Chapter 5 Sol [PDF]

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Applied Statistics and Probability for Engineers, 5th edition

18 January 2010

CHAPTER 5 Section 5-1 5-2.

Let R denote the range of (X,Y). Because

∑ f ( x, y) = c(2 + 3 + 4 + 3 + 4 + 5 + 4 + 5 + 6) = 1 , 36c = 1, and c = 1/36 R

a) P ( X = 1, Y < 4) = f XY (1,1) + f XY (1,2) + f XY (1,3) = b) P(X = 1) is the same as part (a) = 1/4 c) P (Y = 2) = f XY (1,2) + f XY (2,2) + f XY (3,2) = d) P ( X < 2, Y < 2) = f XY (1,1) =

1 36

1 36

1 36

(2 + 3 + 4) = 1 / 4

(3 + 4 + 5) = 1 / 3

(2) = 1 / 18

e)

E ( X ) = 1[ f XY (1,1) + f XY (1,2) + f XY (1,3)] + 2[ f XY (2,1) + f XY (2,2) + f XY (2,3)] + 3[ f XY (3,1) + f XY (3,2) + f XY (3,3)]

) + (3 × 1536 ) = 13 / 6 = 2.167 = (1 × 369 ) + (2 × 12 36 V ( X ) = (1 − 136 ) 2 E (Y ) = 2.167

9 36

+ (2 − 136 ) 2

12 36

+ (3 − 136 ) 2

15 36

= 0.639

V (Y ) = 0.639 f) marginal distribution of X x f X ( x) = f XY ( x,1) + f XY ( x,2) + f XY ( x,3) 1 2 3 g) fY

X

( y) =

1/4 1/3 5/12

f XY (1, y ) f X (1)

y

f Y X ( y)

1 2 3

(2/36)/(1/4)=2/9 (3/36)/(1/4)=1/3 (4/36)/(1/4)=4/9

h) f X Y ( x ) =

f XY ( x,2) and f Y ( 2) = f XY (1,2) + f XY ( 2,2) + f XY (3,2) = fY (2)

x

f X Y ( x)

1 2 3

(3/36)/(1/3)=1/4 (4/36)/(1/3)=1/3 (5/36)/(1/3)=5/12

i) E(Y|X=1) = 1(2/9)+2(1/3)+3(4/9) = 20/9 j) Since fXY(x,y) ≠fX(x)fY(y), X and Y are not independent.

5-1

12 36

= 1/ 3

Applied Statistics and Probability for Engineers, 5th edition 5-7.

18 January 2010

a) The range of (X,Y) is X ≥ 0, Y ≥ 0 and X + Y ≤ 4 . Here X and Y denote the number of defective items found with inspection device 1 and 2, respectively.

y=0 y=1 y=2 y=3 y=4

x=0 x=1 x=2 x=3 x=4 1.10x10-16 2.35x10-14 2.22x10-12 7.88x10-11 1.94x10-19 -16 -13 -11 -9 2.59x10 1.47x10 3.12x10 2.95x10 1.05x10-7 -13 -11 -8 -6 1.29x10 7.31x10 1.56x10 1.47x10 5.22x10-5 2.86x10-11 1.62x10-8 3.45x10-6 3.26x10-4 0.0116 -9 -6 -4 1.35x10 2.86x10 0.0271 0.961 2.37x10

⎡⎛ 4 ⎞ ⎤ ⎡⎛ 4 ⎞ ⎤ f ( x , y ) = ⎢ ⎜⎜ ⎟⎟ ( 0 . 993 ) x ( 0 . 007 ) 4 − x ⎥ ⎢ ⎜⎜ ⎟⎟ ( 0 . 997 ) y ( 0 . 003 ) 4 − y ⎥ ⎣⎝ x ⎠ ⎦ ⎣⎝ y ⎠ ⎦ For x = 1,2,3,4 and y = 1,2,3,4 b) x=0

f(x)

x=1

x=2

⎡⎛ 4 ⎞ ⎤ f ( x , y ) = ⎢ ⎜⎜ ⎟⎟ ( 0 . 993 ) x ( 0 . 007 ) 4 − x ⎥ for x ⎣⎝ ⎠ ⎦ 1.36 x 10-6 2.899 x 10-4 2.40 x 10-9

x=3

x=4

x = 1,2,3,4 0.0274

0.972

c) Because X has a binomial distribution E(X) = n(p) = 4*(0.993)=3.972

d) f Y |2 ( y ) =

f XY (2, y ) = f ( y ) , fx(2) = 2.899 x 10-4 f X (2) y 0 1 2 3 4

fY|1(y)=f(y) 8.1 x 10-11 1.08 x 10-7 5.37 x 10-5 0.0119 0.988

e) E(Y|X=2) = E(Y)= n(p)= 4(0.997)=3.988 f) V(Y|X=2) = V(Y)=n(p)(1-p)=4(0.997)(0.003)=0.0120 g) Yes, X and Y are independent.

5-12.

Let X, Y, and Z denote the number of bits with high, moderate, and low distortion. Then, the joint distribution of X, Y, and Z is multinomial with n =3 and

p1 = 0.01, p2 = 0.04, and p3 = 0.95 . a)

P ( X = 2, Y = 1) = P( X = 2, Y = 1, Z = 0) 3! = 0.0120.0410.950 = 1.2 × 10− 5 2!1!0!

5-2

Applied Statistics and Probability for Engineers, 5th edition b) P ( X = 0, Y = 0, Z = 3) =

18 January 2010

3! 0.0100.0400.953 = 0.8574 0!0!3!

c) X has a binomial distribution with n = 3 and p = 0.01. Then, E(X) = 3(0.01) = 0.03 and V(X) = 3(0.01)(0.99) = 0.0297.

P( X | Y = 2) P (Y = 2) = P( X = 1, Y = 2, Z = 0) + P( X = 0, Y = 2, Z = 1) 3! 3! = 0.01(0.04) 2 0.95 0 + 0.010 (0.04) 2 0.951 = 0.0046 1!2!0! 0!2!1! P( X = 0, Y = 2) ⎛ 3! ⎞ P ( X = 0 | Y = 2) = =⎜ 0.010 0.04 2 0.951 ⎟ 0.004608 = 0.98958 P(Y = 2) ⎝ 0!2!1! ⎠

d) First find

P ( X = 1 | Y = 2) =

P( X = 1, Y = 2) ⎛ 3! ⎞ =⎜ 0.0110.04 2 0.95 0 ⎟ 0.004608 = 0.01042 P (Y = 2) ⎝ 1!2!1! ⎠

E ( X | Y = 2) = 0(0.98958) + 1(0.01042) = 0.01042

V ( X | Y = 2) = E ( X 2 ) − ( E ( X )) 2 = 0.01042 − (0.01042) 2 = 0.01031 3 3

5-13.

Determine c such that c

3

3

∫ ∫ xydxdy = c ∫ y x2

2

0 0

dy = c(4.5

0

0

y2 2

3

)=

0

81 4

c.

Therefore, c = 4/81. 3 2

a) P ( X < 2, Y < 3) =

4 81

∫ ∫ xydxdy =

3

4 81

0 0

(2) ∫ ydy = 814 (2)( 92 ) = 0.4444 0

b) P(X < 2.5) = P(X < 2.5, Y < 3) because the range of Y is from 0 to 3. 3 2.5

P ( X < 2.5, Y < 3) =

4 81

∫

∫ xydxdy =

3

4 81

0 0

2.5 3

c) P (1 < Y < 2.5) =

4 81

∫ ∫ xydxdy = 1 0

(3.125) ∫ ydy = 814 (3.125) 92 = 0.6944 0

2.5

4 81

(4.5) ∫ ydy = 18 81 1

2.5 3

d) P ( X > 1.8,1 < Y < 2.5) =

4 81

∫ ∫ xydxdy =

e) E ( X ) =

4 81

∫∫x

3

2

ydxdy =

0 0

4 81

∫ 9 ydy = 0

4 0

f) P ( X < 0, Y < 4) =

4 81

g) f X ( x ) =

∫f 0

=0.5833

1

(2.88) ∫ ydy =

4 81

1

3

=2

0

4

0

3

XY

( x, y )dy = x 814 ∫ ydy = 814 x(4.5) = 0

5-3

2x 9

2

(2.88) ( 2.52 −1) =0.3733

∫ ∫ xydxdy = 0∫ ydy = 0 0 0

3

2 4 y 9 2

2.5

2.5

4 81

1 1.8

3 3

y2 2

for 0 < x < 3 .

Applied Statistics and Probability for Engineers, 5th edition 4 81 2 9

f (1.5, y ) h) fY 1.5 ( y ) = XY = f X (1.5)

y (1.5) (1.5)

= 92 y for 0 < y < 3.

3

3

2 2 2y3 ⎛2 ⎞ i) E(Y|X=1.5) = ∫ y ⎜ y ⎟dy = ∫ y dy = 9 ⎠ 90 27 0 ⎝

3

=2 0

2

j) P (Y < 2 | X = 1.5) = f Y |1.5 ( y ) = k)

5-19.

f X 2 ( x) =

4 81 2 9

f XY ( x,2) = fY (2)

x(2)

2 1 2 ∫0 9 ydy = 9 y

= 92 x

(2)

18 January 2010

2 0

=

4 4 −0 = 9 9

for 0 < x < 3.

The graph of the range of (X, Y) is

y 5 4 3 2 1 0

1

2

x

4

3

1 x +1

4 x +1

0 0

1 x −1

∫ ∫ cdydx + ∫ ∫ cdydx = 1 1

4

0

1

= c ∫ ( x + 1)dx + 2c ∫ dx = c + 6c = 7.5c = 1 3 2

Therefore, c = 1/7.5=2/15 0.50.5

a) P ( X < 0.5, Y < 0.5) =

∫∫

1 7.5

dydx =

1 30

0 0

0.5x +1

∫∫

b) P ( X < 0.5) =

1 7.5

0.5

dydx =

1 7.5

0 0

∫ ( x + 1)dx =

2 15

( 85 ) = 121

0

c) 1 x +1

E( X ) =

∫∫ 0 0

x 7.5

4 x +1

dydx + ∫

∫

x 7.5 1 x −1

1

=

1 7.5

∫ (x 0

dydx 4

2

+ x)dx +

2 7.5

∫ ( x)dx = 1

d)

5-4

12 15

( 56 ) +

2 7.5

(7.5) =

19 9

Applied Statistics and Probability for Engineers, 5th edition 1 x +1

E (Y ) =

1 7.5

4 x +1

∫ ∫ ydydx + ∫ ∫ ydydx 1 7.5

1 x −1

0 0

1

=

1 7.5

∫

( x +1) 2 2

0

4

dx + 71.5 ∫ ( x +1)

2

− ( x −1) 2 2

= =

1 15

dx

1

1

1 15

18 January 2010

4

∫ ( x + 2 x + 1)dx + ∫ 4 xdx ( ) + (30) = 2

1 15

0

1

7 3

97 45

1 15

e) x +1

∫

f ( x) =

0

1 ⎛ x +1⎞ dy = ⎜ ⎟ for 7.5 ⎝ 7.5 ⎠

0 < x < 1,

x +1

f ( x) =

1 ⎛ x + 1 − ( x − 1) ⎞ 2 dy =⎜ for 1 < x < 4 ⎟= 7 . 5 7 . 5 7 . 5 ⎝ ⎠ x −1

∫

f)

f Y | X =1 ( y ) =

f XY (1, y ) 1 / 7.5 = = 0.5 f X (1) 2 / 7.5

f Y | X =1 ( y ) = 0.5 for 0 < y < 2 2

y y2 g) E (Y | X = 1) = ∫ dy = 2 4 0

2

=1 0

0.5

0.5

h) P (Y < 0.5 | X = 1) =

∫ 0.5dy = 0.5 y 0

5-21.

= 0.25 0

μ = 3.2, λ = 1/3.2 ∞ ∞

P ( X > 5, Y > 5 ) = (1 / 10 .24 ) ∫ ∫ e

−

x y − 3 .2 3 .2

5 5

⎛ −5 = ⎜⎜ e 3 .2 ⎝

⎞ ⎛ − 35.2 ⎟⎜ e ⎟⎜ ⎠⎝ ∞ ∞

⎛ = ⎜⎜ e ⎝

⎞⎛ ⎟⎜ e ⎟⎜ ⎠⎝

x 3 .2

5

−

x y − 3.2 3.2

10 10

10 − 3.2

−

⎛ − 35.2 ⎜e ⎜ ⎝

⎞ ⎟ dx ⎟ ⎠

⎞ ⎟ = 0 .0439 ⎟ ⎠

P ( X > 10, Y > 10) = (1 / 10.24) ∫ ∫ e 10 − 3.2

∞

dydx = 3 .2 ∫ e

∞

dydx = 3.2 ∫ e 10

−

x 3.2

⎛ − 310.2 ⎞ ⎜e ⎟dx ⎜ ⎟ ⎝ ⎠

⎞ ⎟ = 0.0019 ⎟ ⎠

b) Let X denote the number of orders in a 5-minute interval. Then X is a Poisson random variable with λ = 5/3.2 = 1.5625.

P ( X = 2) =

e −1.5625 (1.5625) 2 = 0.256 2!

5-5

Applied Statistics and Probability for Engineers, 5th edition

18 January 2010

For both systems, P ( X = 2) P (Y = 2) = 0.256 = 0.0655 2

c) The joint probability distribution is not necessary because the two processes are independent and we can just multiply the probabilities.

5-28.

a) Let X denote the grams of luminescent ink. Then,

P ( X < 1.14) = P( Z < 1.140.−31.2 ) = P ( Z < −2) = 0.022750 . Let Y denote the number of bulbs in the sample of 25 that have less than 1.14 grams. Then, by independence, Y has a binomial distribution with n = 25 and p = 0.022750. Therefore, the answer is P (Y ≥ 1) = 1 − P (Y = 0) =

( )0.02275 25 0

0

(0.97725) 25 = 1 − 0.5625 = 0.4375 .

b)

P(Y ≤ 5) = P(Y = 0) + P(Y = 1) + P(Y = 2) + ( P(Y = 3) + P(Y = 4) + P(Y = 5)

( )0.02275 (0.97725) + ( )0.02275 (0.97725) =

25 0

0

25

25 3

3

22

( )0.02275 (0.97725) + ( )0.02275 (0.97725) +

25 1

1

24

25 4

4

21

( )0.02275 + ( )0.02275 +

25 2

2

(0.97725) 23

25 5

5

(0.97725) 20

= 0.5625 + 0.3274 + 0.09146 + 0.01632 + 0.002090 + 0.0002043 = 0.99997 ≅ 1 c)

P(Y = 0) =

( )0.02275 25 0

0

(0.97725) 25 = 0.5625

d) The lamps are normally and independently distributed, therefore, the probabilities can be multiplied. Section 5-2 5-29.

E(X) = 1(3/8)+2(1/2)+4(1/8)=15/8 = 1.875 E(Y) = 3(1/8)+4(1/4)+5(1/2)+6(1/8)=37/8 = 4.625

E(XY) = [1 × 3 × (1/8)] + [1 × 4 × (1/4)] + [2 × 5 × (1/2)] + [4 × 6 × (1/8)] = 75/8 = 9.375 σ XY = E ( XY ) − E ( X ) E (Y ) = 9.375 − (1.875)(4.625) = 0.703125

V(X) = 12(3/8)+ 22(1/2) +42(1/8)-(15/8)2 = 0.8594 V(Y) = 32(1/8)+ 42(1/4)+ 52(1/2) +62(1/8)-(37/8)2 = 0.7344

ρ XY =

σ XY 0.703125 = = 0.8851 σ XσY (0.8594)(0.7344)

5-34. Transaction New Order Payment Order Status Delivery Stock Level

Frequency 43 44 4 5 4

Selects(X ) 23 4.2 11.4 130 0

5-6

Updates(Y ) 11 3 0 120 0

Inserts(Z ) 12 1 0 0 0

.

Applied Statistics and Probability for Engineers, 5th edition Mean Value

18.694

12.05

18 January 2010 5.6

(a) COV(X,Y) = E(XY)-E(X)E(Y) = 23*11*0.43 + 4.2*3*0.44 + 11.4*0*0.04 + 130*120*0.05 + 0*0*0.04 - 18.694*12.05=669.0713

(b) V(X)=735.9644, V(Y)=630.7875; Corr(X,Y)=cov(X,Y)/(V(X)*V(Y) )0.5 = 0.9820 (c) COV(X,Z)=23*12*0.43+4.2*1*0.44+0-18.694*5.6 = 15.8416 (d) V(Z)=31; Corr(X,Z)=0.1049 5-35.

From Exercise 5-19, c = 8/81, E(X) = 12/5, and E(Y) = 8/5 3 x

E ( XY ) =

σ xy

3 x

3

3

8 8 8 x3 2 8 x5 2 2 ( ) xy xy dyd x x y dyd x x d x dx = = = 81 ∫0 ∫0 81 ∫0 ∫0 81 ∫0 3 81 ∫0 3

6 ⎛ 8 ⎞⎛ 3 ⎞ ⎜ = ⎜ ⎟⎜ ⎟⎟ = 4 ⎝ 81 ⎠⎝ 18 ⎠ ⎛ 12 ⎞⎛ 8 ⎞ = 4 − ⎜ ⎟⎜ ⎟ = 0.16 ⎝ 5 ⎠⎝ 5 ⎠

E( X 2 ) = 6

E (Y 2 ) = 3

V ( x) = 0.24, V (Y ) = 0.44 0.16 ρ= = 0.4924 0.24 0.44 5-39.

E ( X ) = −1(1 / 4) + 1(1 / 4) = 0 E (Y ) = −1(1 / 4) + 1(1 / 4) = 0 E(XY) = [-1× 0 × (1/4)] + [-1× 0 × (1/4)] + [1× 0 × (1/4)] + [0 × 1× (1/4)] = 0 V(X) = 1/2 V(Y) = 1/2

σ XY = 0 − (0)(0) = 0 ρ XY = σσ σ = XY

X

Y

0 1/ 2 1/ 2

=0

The correlation is zero, but X and Y are not independent, since, for example, if y = 0, X must be –1 or 1. 5-40.

If X and Y are independent, then f XY ( x, y ) = f X ( x) f Y ( y ) and the range of (X, Y) is rectangular. Therefore,

E ( XY ) = ∫∫ xyf X ( x) fY ( y )dxdy = ∫ xf X ( x)dx ∫ yfY ( y )dy = E ( X ) E (Y )

hence σXY=0 Section 5-3 5-43.

a) percentage of slabs classified as high = p1 = 0.05 percentage of slabs classified as medium = p2 = 0.85 percentage of slabs classified as low = p3 = 0.10

5-7

Applied Statistics and Probability for Engineers, 5th edition

18 January 2010

b) X is the number of voids independently classified as high X ≥ 0 Y is the number of voids independently classified as medium Y ≥ 0 Z is the number of with a low number of voids and Z ≥ 0 and X+Y+Z = 20 c) p1 is the percentage of slabs classified as high. d) E(X)=np1 = 20(0.05) = 1 V(X)=np1 (1-p1)= 20(0.05)(0.95) = 0.95 e) P ( X = 1, Y = 17, Z = 3) = 0 Because the point (1,17,3) ≠ 20 is not in the range of (X,Y,Z). f)

P ( X ≤ 1, Y = 17, Z = 3) = P ( X = 0, Y = 17, Z = 3) + P( X = 1, Y = 17, Z = 3) =

20! 0.05 0 0.8517 0.10 3 + 0 = 0.07195 0!17!3!

Because the point (1,17,3) ≠ 20 is not in the range of (X, Y, Z). g) Because X is binomial, P( X ≤ 1) =

( )0.05 0.95 20 0

0

20

+

( )0.05 0.95 20 1

1

19

= 0.7358

h) Because X is binomial E(Y) = np = 20(0.85) = 17 i) The probability is 0 because x+y+z>20

P ( X = 2, Y = 17) . Now, because x+y+z = 20, P(Y = 17) 20! P(X=2, Y=17) = P(X=2, Y=17, Z=1) = 0.05 2 0.8517 0.101 = 0.0540 2!17!1! P( X = 2, Y = 17) 0.0540 = = 0.2224 P( X = 2 | Y = 17) = 0.2428 P(Y = 17)

j) P ( X = 2 | Y = 17) =

k)

⎛ P( X = 0, Y = 17) ⎞ ⎛ P( X = 1, Y = 17) ⎞ ⎟ ⎟⎟ + 1⎜⎜ E ( X | Y = 17) = 0⎜⎜ P(Y = 17) P(Y = 17) ⎟⎠ ⎠ ⎝ ⎝ ⎛ P( X = 2, Y = 17) ⎞ ⎛ P( X = 3, Y = 17) ⎞ ⎟⎟ ⎟⎟ + 3⎜⎜ + 2⎜⎜ P(Y = 17) P(Y = 17) ⎠ ⎠ ⎝ ⎝ ⎛ 0.07195 ⎞ ⎛ 0.1079 ⎞ ⎛ 0.05396 ⎞ ⎛ 0.008994 ⎞ E ( X | Y = 17) = 0⎜ ⎟ + 1⎜ ⎟ + 2⎜ ⎟ + 3⎜ ⎟ ⎝ 0.2428 ⎠ ⎝ 0.2428 ⎠ ⎝ 0.2428 ⎠ ⎝ 0.2428 ⎠ =1 5-45.

a) The probability distribution is multinomial because the result of each trial (a dropped oven) results in either a major, minor or no defect with probability 0.6, 0.3 and 0.1 respectively. Also, the trials are independent b) Let X, Y, and Z denote the number of ovens in the sample of four with major, minor, and no defects, respectively.

P ( X = 2, Y = 2, Z = 0) =

4! 0.6 2 0.3 2 0.10 = 0.1944 2!2!0!

5-8

Applied Statistics and Probability for Engineers, 5th edition c) P( X = 0, Y = 0, Z = 4) =

18 January 2010

4! 0.6 0 0.3 0 0.14 = 0.0001 0!0!4!

d) fXY ( x, y) = ∑ fXYZ ( x, y, z) where R is the set of values for z such that x+y+z = 4. That is, R R

consists of the single value z = 4-x-y and

f XY ( x, y ) =

4! 0.6 x 0.3 y 0.14 − x − y x! y!(4 − x − y )!

for x + y ≤ 4.

e) E ( X ) = np1 = 4(0.6) = 2.4 f) E (Y ) = np 2 = 4(0.3) = 1.2 g)

P ( X = 2 | Y = 2) =

P( X = 2, Y = 2) 0.1944 = = 0.7347 P (Y = 2) 0.2646

⎛ 4⎞ P(Y = 2) = ⎜⎜ ⎟⎟0.3 2 0.7 4 = 0.2646 from the binomial marginal distribution of Y ⎝ 2⎠ h) Not possible, x+y+z = 4, the probability is zero. i) P ( X | Y = 2) = P ( X = 0 | Y = 2), P ( X = 1 | Y = 2), P ( X = 2 | Y = 2)

P( X = 0, Y = 2) ⎛ 4! ⎞ =⎜ 0.6 0 0.3 2 0.12 ⎟ 0.2646 = 0.0204 P(Y = 2) ⎝ 0!2!2! ⎠ P( X = 1, Y = 2) ⎛ 4! ⎞ P ( X = 1 | Y = 2) = =⎜ 0.610.3 2 0.11 ⎟ 0.2646 = 0.2449 P(Y = 2) ⎝ 1!2!1! ⎠ P ( X = 2, Y = 2) ⎛ 4! ⎞ P ( X = 2 | Y = 2) = =⎜ 0.6 2 0.3 2 0.10 ⎟ 0.2646 = 0.7347 P(Y = 2) ⎝ 2!2!0! ⎠ P ( X = 0 | Y = 2) =

j) E(X|Y=2) = 0(0.0204)+1(0.2449)+2(0.7347) = 1.7143 5-49.

Because ρ = 0 and X and Y are normally distributed, X and Y are independent. Therefore, μX = 0.1 mm, σX=0.00031 mm, μY = 0.23 mm, σY=0.00017 mm Probability X is within specification limits is

0.100465 − 0.1 ⎞ ⎛ 0.099535 − 0.1 P(0.099535 < X < 0.100465) = P⎜

⎟ = P ( Z > − 253 ) = 1 − P ( Z < 253 ) ≅ 1 0 . 000345 ⎠ ⎝

Let X denote the average thickness of 10 wafers. Then, E( X ) = 10 and V( X ) = 0.1. a) P (9 < X < 11) = P ( 9 −10 < Z < 0.1

11−10 0.1

) = P(−3.16 < Z < 3.16) = 0.998 .

The answer is 1 − 0.998 = 0.002 b) P ( X > 11) = 0.01 and

σX =

Therefore, P ( X > 11) = P ( Z >

1 n

11−10 1 n

.

) = 0.01 ,

11 − 10 = 2.33 and n = 5.43 which is rounded 1 n

up to 6. c) P ( X > 11) = 0.0005 and

σX =σ

Therefore, P ( X > 11) = P ( Z >

11−10

σ

10

.

) = 0.0005 ,

10

11−10

σ

= 3.29

10

σ = 10 / 3.29 = 0.9612 5-64.

X~N(160, 900) a) Let Y = X1 + X2 + ... + X25, E(Y) = 25E(X) = 4000, V(Y) = 252(900) = 22500

5-10

Applied Statistics and Probability for Engineers, 5th edition

18 January 2010

P(Y > 4300) =

⎛ 4300 − 4000 ⎞ P⎜⎜ Z > ⎟⎟ = P( Z > 2) = 1 − P( Z < 2) = 1 − 0.9773 = 0.0227 22500 ⎠ ⎝ ⎛ x − 4000 ⎞ b) c) P( Y > x) = 0.0001 implies that P⎜⎜ Z > ⎟⎟ = 0.0001. 22500 ⎠ ⎝ 4000 Then x −150 = 3.72 and x = 4558

Section 5-5

5-71.

a) If y = x , then x = 2

y for

f Y ( y ) = f X ( y ) 12 y

x ≥ 0 and y ≥ 0 . Thus,

− 12

=

e−

y

for

2 y

y > 0. b) If y = x y > 0.

1/ 2

, then x = y for x ≥ 0 and y ≥ 0 . Thus, fY ( y ) = f X ( y ) 2 y = 2 ye 2

2

c) If y = ln x, then x = e y for x ≥ 0 . Thus, f Y ( y ) = f X (e )e = e e −∞ < y < ∞. y

5-73.

y

y

−e y

1

for 1 ≤ ln y ≤ 2 . That is, fY ( y ) =

2

1 2 for e ≤ y ≤ e . y

Supplemental Exercises 5-75.

The sum of

∑ ∑ f ( x, y ) = 1 , ( 1 4 ) + ( 1 8 ) + ( 1 8 ) + ( 1 4 ) + ( 1 4 ) = 1 x

and

y

f XY ( x, y ) ≥ 0

a) P ( X < 0.5, Y < 1.5) = f XY (0,1) + f XY (0,0) = 1 / 8 + 1 / 4 = 3 / 8 . b) P ( X ≤ 1) = f XY (0,0) + f XY (0,1) + f XY (1,0) + f XY (1,1) = 3 / 4 c) P (Y < 1.5) = f XY (0,0) + f XY (0,1) + f XY (1,0) + f XY (1,1) = 3 / 4 d) P ( X > 0.5, Y < 1.5) = f XY (1,0) + f XY (1,1) = 3 / 8 e) E(X) = 0(3/8) + 1(3/8) + 2(1/4) = 7/8. V(X) = 02(3/8) + 12(3/8) + 22(1/4) - 7/82 =39/64 E(Y) = 1(3/8) + 0(3/8) + 2(1/4) = 7/8. V(Y) = 12(3/8) + 02(3/8) + 22(1/4) - 7/82 =39/64 f)

f X ( x ) = ∑ f XY ( x, y ) and f X (0) = 3 / 8, f X (1) = 3 / 8, f X (2) = 1/ 4 . y

g)

fY 1 ( y ) =

f XY (1, y ) and fY 1 (0) = f X (1)

1/ 8 3/8

= 1 / 3, fY 1 (1) =

5-11

1/ 4 3/8

= 2/3.

for

= e y − e for y

If y = e , then x = ln y for 1 ≤ x ≤ 2 and e ≤ y ≤ e . Thus, fY ( y ) = f X (ln y ) x

− y2

1 1 = y y

Applied Statistics and Probability for Engineers, 5th edition h)

18 January 2010

E (Y | X = 1) = ∑ yf Y | X =1 ( y ) =0(1 / 3) + 1(2 / 3) = 2 / 3 x =1

i) As is discussed after Example 5-19, because the range of (X, Y) is not rectangular, X and Y are not independent. j) E(XY) = 1.25, E(X) = E(Y)= 0.875 V(X) = V(Y) = 0.6094 COV(X,Y)=E(XY)-E(X)E(Y)= 1.25-0.8752=0.4844

0.4844 = 0.7949 0.6094 0.6094 20! P ( X = 2, Y = 4, Z = 14) = 0.10 2 0.20 4 0.7014 = 0.0631 2!4!14! 0 20 b) P ( X = 0) = 0.10 0.90 = 0.1216 c) E ( X ) = np1 = 20(0.10) = 2 V ( X ) = np1 (1 − p1 ) = 20(0.10)(0.9) = 1.8 f XZ ( x, z ) d) f X | Z = z ( X | Z = 19) f Z ( z) 20! f XZ ( xz ) = 0.1x 0.2 20− x − z 0.7 z x! z!(20 − x − z )! 20! f Z ( z) = 0.3 20 − z 0.7 z z! (20 − z )!

ρ XY =

5-76.

f XZ ( x, z ) (20 − z )! ⎛ 1 ⎞ ⎛ 2 ⎞ (20 − z )! 0.1 x 0.2 20 − x − z = = ⎜ ⎟ ⎜ ⎟ 20 − z f Z ( z) x! (20 − x − z )! 0.3 x! (20 − x − z )! ⎝ 3 ⎠ ⎝ 3 ⎠ x

f X | Z = z ( X | Z = 19)

Therefore, X is a binomial random variable with n=20-z and p=1/3. When z=19,

2 1 and f X |19 (1) = . 3 3 ⎛ 2⎞ ⎛1⎞ 1 e) E ( X | Z = 19) = 0⎜ ⎟ + 1⎜ ⎟ = ⎝ 3⎠ ⎝3⎠ 3 f X |19 (0) =

5-78.

Let X, Y, and Z denote the number of calls answered in two rings or less, three or four rings, and five rings or more, respectively. a)

P( X = 8, Y = 1, Z = 1) =

10! 0.780.2510.051 = 0.0649 8!1!1!

b) Let W denote the number of calls answered in four rings or less. Then, W is a binomial random variable with n = 10 and p = 0.95. Therefore, P(W = 10) =

( )0.95 10 10

10

0.050 = 0.5987 .

c) E(W) = 10(0.95) = 9.5. d)

fZ 8 ( z) =

f XZ (8, z ) 10! 0.70 x 0.25(10 − x − z ) 0.05 z for and f XZ ( x, z ) = x! z!(10 − x − z )! f X (8)

x + z ≤ 10 and 0 ≤ x,0 ≤ z . Then, f Z 8 ( z) =

10! 8!z !( 2 − z )!

0.70 8 0.25 ( 2 − z ) 0.05 z

10! 8!2!

8

0.70 0.30

2

=

( ) ( )

2! 0.25 2 − z 0.05 z z!( 2 − z )! 0.30 0.30

for 0 ≤ z ≤ 2 . That is Z is binomial with n =2 and p = 0.05/0.30 = 1/6.

5-12

20 − x − z

Applied Statistics and Probability for Engineers, 5th edition

18 January 2010

e) E(Z) given X = 8 is 2(1/6) = 1/3. f) Because the conditional distribution of Z given X = 8 does not equal the marginal distribution of Z, X and Z are not independent.

5-82.

a) Let X1, X 2 ,..., X 6 denote the lifetimes of the six components, respectively. Because of independence, P( X1 > 5000, X 2 > 5000,..., X 6 > 5000) = P( X1 > 5000)P( X 2 > 5000)... P( X 6 > 5000)

If X is exponentially distributed with mean θ , then λ = ∞

P( X > x) = ∫ θ1 e − t / θ dt = −e − t / θ x

e

−5 / 8

e

−0.5

e

−0.5

e

−0.25

e

−0.25

e

−0.2

=e

−2.325

∞

1 θ

and

= e − x / θ . Therefore, the answer is

x

= 0.0978 .

b) The probability that at least one component lifetime exceeds 25,000 hours is the same as 1 minus the probability that none of the component lifetimes exceed 25,000 hours. Thus, 1-P(Xa