Eng iSM Chapter 6

8/13/2019 Eng iSM Chapter 6

1/29

104 Chapter 6

CHAPTER 6..Bivariate Probability Distributions and Sampling

Distributions

6.2 a. We know 1 6 and 1 6. x y Each of the 36 outcomes are equally likely, so

1( , )

36 p x y = 1 6, 1 6 x y

b. ( ) ( ) ( ) ( ) ( ) ( )1( 1) (1) 1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 P x p p p p p p= = = + + + + +

=1 1 1 1 1 1 6 1

36 36 36 36 36 36 36 6+ + + + + = =

Similarly, 1 1 11

(2) (3) (6)6

p p p= = = =

Also, 2 2 21

(1) (2) (6)6

p p p= = = =

c. The conditional probability of x, given y, is:

11

2

( , )( | ) ( )

p x y p x y p y=

Since there are 6 levels of y, there are 6 conditional distributions for x.

When y = 1, 12

( , 1)( |1)

(1) p x

p x p

=

12

1(1, 1) 136(1|1)

1(1) 66

p p

p= = =

12

1(2, 1) 136(2|1)

1(1) 66

p p

p= = =


2/29

Bivariate Probability Distributions and Sampling Distributions 105

12

1(6, 1) 136(6|1)

1(1) 66

p p

p= = =

In fact, 1 1( | ) for 1 6, 1 66 p x y x y=

and 21

( | ) for 1 6, 1 66

p y x x y=

d. 1 1( ) ( | ) p x p x y= which implies x and y are independent.

6.4 a. We need to find p( x, y) for X = 0, 1 and Y = 0, 1

p(0, 0) = (.25)(.25) = 0.0625 p(1, 0) = (.75)(.25) = 0.1825 p(0, 1) = (.25)(.75) = 0.1825 p(1, 1) = (.75)(.75) = 0.5625

In table form, we have

b. ( ) 112

( , )|

( ) x y

p x y y

=

( ) 112

(0, 0) 0.06250 | 0 0.2551

(0) 0.2450 p

p p

= = =

( ) 112

(1,0) 0.18251| 0 0.7449

(0) 0.2450 p

p p

= = =

( ) 112

(0,1) 0.18250 |1 0.2450

(1) 0.7450 p

p p

= = =

( ) 112

(1,1) 0.56251|1 0.7550(1) 0.7450

p p p

= = =

c. 1( ) ( , ) y

p x p x y=

( )1 0 (0, 0) (0, 1) 0.0625 0.1825 0.2450 p p p= + = + =

X

0 1

0 0.0625 0.1825Y

1 0.1825 0.5625


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106 Chapter 6

( )1 1 (1, 0) (1, 1) 0.1825 0.5625 0.7450 p p p= + = + =

6.6 a.

3 2 2

0 0 3(0, 0) 073

p

= =

3 2 21 0 2 3 1 1 3

(1, 0)7 35 353

p

= = =

3 2 22 0 1 3 1 2 6

(2, 0)7 35 353

p

= = =

3 2 23 0 0 1 1 1 1

(3, 0)7 35 353

p

= = =

3 2 20 1 2 1 2 1 2

(0, 1)7 35 353

p

= = =

3 2 21 1 1 3 2 2 12

(1, 1)7 35 353

p

= = =

3 2 22 1 0 3 2 1 6

(2, 1)7 35 353

p = = =

p(3, 1) = 0


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3 2 20 2 1 1 1 2 2

(0, 2)7 35 353

p

= = =

3 2 21 2 0 3 1 1 3

(1, 2)7 35 353

p

= = =

p(2, 2) = p(3, 2) = 0

x 0 1 2 3

0 03

35

6

35

1

351

235

1235

635

0 y

22

35

335

0 0

b. 1( ) ( , ) y

p x p x y=

12 2 4

(0) (0, 0) (0, 1) (0, 2) 035 35 35

p p p p= + + = + + =

13 12 3 18

(1) (1, 0) (1, 1) (1, 2)35 35 35 35

p p p p= + + = + + =

16 6 12

(2) (2, 0) (2, 1) (2, 2) 035 35 35

p p p p= + + = + + =

11

(3) (3, 0) (3, 1) (3, 2) 0 035

p p p p= + + = + +

6.8 There are four possible combinations of x and y: (0, 0), (0, 1), (1, 0), (1, 1). Substitutingthese values into the formula for ( , ), x y we obtain the following probabilities:

0 0 2 (0 0) 2(0, 0) p p q q+ += =

x 0 1 2 3

p( x)4

35

1835

1235

135


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108 Chapter 6

0 1 2 (0 1)(0, 1) p p q pq+ += = 1 0 2 (1 0 )(1, 0) p p q pq+ += =

1 1 2 (1 1) 2(1, 1) p p q p+ += =

Whenever 0 1 and 1q p= , we get:

0 ( , ) 1 and ( , ) 1 x y

p x y p x y =

The properties of a bivariate probability distribution are satisfied.

6.10 a.22 10 10 10 10 10

( ) ( , )10 10 5 10 10

y y y y y y y

y y

e e e e e f y f x y dx dx x

y y

= = = = =

This is an exponential distribution with 10 = .

b. The mean DOT estimate is E(Y) = 10.

6.12 a. ( , ) 1 f x y dx dy

=

2 2

0 0 0 0

x x x xce dx dy ce y dx

=

=2

2

0 02

x x cecxe dx

=

= 0 22 2c c

c = =

b.2 2 2

10

0

( ) ( , ) 2 2 2 x x

x x x f x f x y dy e dy e y xe

= = = =

2 2

10 0

( ) 2 0 ( 1) 1 x x f x dx xe dx e

= = = =

c.2

221

( , ) 2 1( | ) 0

( ) 2

x

x

f x y e f y x y x

f x x xe

= = =


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6.14 a. Note ( , ) 1 f x y dx dy+ +

=

Therefore,

11 1 1 2

0 0 0 02

xcxy dx dy cxy dx dy cy dy

+ +

= =

11 12 2 2

0 0 0

1 01

2 2 2 2(2) 4cy cy c

cy dy dy

= = = = =

Since 1 44c

c= =

b.

11 2

10 0

4 4( ) ( , ) 4 2

2 2 xy x

f x f x y dy xy dy x+

= = = = =

11 2

20 0

4 4( ) ( , ) 4 2

2 2 yx y

f y f x y dx xy dx y+

= = = = =

c. 12

( , )( | )

( ) f x y

f x y y

=

From a , 1 2( , ) 4 ; and from ( ) 2 , ( ) 2 f x y xy f x x f y y= = =b,

Therefore, 12

( , ) 4( | ) 2( ) 2

f x y xy x y x f y y= = =

21

( , ) 4( | ) 2

( ) 2 f x y xy

y x y f x x

= = =

6.18 a.3

10

( ) ( ) x

E x xp x=

= = 0(.11) + 1(.29) + 2(.4) + 3(.2) = 1.69 red lights

b. ( ) ( ) ( ) E x y E x E y+ = +

Need to find ( ) E y .3

20

( ) ( ) y

E y xp y=

= = 0(.11) + 1(.25) + 2(.4) + 3(.24) = 1.77 red lights

Therefore, ( ) ( ) ( ) E x y E x E y+ = + = 1.69 + 1.77 = 3.46 red lights.

6.20 a. From Exercise 6.13, c = 1.


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110 Chapter 6

Therefore,if 1 2; 0 1

( , )0 elsewhere x y x y

f x y =

1( ) ( ) E x xf x dx

+

=

We need to find the marginal distribution of x.

11 1 2

10 0 0

1( ) ( , ) ( )

2 2 y

f x f x y dy x y dy xy x

= = = =

22 2 32 2

1 1 1

1 1 1( )

2 2 3 2(2) x

E x x x dx x x dx x = = =

=3 2 3 22 2 1 1 8 4 1 1 7 3 28 9

3 2(2) 3 2(2) 3 4 3 4 3 4 12 12

= + = =

=1912

b. We found the marginal of y which is 23

( ) 0 12

f y y y=

Therefore,

11 1 32 2

0 0 0

3 3 3( )2 2 2(2) 3

y E y y y dy y y dy y = = =

=2 33(1 ) 1 9 4 5

4 3 12 12 12 = =

c.19 5 24

( ) ( ) ( ) 212 12 12

E x y E x E y+ = + = + = =

d.2 1

1 0

( ) ( , ) ( ) E xy xy f x y dy dx xy x y dy dx+ +

= =

=

12 1 2 2 2 32 2

1 0 1 0

( )2 3

x y xy x y xy dy dx dx

=


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=

22 2 3 2

1 12 3 2(3) 3(2)

x x x xdx

=

=3 2 3 22 2 1 1 4 2

6 6 6 6 6 3

= =

6.24 (1) ( ) ( , ) ( , ) 1 E c cf x y dy dx c f x y dy dx c c+ + + +

= = = =

(2) ( )( , ) ( , ) ( , ) E cg x y c g x y f x y dy dx+ +

=

[ ]( , ) ( , ) ( , )c g x y f x y dy dx cE g x y+ +

= =

(3) [ ]1 2( , ) ( , ) ( , )k E g x y g x y g x y+ + +

= [ ]1 2( , ) ( , ) ( , ) ( , )k g x y g x y g x y f x y dy dx+ +

+ + +

= [ ]1 2( , ) ( , ) ( , ) ( , ) ( , ) ( , )k g x y f x y g x y f x y g x y f x y dy dx+ +

+ + +

= 1 2( , ) ( , ) ( , ) ( , ) g x y f x y dy dx g x y f x y dy dx+ + + +

+

( , ) ( , )k g x y f x y dy dx+ +

+

= [ ] [ ] [ ]1 2( , ) ( , ) ( , )k E g x y E g x y E g x y+ + +

6.26 From Exercise 6.5,1

( , )

36

p x y =

1 21 1

( ) , ( ) for , 1, 2, , 66 6

p x p y x y= = =


9/29

112 Chapter 6

To show independence,

1 2( , ) ( ) ( ) p x y p x p y=

1 1 1

36 6 6

=

Thus, x and y are independent.

6.28 To determine if X and Y are independent, we must check 1 2?( , ) ( ) ( ) p x y p x p y .

p(0, 0) = 0.0625 1(0) 0.2450 p = 2 (0) 0.2450 p =

1 2( , ) ( ) ( ) p x y p x p y

X and Y are not independent.

6.30. / 2 / 21

( , )8

x y f x y xe e =

where / 21

( )8

x g x xe =

/ 2( ) yh y e =

x and y are independent

6.32 ( , ) x y x y=

11

( )2

f x x=

23

( )2

y y=

1 21 3

( ) ( ) ( , )2 2

x f y x y f x y =

x and y are not independent

6.34 ( ) ( , ) E xy xy f x y dx dy

=

= 1 2( ) ( ) xy f x f y dx dy

by independence


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= 1 2( ) ( ) x f x y f y dx dy

= 1 2( ) ( ) ( ) ( ) x f x dx y f y dy E x E y

=

6.36 1 2Cov( , ) ( ) x y E xy =

6 6

1 1

( ) ( , ) x y

E xy xy p x y= =

=

=1 1 1 1 441

1(1) 1(2) 1(3) 6(6)36 36 36 36 36 + + + + =

6

1 11

1 1 1 21( ) ( ) 1 2 6

6 6 6 6 x E x x p x

=

= = = + + + =

6

2 21

1 1 1 21( ) ( ) 1 2 6

6 6 6 6 y E y y p y

=

= = = + + + =

441 21 21Cov( , ) 0

36 6 6 x y

= =

6.38 a. From Exercise 6.5, ( ) E x = 1.69 and ( ) E y = 1.77

3 3

0 0

Cov( , ) ( ) ( ) ( ) ( , ) 1.69(1.77) x y

x y E xy E x E y xyp x y= =

= = = 0(0)(.01) + 0(1)(.03) + 0(2)(.05) + 0(3)(.02) + 1(0)(.02) + 1(1)(.06)

+ 1(2)(.12) + 1(3)(.09) + 2(0)(.07) + 2(1)(.10) + 2(2)(.15)+ 2(3)(.08) + 3(0)(.01) + 3(1)(.06) + 3(2)(.08) + 3(3)(.05) 1.69(1.77)

= 2.96 2.9913 = .0313

b.3

2 2 2 2 21 1 1

0

( ) ( ) 1.69 x

E x x p x =

= = = 2 2 2 20 (.11) 1 (.29) 2 (.4) 3 (.2) 2.8561 .8339+ + + =

1 .8339 .9132 = =

32 2 2 2 22 2 2

0

( ) ( ) 1.77 y

E y y p y =

= =


11/29

114 Chapter 6

= 2 2 2 20 (.11) 1 (.25) 2 (.4) 3 (.24) 3.1329 .8771+ + + =

2 .8771 .9365 = =

1 2

Cov( , ) .0313.0366(.9132)(.9365)

x y

= = =

6.40 a. From Exercise 6.13, we calculated:

19 5 2( ) , ( ) , and ( )

12 12 3 E x E y E xy= = =

Therefore,

Cov( , ) ( ) ( ) ( ) x y E xy E x E y=

= 2 19 5 2 95 96 95 13 12 12 3 144 144 144 144 = = =

b.1

Cov( , ) .0069444144

x y = =

[ ]22 2 21 1 1, ( ) ( ) E x E x = =

2 2 2 22 2 2 3

11 1 1

1( ) ( )

2 2 x

E x x f x dx x x dx x dx = = =

=23

4

1

1 16 8 1 1 32 1 314 6 4 6 4 6 12 12 12

x x

= = =

221

31 19 1112 12 144

= =

111

144 =

[ ]22 2 22 2 2, ( ) ( ) E y E y = =

12 2

0

3( )

2 E y y y dy

=


12/29


=11 2 3 4

3

00

3 1 1 10

2 2 4 2 4 4 y y y

y dy = = =

222 2

1 5 11 11

4 12 144 144

= = =

1 2

1 1Cov( , ) 1144 144

11 1111 11144144 144

x y

= = = =

6.42 Cov( , ) ( ) ( ) ( ) x y E xy E x E y=

=1 2 1 2

1 20 0 0 0

( , ) ( ) ( )a a a a

x y x y

xy p x y xp x yp y= = = =

If x and y are independent, then 1 2( , ) ( ) ( ) p x y p x p y=

Thus,1 2 1 2

1 2 1 20 0 0 0

Cov( , ) ( ) ( ) ( ) ( )a a a a

x y x y

x y xy p x p y xp x yp y= = = =

=

=1 2 1 2

1 2 1 20 0 0 0

( ) ( ) ( ) ( ) 0a a a a

x y x y

x p x y p y x p x y p y= = = =

=

6.441 2

0 0 0 0

( ) x y x y E xy xy e e dx dy xe dx ye dy

= = =

1 2 1 2 1 2Cov( , ) ( , ) 0 x y E x y = = =

6.46 From Exercise 6.2,1

( , ) ,36

p x y =

11

( ) for 1, 2, , 66

p x x= =

and 21

( ) for 1, 2, 3, , 66 p y y= =

6

11

1 1 1 1 1 1( ) ( ) 1 2 3 4 5 6 3.5

6 6 6 6 6 6 x E x xp x

=

= = + + + + + =

6

21

1 1 1 1 1 1( ) ( ) 1 2 3 4 5 6 3.5

6 6 6 6 6 6 y E y yp y

=

= = + + + + + =


13/29

116 Chapter 6

( ) ( ) ( ) 3.5 3.5 7 E x y E x E y+ = + = + = 2 21 2( ) 2 Cov( , )V x y x y + = + +

First, find 2 21 2, , Cov( , ) x y

62 2 2 2 2 2 2 2

11

1 1 1 1 1 1( ) ( ) 1 2 3 4 5 6

6 6 6 6 6 6 x E x x p x

=

= = + + + + +

=1 4 9 16 25 36 91

15.16676 6 6 6 6 6 6

+ + + + + = =

[ ]22 2 21 ( ) ( ) 15.1667 (3.5) 2.9167 E x E x = = =

62 2 2 2 2 2 2 2

21

1 1 1 1 1 1( ) ( ) 1 2 3 4 5 6

6 6 6 6 6 6 y E y y p y

=

= = + + + + +

=1 4 9 16 25 36 91

15.16676 6 6 6 6 6 6

+ + + + + = =

[ ]22 2 22 ( ) ( ) 15.1667 (3.5) 2.9167 E y E y = = =

Cov( , ) ( ) ( ) ( ) x y E xy E x E y=

6 6

1 1

1 1 1 441( ) ( , ) 1(1) 1(2) 6(6) 12.25

36 36 36 36 x y E xy xy p x y

= =

= = + + + = =

Cov( , ) ( ) ( ) ( ) x y E xy E x E y= = 12.25 3.5(3.5) = 0

Thus, 2 21 2( ) 2 Cov( , )V x y x y + = + + = 2.9167 + 2.9167 + 2(0) = 5.8334

6.48 f ( x, y) =2 xce , 0 y x <

1 =2 2 2

0 0 0 00

x x x x xc e dydx c e y dx c xe dx = =

c = 2 =

2

0

102 2 2

xe cc c

= =

f ( x, y) =2

2 xe , 0 y x <


14/29


15/29

118 Chapter 6

dv =2

2

2

x x e xe dx y

=

V ( x y) = E ( x y)2 { E ( x y)}2

=

21 1 82 4 2 16 16

= =

6.50 y = x E ( x) = 1 p + 0 (1 p) = p

a. E ( y) = ( ) ( ) E x E x p np= = =

b. p = e s ( ) y = np = ne s

6.52 Let5 for 0 5, which 0

2 22 5 5 15

for 5 10, which implies2 2 2

y y aa

y y a

< < <


16/29


6.560

0 0

00 0

( )w x x

x x

w x w

P w w e dydx e dydx

= +

=

0

00

0 0

x xw x x

w x we y dx e y dx

+

= ( )0

0

00

( ) 0w

x x

w

e x x w dx e x dx

+ +

=0

0

00

w x x

w

w e dx xe dx

+

= ( )0

0

00

00

ww

x x xw

w e xe e dx

+ +

=0 0

00 0 0

w w x x x

ww e xe e

= ( ) ( ) ( )0 0 0 00 00 0 1 1w w w ww e w e e e + =

( ) 1 w F w e =

( )( ) wdF w f w edw= =

6.58 [ ]1 2 3 4( ) 3 2 6 E E y y y y= + +

= 1 2 3 43 ( ) 2 ( ) 6 ( ) ( ) E y E y E y E y + +

= 3(2) + 2(4) + 6( 1) 0 = 4

[ ]1 2 3 4( ) 3 2 6V V y y y y= + +

= 2 2 2 2 2 2 2 21 2 3 4( 3) (2) (6) ( 1) + + + + 2( 3)(2)Cov ( )1 2, y y + 2( 3)(6)Cov ( )1 3, y y +2( 3)( 1)Cov ( )1 4, y y+ 2(2)(6)Cov ( )2 3, y y + 2(2)( 1)Cov ( )2 4, y y +2(6)( 1)Cov ( )3 4, y y

= 9(4) + 4(8) + 36(6) + 1(1) + ( 12)( 1) + ( 36)(1) + (6)12

+ (24)(0)

+ ( 4)(2) + ( 12)(0)= 256


17/29

120 Chapter 6

6.60 a. Let yw e= The range for y is y > 3, which implies 0 < w < 3e = .0497871

0 0 0 0( ) ( ) ( ) ( ). P w w G w F y P y y = = = This implies that 0 0ln . y w=

00

lnln( 3) 3

0 00 0

( ) ( ln )

ww y y

F y F w e dy e e

= = =

= 0 0ln ln3 3 3 3 301w we e e e e w e e = =

3 3( ) F w we e=

3( )( ) dF w

f w edw

= =

b. Let w = y 3. The range for y is y > 3, which implies w > 3 3 or w > 0.

0 0 0 0( ) ( ) ( ) ( ). P w w G w F y P y y = = = This implies 0 0 3 y w= +

( ) ( )0

0

0 0

333( 3) 3 3 3

0 00 0

( ) ( 3) 1

www w y y F y F w e dy e e e e e e

++ +

= + = = = = +

3( ) w F w e e= +

( )( ) w

dF w f w edw

= =

c. Let w =3 y

. The range for y is y > 3, which implies w >33

or w > 1.

0 0 0 0( ) ( ) ( ) ( ). P w w G w F y P y y = = = This implies that 0 03 y w= .

( ) ( )0

0

0 0

333 3( 1)( 3) 3 3 1 2

0 01 1

( ) (3 )

www w y y F y F w e dy e e e e e e e

= = = = = +

3( 1) 2( ) w F w e e = +

3( 1)( )( ) 3 wdF w

f w edw

= =


18/29


6.66 a. The Central Limit Theorem states that the sampling distribution of y is approximatelynormally distributed with a mean of 0.53 y = = and

0.193 0.0272950 y n

= = = .

b. 0.58 0.53( 0.58) ( 1.83) .5000 .4664 .03360.02729

P y P z P Z > = > = > = =

c. Before tensioning:

0.59 0.53( 0.59) ( 2.20) .5000 .4861 .0139

0.02729 P y P z P Z

> = > = > = =

After tensioning:

0.59 0.58( 0.59) ( 0.36) .5000 .1406 .3594

0.02729 P y P z P Z

> = > = > = =

It is more likely that the sample measurements occurred after tensioning.

6.68 Handrubbing

35 y = = ,59

8.34450

yn

= = =

30 35 ( 30) ( .60) .5 .2257 .2743

8.344 P y P z P z

< = < = < = =

(Using Table 5, Appendix B.)

Handwashing

69 y = = ,106

14.99150

yn

= = =

30 69 ( 30) ( 2.60) .5 .4953 .0047

14.991 P y P z P z

< = < = < = =

(Using Table 5, Appendix B.)

This sample of workers probably came from the handrubbing group. The probability ofseeing a mean of 30 or smaller for this group is .2743, while the probability of observing amean of 30 or less for the handwashers is .0047. This would be an extremely rare event if theworkers were handwashers because the probability is so small. It would not be a veryunusual event for the handrubbers because the probability is not that small.


19/29


20/29


21/29

124 Chapter 6

6.82 Let Y = the number of the 70 major Denver bridges that have inspection ratings of 4 or below.Then Y has a binomial probability distribution with n = 70 and p = .09.

a. Using the normal approximation to the binomial, the distribution of y is approximatelynormal with mean = np = 70(.09) = 6.3 and standard deviation

70(.09)(.91) 2.3944npq = = =

34.5 6.3( 35) ( 11.78) 0

2.3944 P Y P Z P Z

> = >

b. Because the probability of observing such an outcome is so small, we would infer thatthe 9% forecast is too low.

6.84 Let y = the number of the 500 drywall installers that are women. Then y has a binomial probability distribution with n = 500 and p = .01.

a. Using the normal approximation to the binomial, the distribution of y is approximatelynormal with = np = 500 .01 = 5 and = 500(.01)(.99)npq = = 2.225.

100.5 5( 100) ( 42.92) 0

2.225 P y P z P z

> > = >

b.5.5 5

( 5) ( .22) .5 (0 .22)2.225

P y P z P z P z = = +

= .5 + .0871 = .5871

6.86 From Theorem 6.11,2

22

( 1)n s

= has a chi-square distribution with = n 1 = 10 1 = 9

degrees of freedom.

a. ( ) ( )2 2 2(10 1)14.414.4 14.4 ) .1009 P s P P > = > = >

b. ( ) ( )2 2 2(10 1)33.333.3 33.3 .0059 P s P P > = > = > = > = >

6.88 a. From Theorem 6.11,2

22

( 1)

n s

= has a chi-square distribution with = n 1 = 10

1 = 9 degrees of freedom.

b. 2 2 2(10 1)16.92

( 16.92) ( 1.5228) .005100

P s P P < = < = <


22/29


c. From the data set, we find s2 = 98.158.

( ) ( )2 2 2(10 1)98.15898.158 8.83424 .100100 P s P P > = > = > >

This probability does not contradict the assumption that 2

= 100.

6.902

1 121

( 1)n s

has a 2 distribution with 1 1 1n = degrees of freedom.

22 2

22

( 1)n s

has a 2 distribution with 2 2 1n = degrees of freedom.

21 1

1 2 221 11

2 2 22 2 2 222

2

( 1)/( 1)

( 1) /( 1)

n sn

s F

n s sn

= =

has an F distribution with 1 1 1n = and 2 2 1n = degrees of freedom.

6.92 Let A = 1 2 1 2

2

1 2

( ) ( )

1 1

y y

n n

+

Then A has a standard normal distribution.

Let B = 2 21 2( 2) /n n s + 2

1 1 2

Then has a distribution with

2 degrees of freedom.

B

n n = +

Let1 2/( 2) A

t B n n

=+

=

1 2 1 2

2

1 2 1 2 1 22 2

2 2 21 2

1 21 2

( ) ( )

1 1

( ) ( )

( 2) / 1 1/

( 2)

y y

n n y y

n n s s

n nn n

+ = +

+ +

= 1 2 1 2 1 2 1 2

2

1 2 1 2

( ) ( ) ( ) ( )

1 1 1 1

y y y y

s sn n n n

=

+ +


23/29


24/29


12

(20, 1) .002(20 |1) .0190

(1) .105 p

p p

= = =

12

(90, 1) 0(90 |1) 0

(1) .105 p

p p

= = =

The conditional probability distribution of x, given y = 1 is given in the table:

x 0 10 20 30 40 50 60 70 80 90

1( |1) p x .0095 .0190 .0190 .2381 .3810 .2381 .0476 .0476 0 0

The other conditional distributions are found in a similar manner and appear below:

x 0 10 20 30 40 50 60 70 80 90

1( | 2) p x .0123 .0123 .0247 .1852 .2469 .1852 .1234 .0741 .0741 .0617

x 0 10 20 30 40 50 60 70 80 90

1( | 3) p x 0 0 0 .0704 .1408 .2254 .1408 .2254 .1127 .0845

x 0 10 20 30 40 50 60 70 80 901( | 4) p x 0 .0172 .0345 .0862 .1724 .4310 .1724 .0517 .0172 .0172

x 0 10 20 30 40 50 60 70 80 901( | 5) p x 0 .0260 .0649 .0649 .2597 .3896 .1948 0 0 0

c. Find( 40 | 3) P x y =

= 1 1 1 1 1 1(40 | 3) (50 | 3) (60 | 3) (70 | 3) (80 | 3) (90 | 3) p p p p p p+ + + + +

= .1408 + .2254 + .1408 + .2254 + .1127 + .0845 = .9296

d.5

21

( ) ( ) 1(.105) 2(.405) 3(.355) 4(.058) 5(.077) y

E y yp y=

= = + + + + = 2.597

e. Using the formulas presented in this chapter, we find:

( ) 51.86( ) 2.597 Cov( ) ( ) ( ) ( ) 139.61 51.86(2.597) 4.92958( ) 139.61

E x

E y xy E xy E x E y

E xy

== = = ==

x and y are correlated They are not independent.

f. Additional Profit = $25,000 x ( ) 51.86, ( ) 334.34 E x V x= =

E (additional profit) = $25,000 ( ) $25,000(51.86) $1,296,500 E x = = V (additional profit) = 2 2($25,000) ( ) ($25,000) (334.34)V x = Standard deviation = variance $457,124=


25/29

128 Chapter 6

6.96 a. By Theorem 6.9, the sampling distribution of y is approximately normal with5

15 and 1.11820

y yn

= = = = =

b.6 15

7.421.118

x

y

y z

= = =

( 6) ( .742) P y P z = = .5 ( 7.42 0) P z .5 .5 (from Table 5, Appendix B)= 0

c. Since we should not see this type of result if = 15, we believe the mean time to fallasleep for the men taking melatonin is less than 15 minutes.

6.98 a. ( .3) (.3, 0) (.3, 10) (.3, 20) (.3, 30) P x p p p p= = + + +

= .07 + .05 + .03 + .02 = .17

b.1

(.3, 20) .03( 20 | .3) .1765

(.3) .17 p

P y x p

= = = = =

c. Using the formulas presented in this chapter:

( ) .207 E x = ( ) 10.0 E y = ( ) 1.76 E xy =

Cov( ) ( ) ( ) ( ) 1.76 (.207)(10) .31 xy E xy E x E y= = =

x and y are correlated.

d. Since x and y are correlated, they cannot be independent.

6.100 a. .000005 y = =

/ .000002 / 50 .00000028 y n = = =

b. The Central Limit Theorem guarantees that the sampling distribution of y isapproximately normal.

.0000053 .000005( .0000053).00000028

P y P z > = >

= ( 1.06) .5 (0 1.06) P z P z > = < < = .5 .3554 = .1446


26/29


6.102 From Theorem 6.11,2

22

( 1)

n s

= has a chi-square distribution with 1n = degrees of

freedom.

( )2 2 2

2

(32 1)(1,311)( 1,311) 53.66

(27.52) P s P P

> = > = >

< .005

6.104 The number of defectives, y, in the sample of 200 has a binomial distribution with n = 200and p = .08. The lot will be rejected if more than 6% (12/200) of the sample proves to bedefective.

P (lot rejected) =(12 .5)

( 12) np

P y P z npq

+ > >

=(12.5 200(.08)

( .091)200(.08)(.92)

P z P z > = >

= .5 + ( 0.91 0) P z < < = .5 + .3186 = .8186

6.106 a. By Theorem 6.9, the sampling distribution of y has an approximate normal distribution

with mean y = =1,312 and standard deviation / 422/ 100 y n = = = 42.2.

b.1,418 1,312

( 1,418) ( 2.51)42.2

P y P z P z = =

= .5 P (0 < z < 2.51) = .5 .4940 = .0060

c. If the actual mean is 1.418, y = we would believe the population average stated,

= 1,312, is incorrect.

d. The sampling distribution of2

2( 1)n s

is chi-square with ( 1)n = = (100 1) = 99

degrees of freedom.

e. 2 2 2 22(100 1)250,000

( 500 ) ( 138.979)422

P s P P > = > = >

Using Table 8, Appendix B, this probability is between .005 and .01. Since2 2.005 ( 500 ) .01, P s< > < then .005 ( 500) .01. P s< > <

6.108 a. If y is an exponential random variable, then = E ( y) = = 60. The standard deviationof y is = = 60.

Then, ( ) E y = y = = 60;

( )V y =2 2

2 60100 y n

= = = 36


27/29

130 Chapter 6

b. Because the sample size is fairly large, the Central Limit Theorem says that thesampling distribution of is approximately normal.

c. P ( y 30) =

36

6030 z P = P ( z 5.0) .5 .5 = 0

6.110 Let 2(2 3).c y= + The range of the y is 1 5 y which implies the range of the c is5 53.c

0 0 0 0( ) ( ) ( ) ( ). P c c C c F y P y y = = = This implies that 03

2c

y =

( 3) / 2( 3) / 2

01 1

3 1 1 1 3 1( )

2 4 4 4 2 4

cc

c c F y F dy y

= = = =

1 3 1( )

4 2 4c

F c =

32( ) 1( ) ( 3)

8 2dF c

f c cdc

= =

20 32

5

1( 20) 1 ( 20) 1 ( 3)

8 2 P c P c c dc = =

=20

5

1 3 1 17 21 1

4 2 4 2 2c =

= 1 .4789 = .5211

6.112 a. ( ) ( ) ( ) 2(1)(1)Cov( , ) ( ) ( ) 0 ( ) ( )V x y V x V y x y V x V y V x V y+ = + + = + + = +

( ) ( ) ( ) 2(1)( 1)Cov( , ) ( ) ( ) 0 ( ) ( )V x y V x V y x y V x V y V x V y = + + = + + = +

( ) ( )V x y V x y + =

b. [ ]Cov ( ), ( ) x y x y+ = (1)(1)Cov( , ) (1)(1)Cov( , ) (1)( 1)Cov( , ) (1)( 1)Cov( , ) x x y x x y y y+ + + = Cov( , ) Cov( , ) Cov( , ) Cov( , ) x x x y x y y y+ = Var( x) Var( y)


28/29


6.114 a. Let .w y= The range for y is 0 1 y which implies 0 1w or 0 1.w

0 0 0 0( ) ( ) ( ) ( ). P w w w w F y P y y = = = This implies2

0 0 . y w=

( )20 2

02 2 3 6 60 0 0 000

( ) 3 0

ww

F y F w y dy y w w= = = = =

60( ) F w w=

50

( )( ) 6

dF w f w w

dw= =

b. Let w = 3 y. The range for w is 3 1 3 0 or 2 3.w w

0 0 0 0( ) ( ) ( ) ( ). P w w w w F y P y y = = = This implies 0 0 3. y w=

( ) ( ) ( )0

03

3 3 32 30 0 0 00

0

( ) 3 3 3 0 3w

w F y F w y dy y w w

= = = = =

3( ) ( 3) F w w=

2( )( ) 3( 3)dF w

f w wdw

= =

c. Let ln( ).w y= The range of w is ln(3) ln(2).w

0 0 0 0( ) ( ) ( ) ( ). P w w w w F y P y y = = = This implies 00 .w y e =

( ) ( )0 0

0 0 03 32 3

00

( ) 3 0ww ew w w F y F e y dy y e e

= = = = =

3( ) w F w e =

3( )( ) 3 wdF w

f w edw

= =

6.116 Let y have an exponential distribution with = .

Let w = 2 y/ y = w /2

( ) 2/ ( ) 2 / 2 E w E y = = =


29/29

/

0( )

0 elsewhere

ye y

f y

=

/ 2

0( )

0 elsewhere

wew f w

=

/ 21 0( )

0 elsewhere

we w f w

=

which is the density function of a 2 random variable with v = 2 degrees of freedom.

6.120 To use Theorem 6.7, we must first find the cumulative distribution function of y.

( )2 2 211

( ) 2( 1) 2 2 (1 2) ( 1) y

y F y t dt t t y y y= = = =

Let 2( ) ( 1) . g F y y= = By Theorem 6.7, g has a uniform density function over the interval0 1. g

2( 1) ( 1) 1 g y g y y g = = = +

By selecting random numbers from the random number table and substituting them in for g ,we will get random selections for y.

6.122

2

2(ln )

21 0( ) 20 otherwise

y

e y f y y

>=

1ln , x y dx dy x

y= = < <

Thus, by substitution,

2

2

( )

21( )2

x

f x e x

= < <

Then x has a normal distribution with mean and variance 2 .

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Eng iSM Chapter 6

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