492(c) µ = E(X) = np = 20*0.13 = 2.6

σ2 = V(X) = np(1-p) = 2.262

σ = )(XV = 1.504

Section 3-9.1

3-103. a)

b) 9997.0!3

)3.0(e!2

)3.0(e!1

)3.0(ee)3X(P33.023.03.0

3.0 ≤−−−

−

c) 0!6

)3.0(e)6X(P63.0

−

d) 0333.0!2

)3.0(e)2X(P23.0

−

3-104. a) 00674.0!05)0( 5

05

−−

eeXP

b) )3X(P)2X(P)1X(P)0X(P)3X(P ≤

2650.0

!35e

!25e

!15ee

3525155

−−−

−

c) 146.0!65e)6X(P

65

−

d) 0363.0!95e)9X(P

95

−

3-105. P(X = 0) = exp(-). Therefore, = ln(0.02) = 3.912. Consequently, E(X) = V(X) = 3.912.

3-106. P(X = 0) = exp(-). Therefore, = ln(0.04) = 3.219. Consequently, E(X) = V(X) = 3.219.

3-107. a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with

= 20. 0844.0!18

20e)18X(P1820

−

.

b) = 10 for a thirty minute period. 0103.0!310e

!210e

!110ee)3X(P

31021011020 ≤

−−−−

c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable

with = 40. 0185.0!30

40e)30Y(P3040

−

d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable

with = 10. 1251.0!10

10e)10W(P1010

−

3-108. a) Let X denote the number of tremors in a 12 month period. Then, X is a Poisson random

variable with = 6. −

!106

)10(106e

WP 0.0413

b) =2(6) = 12 for a two year period. Let Y denote the number of tremors in a two year

period. −

!1812)18(

1812eYP 0.0255

c) = (1/12)6 = 0.5 for a one month period. Let W denote the number of tremors in a

one-month period. −

!05.0)0(

05.0eWP 0.6065

d) = (1/2)6 = 3 for a six month period. Let V denote the number of tremors in a six- month period.

0839.09161.01

!53

!43

!33

!23

!13

!031

)5(1)5(534333231303

−

⎥⎥⎦

⎤⎢⎢⎣

⎡−

≤−>−−−−−− eeeeee

VPVP

3-109. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with = 10. P X e( ) . 0 4 54 1010 5

b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with = 1. 6321.0e1)0Y(P1)1Y(P 1 −−≥ − .

c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway.

3-110. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random variable with x = 0.05/ft2. Let Y be a Poisson random variable with y =0.5/10ft2. Then, the probability there are no surface flaws in the auto’s interior is P Y e( ) .. −0 0 60650 5 .

b) Let Y denote the number of cars with no flaws, P Y e e( ) ..= = = =− −0 0 006740 5 10 5

c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part a), the probability a car contains surface flaws is 10.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935.

P W P W P W( ) ( ) ( )

. ( . ) . ( . )

.

1 0 1

100

0 3935 0 6065101

0 3935 0 6065

0 0504

0 10 1 9

3-111. a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with = 0.32. 7261.0e)0X(P 32.0 −

b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with = 0.96.

0731.09269.01!2

)96.0(!1

)96.0(1)2(1)3(296.0196.0

96.0 −⎥⎦⎤

⎢⎣⎡

−≤−≥−−

− eeeYPYP

3-112. a) = 3.2, P(X = 5) = 0.1140b) W = 6.4, P(W = 8) = 0.1160c) Y = 9.6, P(Y = 0) = 0.0001

3-113. Let X denote number of calls exceeding the maximum design constraint. X follows a Poissondistribution with = 9. 2941.0706.01)10X(P1)10X(P −≤−> .

3-114. a) Let X denote the number of flaws in 25 square yards. Then, X is a Poisson random variable with = 25(0.01) = 0.25. P(X = 0) = e-0.25 = 0.7788.

b) Let Y denote the number of flaws in one square yard, then P(Y = 0) = e-0.01 = 0.99

c) P(Y = 0) = 0.99. Let V denote the number of square yards out of 10 that contain no flaws. Then, V is a binomial random variable with n = 10 and p = 0.99.

P(V 8) = P(V = 8) + P(V = 9) + P(V = 10) = 0.9999

3-115. For one hour = 10

a)

b) For a thirty minute period, = 5

3-116. a) λ=14.4, P(X=0)=6*10-7

b) λ=14.4/5=2.88, P(X=0)=0.056c) λ=14.4*7*28.35/225=12.7 P(X≥1)=0.999997d) P(X≥28.8) =1-P(X ≤ 28) = 0.0005. Unusual.

3-117. (a) λ=0.61. P(X≥1)=0.4566(b) λ=0.61*5=3.05, P(X=0)= 0.047.

Section 3-9.2

3-118. a)

b)

c)

d)

e) and x = 0.0171

3-119. If E(X) = 5, then =0.20

a)

b)

c)

d) and x = 14.9787.

3-120. Let X denote the time until the first count. Then, X is an exponential random variable with = 3 counts per minute.

a)

b)

c)

d) E(X) = 1/3 minutes.

e) V(X) = 1/32 = 1/9 minutes2 and =1/3 minutes.

f) P(X < x) = 0.95 and P(X < x) = .

Therefore, x = 0.9986 minutes.

3-121. Let X denote the time until the first call. Then, X is exponential and

calls/minute.

a)

b) The probability of at least one call in a 10-minute interval equals one minus the probability of zero calls in a 10-minute interval and that is P(X > 10).

.

Therefore, the answer is 1 0.4346 = 0.5654.

c) 0.2246

d) P(X < x) = 0.90 and . Therefore, x = 27.63 minutes.

3-122. X follows an exponential distribution with = 50.

a) 0.8111

b) 0.101

c) E(X) = = 50 in 1 hour In a twenty-minute interval, E(X) = 50/3 = 16.667.

d) P(X=2) = 0

3-123. Let X denote the distance between major cracks. Then, X is an exponential random variable with = 1/E(X) = 0.2 cracks/mile.

a) P(X > 10) =

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 10(0.2) = 2 cracks per 10 miles.

P(Y = 2) = e

2 222

0 2707!

.

c) X = 1/ = 5 miles

d) P X e dx e e ex x( ) . .. . .12 15 0 2 0 04090 2 0 2

12

15

12

15 2 4 3< <

e) P(X > 5) =

e ex0 2

5

1 0 3679. . . By independence of the intervals in a Poisson process, the answer is

0.36792 = 0.1353. Alternatively, the answer can also be found as P(X > 10) = e-2 = 0.1353. The probability does depend on whether or not the lengths of highway are consecutive.

f) By the memoryless property, the answer is P(X > 10) = 0.1353 from part b.

3-124. Let X denote the time until a failure occurs. Then, X is an exponential random variable and = 1/4.

a) P(X < 1) = 2212.0e1dxe1

0

1

0

4/x4/x41 − −−

b) P(X < 2) = 3935.0e1dxe2

0

2

0

4/x4/x41 − −−

c) P(X < 4) = 6321.0e1dxe4

0

4

0

4/x4/x41 − −−

d) P(X < b) = 0.03

122.0b

)97.0ln(4b

97.0e

03.0e1edxe

4/b

4/bb

0

b

0

4/x4/x41

−

−−

−

−−−

Where b = 0.122 years is equivalent to 1.5 months.e) The mean time to failure in this case is E(X) = . Solve the following for .

3.33

03.01

)97.0ln(197.0e

03.0e1edxe

/1

/11

0

1

0

/x/x1

−

−−

−

−−−

.

3-125. Let X denote the time until a message is received. Then, X is an exponential random variable and =1/2.

a) P(X > 2) = 12

2 2

2 2

1 0 3679e dx e ex x

/ / .

b) The same as part a.

c) E(X) = 2 hours.

3-126. Let X denote the number of calls in 30 minutes. Because the time between calls is an exponential random variable, X is a Poisson random variable with 1 01/ ( ) .E X calls per minute 3 calls per 30 minutes.

a) P(X > 3) = 1 3 1 0 35283 0 3 1 3 2 3 330

31

32

33

P X e e e e( ) .! ! ! !

b) P(X = 0) = e

3 030 0 04979! .

c) Let Y denote the time between calls in minutes. Then, P Y x( ) . 0 01 and

P Y x e dt e et t

xx

x( ) . . . .

01 0 1 0 1 0 1 .

Therefore, e x 0 1 0 01. . and x = 46.05 minutes.

d) 612

120 120

1.01.0 1014.61.0)120( −−∞ ∞

−− ×−≥ eedteYP tt

e) The probability of no calls in one-half hour is (from part b) e-3 = 0.04979. Therefore, for four non-overlapping one-half hour intervals, the probability of no calls is (e-3)4 = (0.04979)4 = 6.14 10-6.

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