Chapter 5 Sol

(1)

CHAPTER 5

Section 5-1

5-2. Let R denote the range of (X,Y). Because

1 )

6

5

4

5

4

3

4

3

2 (

)

,

(

=

+

=

∑

f

x

y

c

R , 36c = 1, and c = 1/36 a)

(

₌

1 ,

_<

4 )

₌

(

1 ,

1 )

₊

(

1 ,

2 )

₊

(

1 ,

3 )

₌

₃₆1

(

2 ₊

3 ₊

4 )

₌

1 /

4

XY XY XY

f

Y

X

P

b) P(X = 1) is the same as part (a) = 1/4

c)

(

=

2 )

=

(

1 ,

2 )

+

(

2 ,

2 )

+

(

3 ,

2 )

=

₃₆1

(

3 +

4 +

5 )

=

1 /

3

XY XY XY

f

Y

P

d)

(

<

2 ,

<

2 )

=

(

1 ,

1 )

=

₃₆1

(

2 )

=

1 /

18

XY

f

Y

X

P

e)

[

] [

]

[

]

(

1 ) (

2 ) (

3 )

13 /

6

2 .

167 )

3 ,

3 (

)

2 ,

3 (

)

1 ,

3 (

3 )

3 ,

2 (

)

2 ,

2 (

)

1 ,

2 (

2 )

3 ,

1 (

)

2 ,

1 (

)

1 ,

1 (

1 )

(

36 15 36 12 369

+

×

+

×

=

×

=

+

=

XY XY XY XY XY XY XY XY XY

f

X

E

639 .

0 )

3 (

)

2 (

)

1 (

)

(

2 ₃₆15 6 13 36 12 2 6 13 369 2 6 13

+

−

+

−

=

−

=

X

V

639 .

0 )

(

167 .

2 )

(

=

Y

V

Y

E

f) marginal distribution of X x

_f

₍

_x

₎

_f

₍

_x

_,

₁

₎

_f

₍

_x

_,

₂

₎

_f

₍

_x

_,

₃

₎

XY XY XY X

=

+

1 1/4 2 1/3 3 5/12 g)

)

1 (

)

,

1 (

)

(

X XY X Y

f

y

f

y

f

=

y f_{Y X}( )y 1 (2/36)/(1/4)=2/9 2 (3/36)/(1/4)=1/3 3 (4/36)/(1/4)=4/9 h)

)

2 (

)

2 ,

(

)

(

Y XY Y X

f

x

f

x

f

=

and

(

2 )

₌

(

1 ,

2 )

₊

(

2 ,

2 )

₊

(

3 ,

2 )

₌

₃₆12

₌

1 /

3

XY XY XY Y

f

x f_{X Y}( )x 1 (3/36)/(1/3)=1/4 2 (4/36)/(1/3)=1/3 3 (5/36)/(1/3)=5/12 i) E(Y|X=1) = 1(2/9)+2(1/3)+3(4/9) = 20/9

(2)

5-7. 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. x=0 x=1 x=2 x=3 x=4 y=0 1.94x10-19 1.10x10-16 2.35x10-14 2.22x10-12 7.88x10-11 y=1 2.59x10-16 1.47x10-13 3.12x10-11 2.95x10-9 1.05x10-7 y=2 1.29x10-13 7.31x10-11 1.56x10-8 1.47x10-6 5.22x10-5 y=3 2.86x10-11 1.62x10-8 3.45x10-6 3.26x10-4 0.0116 y=4 2.37x10-9 1.35x10-6 2.86x10-4 0.0271 0.961 For x = 1,2,3,4 and y = 1,2,3,4 b) x=0 x=1 x=2 x=3 x=4 f(x) 2.40 x 10-9 1.36 x 10-6 2.899 x 10-4 0.0274 0.972

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

d)

(

)

2 (

)

,

2 (

)

(

2 |

f

y

f

y

f

y

f

X XY Y

=

, fx(2) = 2.899 x 10 -4 y fY|1(y)=f(y) 0 8.1 x 10-11 1 1.08 x 10-7 2 5.37 x 10-5 3 0.0119 4 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

95 .

0 ,

04 .

0 ,

01 .

0

₂ ₃ 1

=

p

=

and

p

=

p

. a) ₂ ₁ ₀ ₅

10

2 .

1

95 .

0

04 .

0

01 .

0 !

1 !

2 !

3 )

0 ,

1 ,

2 (

)

1 ,

2 (

−

×

=

Y

P

X

Y

Z

X

P

⎥

⎦

⎤

⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎥

⎦

⎤

⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

x −x y −y

y

x

y

x

f

(

,

)

4 (

0 .

993 )

(

0 .

007 )

4

4 (

0 .

997 )

(

0 .

003 )

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

(3)

b)

0 .

01

0 .

04

0 .

95

0 .

8574

!

3 !

0 !

3 )

3 ,

0 ,

0 (

₌

0 0 3

₌

Z

Y

X

P

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. d) First find

P

(

X

|

Y

=

2 )

0046

.

0

95 .

0 )

04 .

0 (

01 .

0 !

1 !

2 !

0 !

3

95 .

0 )

04 .

0 (

01 .

0 !

2 !

1 !

3 )

1 ,

2 ,

0 (

)

0 ,

2 ,

1 (

)

2 (

1 2 0 0 2

₊

₌

=

+

=

P

X

Y

Z

P

X

Y

Z

Y

P

98958

.

0 004608

.

0

95 .

0

04 .

0

01 .

0 !

1 !

2 !

0 !

3 )

2 (

)

2 ,

0 (

)

2 |

0 (

0 2 1

_⎟

₌

⎠

⎞

⎜

⎝

⎛

=

Y

P

Y

X

P

Y

X

P

01042

.

0 004608

.

0

95 .

0

04 .

0

01 .

0 !

1 !

2 !

1 !

3 )

2 (

)

2 ,

1 (

)

2 |

1 (

1 2 0

_⎟

₌

⎠

⎞

⎜

⎝

⎛

=

Y

P

Y

X

P

Y

X

P

)

2 |

(

X

Y

=

E

=

0 (

0 .

98958

)

+

1 (

0 .

01042

)

=

0 .

01042

01031

.

0 )

01042

.

0 (

01042

.

0 ))

(

)

(

)

2 |

(

₌

2

₋

2

₌

₋

2

₌

X

E

X

E

Y

X

V

5-13. Determine c such that

(

4 .

5 )

81₄

.

3 0 2 3 0 3 0 3 0 2 3 0 2 2

c

dy

y

c

xydxdy

c

∫

=

∫

x

=

y

=

Therefore, c = 4/81. a)

(

2 ,

3 )

(

2 )

₈₁4

(

2 )(

₂9

)

0 .

4444

3 0 3 0 814 2 0 814

=

<

Y

_∫

xydxdy

_∫

ydy

X

P

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

6944

.

0 )

125 .

3 (

)

125 .

3 (

)

3 ,

5 .

2 (

₂9 814 3 0 3 0 814 5 . 2 0 814

=

<

Y

∫

xydxdy

∫

ydy

X

P

c)

(

1

2 .

5 )

(

4 .

5 )

2.5

0 .

5833

1 2 81 18 5 . 2 1 5 . 2 1 814 3 0 814 2

=

<

_∫

y

ydy

xydxdy

Y

P

d)

(

1 .

8 ,

1

2 .

5 )

(

2 .

88 )

(

2 .

88 )

0 .

3733

5 . 2 1 5 . 2 1 2 ) 1 5 . 2 ( 814 814 3 8 . 1 814 2

∫

=

∫

=

<

>

−

ydy

xydxdy

Y

X

P

e)

(

)

9

3

2

0 2 94 3 0 3 0 814 3 0 2 814 2

=

∫

y

ydy

ydxdy

x

X

E

f)

<

=

∫∫

=

∫

=

4 0 4 0 0 0 814

0

0 )

4 ,

0 (

X

Y

xydxdy

ydy

P

g)

(

)

(

,

)

(

4 .

5 )

2₉

0

3

81 4 3 0 81 4 3 0

<

=

_∫

f

x

y

dy

x

_∫

ydy

x

for

x

f

x XY X .

(4)

h)

y

f

y

f

y

f

X XY Y 9 2 9 2 81 4 5 . 1

₍

₁

_.

₅

₎

)

5 .

1 (

)

5 .

1 (

)

,

5 .

1 (

)

(

=

for 0 < y < 3. i) E(Y|X=1.5) =

2

27

2

9

2

9

2

3 0 3 0 3 2 3 0

=

⎟

⎠

⎞

⎜

⎝

⎛

∫

y

dy

y

dy

y

j)

9

4

0

9

4

9

1

9

2 )

(

)

5 .

1 |

2 (

2 0 2 2 0 5 . 1 |

=

−

=

<

X

f

y

∫

ydy

y

Y

P

Y k)

x

f

x

f

x

f

Y XY X 9 2 9 2 81 4 2

₍

₂

₎

)

2 (

)

2 (

)

2 ,

(

)

(

=

for 0 < x < 3.

5-19. The graph of the range of (X, Y) is

0 1 2 3 4 5 1 2 3 4 y x

1

5 .

7

6

2 )

1 (

1

2 3 4 1 1 0 4 1 1 1 1 0 1 0

=

+

=

+

=

+

∫

∫ ∫

+ − +

c

dx

c

dx

x

c

cdydx

x x x Therefore, c = 1/7.5=2/15 a) ₃₀1 5 . 0 0 5 . 0 0 5 . 71

)

5 .

0 ,

5 .

0 (

X

<

Y

<

=

∫ ∫

dydx

=

P

b) ₁₂1 8 5 152 5 . 0 0 5 . 71 5 . 0 0 1 0 5 . 71

(

1 )

(

)

5 .

0 (

X

<

=

∫ ∫

+

dydx

=

∫

x

+

dx

=

P

x c) 9 19 5 . 7 2 6 5 15 12 4 1 5 . 7 2 1 0 2 5 . 7 1 4 1 1 1 5 . 7 1 0 1 0 5 . 7

)

5 .

7 (

)

(

)

(

)

(

)

(

=

+

=

+

=

+

=

∫

∫ ∫

+ − +

dx

x

dx

x

dydx

X

E

x x x x x d)

(5)

( )

₄₅97 151 3 7 151 4 1 151 1 0 2 151 4 1 2 ) 1 ( ) 1 ( 5 . 71 1 0 2 ) 1 ( 5 . 71 4 1 1 1 5 . 71 1 0 1 0 5 . 71

30

4 )

1

2 (

)

(

2 2 2

=

+

=

+

=

+

=

+

=

∫

∫ ∫

− − + + + − +

xdx

dx

x

dx

ydydx

Y

E

x x x x x x e)

4

1 for

5 .

7

2

5 .

7 )

1 (

1

5 .

7

1 )

(

,

1

0 for

5 .

7

1

5 .

7

1 )

(

1 1 1 0

<

=

⎟

⎠

⎞

⎜

⎝

⎛

+

−

=

<

⎟

⎠

⎞

⎜

⎝

⎛ +

=

∫

+ − +

x

dy

x

f

x

dy

x

f

x x x f)

2

0 for

5 .

0 )

(

5 .

0

5 .

7 /

2

5 .

7 /

1 )

1 (

)

,

1 (

)

(

1 | 1 |

<

=

= =

y

f

y

f

y

f

X Y X XY X Y g)

=

∫

=

2 0 2 0 2

1

4

2 )

1 |

(

Y

X

y

dy

y

E

h)

(

0 .

5 |

1 )

0 .

5

0 .

5

0 .

25

5 . 0 0 5 . 0 0

=

<

X

∫

dy

y

Y

P

5-21. μ = 3.2, λ = 1/3.2

0439

.

0

2 .

3 )

24 .

10 /

1 (

)

5 ,

5 (

2 . 3 5 2 . 3 5 2 . 3 5 5 5 2 . 3 5 2 . 3 2 . 3

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

>

− − − ∞ ∞ − ∞ − −

∫

e

dx

e

dydx

e

Y

X

P

x y x

0019

.

0

2 .

3 )

24 .

10 /

1 (

)

10 ,

10 (

2 . 3 10 2 . 3 10 2 . 3 10 10 10 2 . 3 10 2 . 3 2 . 3

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

=

>

− − − ∞∞ ₋ ₋ ∞ ₋

∫

e

dx

e

dydx

e

Y

X

P

x y x

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.

256 .

0 !

2 )

5625

.

1 (

)

2 (

2 5625 . 1

=

e

−

X

P

(6)

For both systems,

P

(

X

=

2 )

P

(

Y

=

2 )

=

0 .

256

2

=

0 .

0655

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,

022750

.

0 )

2 (

)

(

)

14 .

1 (

_X

_<

₌

_P

_Z

_<

1.14₀_.−₃1.2

₌

_P

_Z

_<

₋

₌

P

.

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

(

1 )

1 (

0 )

( )

25

0 .

02275

0

(

0 .

97725

)

25

1

0 .

5625

0 .

4375

0

=

−

=

−

=

≥

P

Y

P

. b)

( )

1 99997

.

0 0002043

.

0 002090

.

0 01632

.

0 09146

.

0 3274

.

0 5625

.

0 )

97725

.

0 (

02275

.

0 )

97725

.

0 (

02275

.

0 )

97725

.

0 (

02275

.

0 )

97725

.

0 (

02275

.

0 )

97725

.

0 (

02275

.

0 )

97725

.

0 (

02275

.

0 )

5 (

)

4 (

)

3 (

(

)

2 (

)

1 (

)

0 (

)

5 (

20 5 25 5 21 4 25 4 22 3 25 3 23 2 25 2 24 1 25 1 25 0 25 0

≅

=

+

=

+

=

+

=

+

=

+

=

+

=

+

=

≤

P

Y

P

Y

P

Y

P

Y

P

Y

P

Y

P

. c)

P

(

Y

=

0 )

=

( )

₀25

0 .

02275

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

9.375 =

75/8

=

(1/8)]

6 [4

+

(1/2)]

5 [2

+

(1/4)]

4 [1

+

(1/8)]

3 [1

=

E(XY)

×

703125

.

0 )

625 .

4 )(

875 .

1 (

375 .

9 )

(

)

(

)

(

−

=

−

=

E

XY

E

X

E

Y

XY

σ

V(X) = 12_{(3/8)+ 2}2_{(1/2) +4}2_(1/8)-(15/8)2 _{= 0.8594} V(Y) = 32_{(1/8)+ 4}2_{(1/4)+ 5}2_{(1/2) +6}2_(1/8)-(37/8)2 _{= 0.7344}

8851

.

0 )

7344

.

0 )(

8594

.

0 (

703125

.

0 ₌

=

Y X XY XY

σ

ρ

5-34. Transaction Frequency Selects(X ) Updates(Y ) Inserts(Z ) New Order 43 23 11 12 Payment 44 4.2 3 1 Order Status 4 11.4 0 0 Delivery 5 130 120 0 Stock Level 4 0 0 0

(7)

Mean Value 18.694 12.05 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

4

18

3

81

8

3

81

8

3

81

8

81

8 )

(

81

8 )

(

6 3 0 3 0 3 0 5 3 0 2 3 0 2 2 0

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

∫

xy

dyd

x

∫

x

y

dyd

x

∫

x

d

x

∫

x

d

x

XY

E

x x

4924

.

0

44 .

0

24 .

0

16 .

0

44 .

0 )

(

,

24 .

0 )

(

3 )

(

6 )

(

16 .

0

5

8

5

12

4

2 2

=

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

−

=

ρ

σ

Y

V

x

V

Y

E

X

E

xy 5-39.

E

(

X

)

=

−

1 (

1 /

4 )

+

1 (

1 /

4 )

=

0

0 )

4 /

1 (

1 )

4 /

1 (

1 )

(

Y

=

−

+

=

E

0 (1/4)]

1 [0

+

(1/4)]

0 [1

+

(1/4)]

0 [-1

+

(1/4)]

0 [-1

=

E(XY)

×

=

V(X) = 1/2 V(Y) = 1/2

0

2 /

1

2 /

1

0

0 )

0 )(

0 (

0 =

=

−

=

Y X XY XY XY σ σ σ

ρ

σ

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,

)

(

)

(

)

(

)

(

)

(

)

(

)

(

XY

xyf

x

f

y

dxdy

xf

x

dx

yf

y

dy

E

X

E

Y

E

=

∫∫

X Y

=

∫

X

∫

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

(8)

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)

07195

.

0

10 .

0

85 .

0

05 .

0 !

3 !

17 !

0 !

20 )

3 ,

17 ,

1 (

)

3 ,

17 ,

0 (

)

3 ,

17 ,

1 (

3 17 0

₊

₌

=

+

=

≤

Y

Z

P

X

Y

Z

P

X

Y

Z

X

P

Because the point

(

1 ,

17 ,

3 )

≠

20

is not in the range of (X, Y, Z).

g) Because X is binomial,

P

(

X

≤

1 )

=

( )

₀20

0 .

05

0

0 .

95

20

+

( )

₁20

0 .

05

1

0 .

95

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

j)

)

17 (

)

17 ,

2 (

)

17 |

2 (

=

Y

P

Y

X

P

Y

X

P

. Now, because x+y+z = 20,

P(X=2, Y=17) = P(X=2, Y=17, Z=1) =

0 .

05

0 .

85

0 .

10

0 .

0540

!

1 !

17 !

2 !

20

2 17 1

₌

2224

.

0 2428

.

0 0540

.

0 )

17 (

)

17 ,

2 (

)

17 |

2 (

=

Y

P

Y

X

P

Y

X

P

k)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

)

17 (

)

17 ,

3 (

3 )

17 (

)

17 ,

2 (

2 )

17 (

)

17 ,

1 (

1 )

17 (

)

17 ,

0 (

0 )

17 |

(

Y

P

Y

X

P

Y

P

Y

X

P

Y

P

Y

X

P

Y

P

Y

X

P

Y

X

E

1 2428

.

0 008994

.

0

3 2428

.

0 05396

.

0

2 2428

.

0 1079

.

0

1 2428

.

0 07195

.

0

0 )

17 |

(

=

⎟

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

=

Y

X

E

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.

1944

.

0

1 .

0

3 .

0

6 .

0 !

2 !

4 )

0 ,

2 ,

2 (

₌

2 2 0

₌

Z

Y

X

P

(9)

c)

0 .

6

0 .

3

0 .

1

0 .

0001

!

4 !

0 !

4 )

4 ,

0 ,

0 (

₌

0 0 4

₌

Z

Y

X

P

d) f_XY x y f_XYZ x y z R

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

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

.

4

1 .

0

3 .

0

6 .

0 )!

4 (

!

4 )

,

(

4

₊

_≤

−

=

− −

y

x

for

y

x

y

x

y

x

f

_XY x y x y e)

E

(

X

)

= np

₁

=

4 (

0 .

6 )

=

2 .

4

f)

E

(

Y

)

= np

₂

=

4 (

0 .

3 )

=

1 .

2

g)

P

(

X

=

2 |

Y

=

2 )

=

0 .

7347

2646

.

0 1944

.

0 )

2 (

)

2 ,

2 (

=

Y

P

Y

X

P

2646

.

0

7 .

0

3 .

0

2

4 )

2 (

2 4

₌

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

Y

P

from the binomial marginal distribution of Y

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 )

0204

.

0 2646

.

0

1 .

0

3 .

0

6 .

0 !

2 !

0 !

4 )

2 (

)

2 ,

0 (

)

2 |

0 (

0 2 2

_⎟

₌

⎠

⎞

⎜

⎝

⎛

=

Y

P

Y

X

P

Y

X

P

2449

.

0 2646

.

0

1 .

0

3 .

0

6 .

0 !

1 !

2 !

1 !

4 )

2 (

)

2 ,

1 (

)

2 |

1 (

1 2 1

_⎟

₌

⎠

⎞

⎜

⎝

⎛

=

Y

P

Y

X

P

Y

X

P

7347

.

0 2646

.

0

1 .

0

3 .

0

6 .

0 !

2 !

4 )

2 (

)

2 ,

2 (

)

2 |

2 (

2 2 0

_⎟

₌

⎠

⎞

⎜

⎝

⎛

=

Y

P

Y

X

P

Y

X

P

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

0.8664

)

5 .

1 (

)

5 .

1 (

)

5 .

1

5 .

1 (

00031

.

0

1 .

0 100465

.

0 00031

.

0

1 .

0 099535

.

0 )

100465

.

0 099535

.

0 (

=

−

<

−

<

=

<

−

=

⎟

⎠

⎞

⎜

⎝

⎛

−

<

−

=

<

Z

P

Z

P

Z

P

Z

P

X

P

Probability that Y is within specification limits is

0.9545

)

2 (

)

2 (

)

2

2 (

00017

.

0

23 .

0 23034

.

0 00017

.

0

23 .

0 22966

.

0 )

23034

.

0 22966

.

0 (

=

−

<

−

<

=

<

−

=

⎟

⎠

⎞

⎜

⎝

⎛

−

_<

−

=

<

Z

P

Z

P

Z

P

Z

P

X

P

Probability that a randomly selected lamp is within specification limits is (0.8664)(0.9594) = 0.8270

(10)

∫

∫ ∫

∞ ∞ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − ∞ ∞ − ∞ ∞ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ₊ − − ∞ ∞ − ∞ ∞ −

⎥

⎦

⎤

⎢

⎣

⎡

⎥

⎦

⎤

⎢

⎣

⎡

=

⎥

⎦

⎤

⎢

⎣

⎡

=

dy

e

dx

e

dxdy

e

dxdy

y

x

f

Y Y Y X X X Y Y X X Y X y x y x XY 2 2 ) ( 2 1 2 2 ) ( 2 1 2 2 ) ( 2 2 ) ( 2 1 2 1 2 1 2 1

)

,

(

σ

μ π μ π μ μ π

and each of the last two integrals is recognized as the integral of a normal probability density function from −∞ to ∞. That is, each integral equals one. Since

f

_XY

(

x

,

y

)

=

f

(

x

)

f

(

y

)

then

X and Y are independent.

Section 5-4

5-54. a) E(2X + 3Y) = 2(0) + 3(10) = 30 b) V(2X + 3Y) = 4V(X) + 9V(Y) = 97

c) 2X + 3Y is normally distributed with mean 30 and variance 97. Therefore,

P

(

2 X

+

3 Y

<

30 )

=

P

(

Z

<

30−₉₇30

)

=

P

(

Z

<

0 )

=

0 .

5

d)

P

(

2 X

+

3 Y

<

40 )

=

P

(

Z

<

40−₉₇30

)

=

P

(

Z

<

1 .

02 )

=

0 .

8461

5-59. a) X∼N(0.1, 0.00031) and Y∼N(0.23, 0.00017) Let T denote the total thickness.

Then, T = X + Y and E(T) = 0.33 mm,

V(T) =

0 .

00031

2

+

0 .

00017

2

=

1 .

25 x

10

−7

mm

2, and

σ

_T

=

0 .

000354

mm.

0 )

272 (

000354

.

0

33 .

0 2337

.

0 )

2337

.

0 (

⎟

=

<

−

≅

⎠

⎞

⎜

⎝

⎛

_<

−

=

<

P

Z

P

Z

T

P

b) 1 ) 253 ( 1 ) 253 ( 000345 . 0 33 . 0 2405 . 0 ) 2405 . 0 ( ⎟ = > − = − < ≅ ⎠ ⎞ ⎜ ⎝ ⎛ _> − = > P Z P Z P Z T P

5-63. Let X denote the average thickness of 10 wafers. Then, E(X) = 10 and V(X) = 0.1.

a)

(

9

11 )

(

)

(

3 .

16

3 .

16 )

0 .

998

1 . 0 10 11 1 . 0 10 9

_<

₌

₋

_<

₌

=

<

_X

_P

−

_Z

−

_P

_Z

P

. The answer is 1 − 0.998 = 0.002 b)

P

₍

X

>

₁₁

₎

=

₀

_.

₀₁

and

σ

_X

=

1 _n_. Therefore,

(

>

11 )

=

(

>

11₁−10

)

=

0 .

01

n

Z

P

X

P

, 11 10₁− n

= 2.33 and n = 5.43 which is rounded up to 6. c)

(

>

11 )

=

0 .

0005

σ

=

σ ₁₀ X

and

X

P

. Therefore,

(

11 )

(

)

0 .

0005

10 10 11

=

>

=

>

− σ

Z

P

X

P

, 10 10 11 σ− = 3.29

9612

.

0

29 .

3 /

10 =

=

σ

5-64. X~N(160, 900)

(11)

P(Y > 4300) =

0227

.

0 9773

.

0

1 )

2 (

1 )

2 (

22500

4000

4300

₌

_>

₌

₋

_<

₌

₋

₌

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

_>

−

Z

P

Z

P

Z

P

b) c) P(Y > x) = 0.0001 implies that

0 .

0001

.

22500

4000 =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

_>

x

−

Z

P

Then x−₁₅₀4000_{= 3.72 and}

_x

=

₄₅₅₈

Section 5-5

5-71. a) If

y

=

x

2, then

x

=

y

for x≥ 0 and y≥ 0. Thus,

y

e

y

f

y

f

y X Y

2 )

(

)

(

2 1 2 1 − −

=

for y > 0.

b) If

y

=

x

1/2, then

x

=

y

2 for

x

≥

0

and

y

≥

0

. Thus,

(

)

(

2

)

2

y2 X

Y

y

f

y

ye

f

=

− for

y > 0.

c) If y = ln x, then x=ey for

x

≥

0

. Thus, y y y ey y ey X Y

y

f

e

f

(

)

=

(

)

=

−

=

− for

∞

<

∞

−

y

. 5-73. If x

e

y

=

, then x = ln y for 1

≤

x

≤

2 and

e

1

≤

y

≤

e

2. Thus,

y

f

y

f

_Y

(

)

=

_X

(ln

)

1 =

1

for 1

≤

ln

y

≤

2

. That is,

y

f

_Y

(

)

=

1

for

e

≤

y

≤

e

2. Supplemental Exercises 5-75. The sum of

∑∑

=

x y

y

x

f

(

,

)

1

,

( )

1 ₄

+

( ) ( )

1 ₈

+

1 ₈

+

( ) ( )

1 ₄

+

1 ₄

=

1

and

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) + 1}2_{(3/8) + 2}2_{(1/4) - 7/8}2 _=39/64 E(Y) = 1(3/8) + 0(3/8) + 2(1/4) = 7/8. V(Y) = 12_{(3/8) + 0}2_{(3/8) + 2}2_{(1/4) - 7/8}2 _=39/64 f)

(

)

=

∑

(

,

)

X

(

0 )

=

3 /

8 ,

X

(

1 )

=

3 /

8 ,

X

(

2 )

=

1 /

4

y XY X

x

f

x

y

and

f

. g)

(

)

=

XY

(

1 ,

)

(

0 )

=

1/8

=

1 /

3 ,

(

1 )

=

1/4

=

2 /

3 f

f

and

y

f

y

f

.

(12)

h)

(

|

1 )

(

)

0 (

1 /

3 )

1 (

2 /

3 )

2 /

3

1 | 1

=

+

=

∑

= = x X Y

y

yf

X

Y

E

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

7949

.

0 6094

.

0 6094

.

0 4844

.

0 ₌

=

XY

ρ

5-76.

0 .

10

0 .

20

0 .

70 0.0631

!

14 !

4 !

2 !

20 )

14 ,

4 ,

2 (

₌

2 4 14

₌

Z

Y

X

P

b)

_P

(

_X

=

0 )

=

0 .

10

0

0 .

90

20

=

0.1216

c)

E

(

X

)

= np

₁

=

20 (

0 .

10 )

=

2 V

(

X

)

=

np

₁

(

1 −

p

₁

)

=

20 (

0 .

10 )(

0 .

9 )

=

1 .

8

d)

)

(

)

,

(

)

19 |

(

|

z

f

z

x

f

Z

X

f

Z XZ z Z X =

=

z z x x XZ

z

x

z

x

xz

f

0 .

1

0 .

2

0 .

7 )!

20 (

!

20 )

(

20− −

−

=

z z Z

z

f

0 .

3

0 .

7 )!

20 (

!

20 )

(

20−

−

=

z x x z z x x Z XZ z Z X

z

x

z

x

z

f

z

x

f

Z

X

f

− − − − − =

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

−

=

−

=

20 20 20 |

3

2

3

1 )!

20 (

!

)!

20 (

3 .

0

2 .

0

1 .

0 )!

20 (

!

)!

20 (

)

(

)

,

(

)

19 |

(

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

3

2 )

0 (

19 |

=

X

f

and

3

1 )

1 (

19 |

=

X

f

. e)

3

1

3

1

3

2

0 )

19 |

(

⎟

=

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

=

Z

X

E

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)

0 .

7

0 .

25

0 .

05

0 .

0649

!

1 !

8 !

10 )

1 ,

8 (

₌

8 1 1

₌

Z

Y

X

P

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) =

( )

10

0 .

95

10

0 .

05

0

0 .

5987

10

=

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

)

8 (

)

,

8 (

)

(

8 X XZ Z

f

z

f

z

f

=

and x x z z XZ

z

x

z

x

z

x

f

0 .

70

0 .

25

0 .

05 )!

10 (

!

10 )

,

(

(10− − )

−

=

for

x

+

z

≤

10 and

0 ≤

x

,

0 ≤

z

. Then, _z _z

( ) ( )

z z z z z z Z

z

f

₈ ₂ _!₍₂2! _)! 0₀._.25₃₀ 2 ₀0._.₃₀05 ! 2 ! 810! ) 2 ( 8 )! 2 ( ! ! 8 10! 8

30 .

0

70 .

0

05 .

0

25 .

0

70 .

0 )

(

₋ − − −

₌

=

(13)

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 X X₁, ₂,...,X₆ denote the lifetimes of the six components, respectively. Because of

independence,

P X( ₁>5000,X₂>5000,...,X₆>5000)=P X( ₁>5000) (P X₂>5000)... (P X₆>5000)

If X is exponentially distributed with mean θ, then λ=_θ1_and

₍

₎

_θ1 /θ /θ x/θ x t x t

e

dt

e

x

X

P

− ∞ − ∞ −

₌

₋

₌

=

>

∫

. Therefore, the answer is

e

−5/8

e

−0.5

e

−0.5

e

−0.25

e

−0.25

e

−0.2

=

e

−2.325

=

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<25,000, X2<25,000, …, X6<25,000)=1-P(X1<25,000)…P(X6<25,000)

=1-(1-e-25/8_)(1-e-2.5_)(1-e-2.5_)(1-e-1.25_)(1-e-1.25_)(1-e-1_{)=1-.2592=0.7408}

5-85. a) Because

0 .

25 ,

25 . 5 75 . 4 25 . 18 75 . 17

c

cdydx

=

∫

c = 4. The area of a panel is XY and P(XY > 90) is the shaded area times 4 below,

5.25 4.75 17.25 18.25 That is,

4

5 .

25

4 (

5 .

25

90 ln

18.25

)

0 .

499

75 . 17 25 . 18 75 . 17 90 25 . 5 / 90 25 . 18 75 . 17

=

−

=

−

=

_∫

∫

dydx

_x

dx

x

b) The perimeter of a panel is 2X+2Y and we want P(2X+2Y >46)

5 .

0 )

2

75 .

17 (

4 )

75 .

17 (

4 )

23 (

25 .

5

4

25 . 18 75 . 17 2 25 . 18 75 . 17 25 . 18 75 . 17 25 . 5 23 25 . 18 75 . 17

=

+

−

=

+

−

=

−

=

∫

−

x

dx

x

dx

x

dydx

x

5-86. a) Let X denote the weight of a piece of candy and X∼N(0.1, 0.01). Each package has 16 candies, then P is the total weight of the package with 16 pieces and E(

P

) = 16(0.1)=1.6 ounces and V(P) = 162_(0.012_{)=0.0256 ounces}2

(14)

c) Let Y equal the total weight of the package with 17 pieces, E(Y) = 17(0.1)=1.7 ounces and V(Y) = 172_(0.012_{)=0.0289 ounces}2

2776

.

0 )

59 .

0 (

)

(

)

6 .

1 (

_Y

_<

₌

_P

_Z

_<

1.₀6_.−₀₂₈₉1.7

₌

_P

_Z

_<

₋

₌

P

.

5-92. Let T denote the total thickness. Then, T = X1 + X2 + X3 and

a) E(T) = 0.5+1+1.5 =3 mm V(T)=V(X1)+V(X2) +V(X3)+2Cov(X1X2)+ 2Cov(X2X3)+ 2Cov(X1X3)=0.01+0.04+0.09+2(0.014)+2(0.03)+ 2(0.009)=0.246mm2 where Cov(XY)=ρσXσY b)

(

6 .

10 )

0

246 .

0

3

5 .

1 )

5 .

1 (

⎟

=

<

−

≅

⎠

⎞

⎜

⎝

⎛

−

<

=

<

P

Z

P

Z

T

P

5-93. Let X and Y denote the percentage returns for security one and two respectively. If ½ of the total dollars is invested in each then ½X+ ½Y is the percentage return. E(½X+ ½Y) = 0.05 (or 5 if given in terms of percent)

V(½X+ ½Y) = 1/4 V(X)+1/4V(Y)+2(1/2)(1/2)Cov(X,Y) where Cov(XY)=ρσXσY=-0.5(2)(4) = -4

V(½X+ ½Y) = 1/4(4)+1/4(6)-2 = 3

Also, E(X) = 5 and V(X) = 4. Therefore, the strategy that splits between the securities has a lower standard deviation of percentage return than investing 2million in the first security.