CHAPTER 5 Discrete Probability Distributions€¦ · CHAPTER 5 Discrete Probability Distributions...

5-1 Copyright ©2015 Pearson Education, Inc.

CHAPTER 5 Discrete Probability Distributions

5.1 ( ) ( )5 1.0 0.06 0.11 0.24 0.27 0.20 1.0 0.88 0.12P x = = − + + + + == − =

5.2 a) µ = (1)(0.18) + (2)(0.25) + (3)(0.35) + (4)(0.22) = 2.61

b) 27.850 (2.61)σ = − = 1.019

5.3 a) µ = (10)(0.10) + (15)(0.30) + (20)(0.20) + (25)(0.30) + (30)(0.10) = 20.0

b) 2435 (20.0)σ = − = 5.916

5.4 A – not a discrete probability distribution (the sum of the probabilities of all outcomes is greater than 1) B – discrete probability distribution C – discrete probability distribution D – not a discrete probability distribution (probability of one outcome is greater than 1) A – not a discrete probability distribution (the sum of the probabilities of all outcomes is less than 1) 5.5

a) µ = 88 40 42 20 7 3

(0) (1) (2) (3) (4) (5)200 200 200 200 200 200

+ + + + + =

= (0)(0.44) + (1)(0.20) + (2)(0.21) + (3)(0.10) + (4)(0.035) + (5)(0.015) = 1.135

b) 22.875 (1.135)σ = − = 1.260

5.6 a) Stanton:

( ) ( ) ( ) ( ) ( )

4 12 8 36 20(1) (2) (3) (4) (5)

80 80 80 80 80

(1) 0.05 (2) 0.15 (3) 0.10 (4) 0.45 (5) 0.25 3.7

μ = + + + + =

= + + + + =

Newark:

( ) ( ) ( ) ( ) ( )

12 12 18 15 18(1) (2) (3) (4) (5)

75 75 75 75 75

(1) 0.16 (2) 0.16 (3) 0.24 (4) 0.20 (5) 0.24 3.2

μ = + + + + =

= + + + + =

5-2 Chapter 5

Copyright ©2015 Pearson Education, Inc.

b) Stanton: 215.00 (3.7)σ = − = 1.14; Newark: 212.16 (3.2)σ = − =1.36

c) Stanton’s fast-food restaurants have higher average customer satisfaction rating than ones in Newark. Customer satisfaction ratings for Stanton’s fast-food restaurants are more consistent when compared with ones in Newark. 5.7 a) µ = (3)(0.23) + (4)(0.57) + (5)(0.14) + (6)(0.06) = 4.03

b) 216.850 (4.03)σ = − = 0.78

5.8 EMV = ($4,000)(0.30) + ($20,000)(0.45) + ($36,000)(0.25) = $19,200 5.9 Profit = Revenue– Costs Revenue is based on the number loaves sold, which is the minimum of the supply of loaves and the eventual demand.

Demand Profit (Bake 25 loaves) Probability 25 (25)($6) – (25)($3) = $75 0.35 50 (25)($6) – (25)($3) = $75 0.25 75 (25)($6) – (25)($3) = $75 0.40

EMV (25) = ($75)(0.35) + ($75)(0.25) + ( $75)(0.40) = $75

Demand Profit (Bake 50 loaves) Probability

25 (25)($6) – (50)($3) = $0 0.35 50 (50)($6) – (50)($3) = $150 0.25 75 (50)($6) – (50)($3) = $150 0.40

EMV (50) = ($0)(0.35) + ($150)(0.25) + ( $150)(0.40) = $97.50

Demand Profit (Bake 75 loaves) Probability

25 (25)($6) – (75)($3) = -$75 0.35 50 (50)($6) – (75)($3) = $75 0.25 75 (75)($6) – (75)($3) = $225 0.40

EMV (75) = (-$75)(0.35) + ($75)(0.25) + ( $225)(0.40) = $82.50 The most profitable is to bake 50 loaves every morning.

Discrete Probability Distributions 5-3


5.10

a) 2 57!(2,7) (0.6) (0.4) 0.0774

(7 2)!2!P = =

−

b) 0 7 1 67! 7!(0,7) (1,7) (0.60) (0.40) (0.60) (0.40) 0.0188

(7 0)!0! (7 1)!1!P P+ = + =

− −

c) 6 1 7 07! 7!(6,7) (7,7) (0.60) (0.40) (0.60) (0.40) 0.1586

(7 6)!6! (7 7)!7!P P+ = + =

− −

5.11

a) 3 58!(3,8) (0.35) (0.65) 0.2786

(8 3)!3!P = =

−

b)

( )

0 8

1 7

2 6

8!(0,8) (0.35) (0.65) 0.0319

(8 0)!0!

8!(1,8) (0.35) (0.65) 0.1373

(8 1)!1!

8!(2,8) (0.35) (0.65) 0.2587

(8 2)!2!

3 0.0319 0.1373 0.2587 0.4279

P

P

P

P x

= =−

= =−

= =−

< = + + =

c)

( )

6 2

7 1

8 0

8!(6,8) (0.35) (0.65) 0.0217

(8 6)!6!

8!(7,8) (0.35) (0.65) 0.0033

(8 7)!7!

8!(8,8) (0.35) (0.65) 0.0002

(8 8)!8!

6 0.0217 0.0033 0.0002 0.0252

P

P

P

P x

= =−

= =−

= =−

≥ = + + =

5.12

a) 4 15!(4,5) (0.25) (0.75) 0.0146

(5 4)!4!P = =

−

b) 4 37!(4,7) (0.25) (0.75) 0.0577

(7 4)!4!P = =

−

c) 4 610!(4,10) (0.25) (0.75) 0.1460

(10 4)!4!P = =

−

5-4 Chapter 5


5.13 (15)(0.65) 9.75; (15)(0.65)(0.35) 1.847μ σ= = = =

5.14 ( 14) 0.1091 0.1746 0.2182 0.2054 0.1369 0.0576 0.0115 0.9133P x ≥ = + + + + + + =

5.15

a) 3 710!(3,10) (0.28) (0.72) 0.2642

(10 3)!3!P = =

−

b)

( )

0 10

1 9

2 8

3 7

10!(0,10) (0.28) (0.72) 0.0374

(10 0)!0!

10!(1,10) (0.28) (0.72) 0.1456

(10 1)!1!

10!(2,10) (0.28) (0.72) 0.2548

(10 2)!2!

10!(3,10) (0.28) (0.72) 0.2642

(10 3)!3!

3 0.0374 0.1456 0.2548 0.2642 0.

P

P

P

P

P x

= =−

= =−

= =−

= =−

≤ = + + + = 7020

c)

0 10

1 9

2 8

3 7

4 6

10!(0,10) (0.28) (0.72) 0.0374

(10 0)!0!

10!(1,10) (0.28) (0.72) 0.1456

(10 1)!1!

10!(2,10) (0.28) (0.72) 0.2548

(10 2)!2!

10!(3,10) (0.28) (0.72) 0.2642

(10 3)!3!

10!(4,10) (0.28) (0.72)

(10 4)!4!

P

P

P

P

P

= =−

= =−

= =−

= =−

= =−

( ) ( )( )

0.1798

5 1 5

5 1 0.0374 0.1456 0.2548 0.2642 0.1798 0.1182

P x P x

P x

≥ = − <

≥ = − − − − − =

d) (10)(0.28) 2.8; (10)(0.28)(0.72) 1.42μ σ= = = =



e)

5.16

a) 7 18!(7,8) (0.90) (0.10) 0.3826

(8 7)!7!P = =

−

b) ( ) 8 08!7 (8,8) (0.90) (0.10) 0.4305

(8 8)!8!P x P> = = =

−

c)

( )

( )

6 2

7 1

8 0

8!(6,8) (0.90) (0.10) 0.1488

(8 6)!6!

8!(7,8) (0.90) (0.10) 0.3826

(8 7)!7!

8!(8,8) (0.90) (0.10) 0.4305

(8 8)!8!

6

6

( 6) 1

1 1 0.1488 0.3826 0.4305 0.0381

P

P

P

x

x

P x PP

= =−

= =−

= =−

≥

≥

< = −− = − − − =

5-6 Chapter 5


d) 4 48!(4,8) (0.90) (0.10) 0.0046

(8 4)!4!P = =

−

The probability of 4 out of 8 customers being satisfied is very low assuming that 90% of the customer base is satisfied. Based on this sample, it is not likely that 90% of the customers are satisfied. 5.17

a) 2 911!(2,11) (0.25) (0.75) 0.2581

(11 2)!2!P = =

−

b)

( )

0 11

1 10

2 9

11!(0,11) (0.25) (0.75) 0.0422

(11 0)!0!

11!(1,11) (0.25) (0.75) 0.1549

(11 1)!1!

11!(2,11) (0.25) (0.75) 0.2581

(11 2)!2!

3 0.0422 0.1549 0.2581 0.4552

P

P

P

P x

= =−

= =−

= =−

< = + + =

c)

0 11

1 10

2 9

3 8

4 7

11!(0,11) (0.25) (0.75) 0.0422

(11 0)!0!

11!(1,11) (0.25) (0.75) 0.1549

(11 1)!1!

11!(2,11) (0.25) (0.75) 0.2581

(11 2)!2!

11!(3,11) (0.25) (0.75) 0.2581

(11 3)!3!

11!(4,11) (0.25) (0.75)

(11 4)!4!

P

P

P

P

P

= =−

= =−

= =−

= =−

=−

( )

5 6

0.1721

11!(5,11) (0.25) (0.75) 0.0803

(11 5)!5!

5 1.0 0.0422 0.1549 0.2581 0.2581 0.1721 0.0803 0.0343

P

P x

=

= =−

> = − − − − − − =

d) (11)(0.25) 2.75; (11)(0.25)(0.75) 1.436μ σ= = = =



e)

5.18

a) 3 69!(3,9) (0.35) (0.65) 0.2716

(9 3)!3!P = =

−

b)

( )

0 9

1 8

2 7

3 6

9!(0,9) (0.35) (0.65) 0.0207

(9 0)!0!

9!(1,9) (0.35) (0.65) 0.1004

(9 1)!1!

9!(2,9) (0.35) (0.65) 0.2162

(9 2)!2!

9!(3,9) (0.35) (0.65) 0.2716

(9 3)!3!

4 0.0207 0.1004 0.2162 0.2716 0.6089

P

P

P

P

P x

= =−

= =−

= =−

= =−

< = + + + =

c)

6 3

7 2

9!(6,9) (0.35) (0.65) 0.0424

(9 6)!6!

9!(7,9) (0.35) (0.65) 0.0098

(9 7)!7!

(6,9) (7,9) 0.0424 0.0098 0.0522

P

P

P P

= =−

= =−

+ = + =

5-8 Chapter 5


d) (9)(0.35) 3.15; (9)(0.35)(0.65) 1.43μ σ= = = =

e)

5.19 a)

1 1112!

(1,12) (0.14) (0.86) 0.3197(12 1)!1!

P = =−

b)

( )

0 12

1 11

2 10

3 9

12!(0,12) (0.14) (0.86) 0.1637

(12 0)!0!

12!(1,12) (0.14) (0.86) 0.3197

(12 1)!1!

12!(2,12) (0.14) (0.86) 0.2863

(12 2)!2!

12!(3,12) (0.14) (0.86) 0.1553

(12 3)!3!

4 0.1637 0.3197 0.2863 0.1553

P

P

P

P

P x

= =−

= =−

= =−

= =−

< = + + + = 0.9250



c)

0 12

1 11

2 10

12!(0,12) (0.14) (0.86) 0.1637

(12 0)!0!

12!(1,12) (0.14) (0.86) 0.3197

(12 1)!1!

12!(2,12) (0.14) (0.86) 0.2863

(12 2)!2!

( 2) 1 0.1637 0.3197 0.2863 0.2303

P

P

P

P x

= =−

= =−

= =−

> = − − − =

d) (12)(0.14) 1.68; (12)(0.14)(0.86) 1.202μ σ= = = = e)

5.20

a) 0 1515!(0,15) (0.06) (0.94) 0.3953

(15 0)!0!P = =

−

b)

( )

0 15

1 14

2 13

15!(0,15) (0.06) (0.94) 0.3953

(15 0)!0!

15!(1,15) (0.06) (0.94) 0.3785

(15 1)!1!

15!(2,15) (0.06) (0.94) 0.1691

(15 2)!2!

3 0.3953 0.3785 0.1691 0.9429

P

P

P

P x

= =−

= =−

= =−

< = + + =

5-10 Chapter 5


c)

0 15

1 14

15!(0,15) (0.06) (0.94) 0.3953

(15 0)!0!

15!(1,15) (0.06) (0.94) 0.3785

(15 1)!1!

( 1) 1 0.3953 0.3785 0.2262

P

P

P x

= =−

= =−

> = − − =

d)

e) 4 1115!(4,15) (0.06) (0.94) 0.0090

(15 4)!4!P = =

−

The probability of 4 out of 15 customers making a purchase on the web site is very low assuming that 6% of the customers make a purchase. Based on this sample, it is not likely that 6% of the customers make purchases. 5.21

a) 9 514!(9,14) (0.70) (0.30) 0.1963

(14 9)!9!P = =

−

b) 14 014!(14,14) (0.70) (0.30) 0.0068

(14 14)!14!P = =

−



c)

12 2

13 1

14 0

14!(12,14) (0.70) (0.30) 0.1134

(14 12)!12!

14!(13,14) (0.70) (0.30) 0.0407

(14 13)!13!

14!(14,14) (0.70) (0.30) 0.0068

(14 14)!14!

( 11) 1 0.1134 0.0407 0.0068 0.8391

P

P

P

P x

= =−

= =−

= =−

≤ = − − − =

d) .715(14)(0.70) 9.8; (14)(0.70)(0.30) 1μ σ= = = = e)

5.22

a) ( )( )3 3.03.0 2.71828

(3) 0.22403!

P−

= =

5-12 Chapter 5


b)

( )( )

( )( )

( )( )

0 3.0

1 3.0

2 3.0

3.0 2.71828(0) 0.0498

0!

3.0 2.71828(1) 0.1494

1!

3.0 2.71828(2) 0.2240

2!( 2) 0.0498 0.1494 0.2240 0.4232

P

P

P

P x

−

−

−

= =

= =

= =

≤ = + + =

c)

( )( )

( )( )

( )( )

( )( )

( ) ( )( )

( ) ( )( )

0 3.0

1 3.0

2 3.0

4 3.0

3 3.0

3.0 2.71828(0) 0.0498

0!

3.0 2.71828(1) 0.1494

1!

3.0 2.71828(2) 0.2240

2!

(3) 0.2240

3.0 2.718284 0.1680

4!4 4

4

3.0 2.71828

3!

1

1 0.0498 0.1494 0.2240 0.2240 0.1680 0.18

P

P

P

P

P

x x

x

P PP

−

−

−

−

−

= =

= =

= =

= =

= =

> = ≤

>

−= − − − − − = 48

5.23

a) ( )( )5 4.74.7 2.71878

(5) 0.17385!

P−

= =



b)

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

0 4.7

1 4.7

2 4.7

3 4.7

4 4.7

5 4.7

6 4.7

4.7 2.71878(0) 0.0091

0!

4.7 2.71878(1) 0.0427

1!

4.7 2.71878(2) 0.1005

2!

4.7 2.71878(3) 0.1574

3!

4.7 2.71878(4) 0.1849

4!

4.7 2.71878(5) 0.1738

5!

4.7 2.71878(6) 0.1

6!

P

P

P

P

P

P

P

−

−

−

−

−

−

−

= =

= =

= =

= =

= =

= =

= =

( )

362

6

1 0.0091 0.0427 0.1005 0.1574 0.1849 0.1738 0.1362

( 6) 1

( 6)

( 6) 0.1954

xP x PP xP x

≤= − − − − − − −

> = −>> =

c)

( )( )

( )( )

( )( )

( )( )

0 4.7

1 4.7

2 4.7

3 4.7

4.7 2.71878(0) 0.0091

0!

4.7 2.71878(1) 0.0427

1!

4.7 2.71878(2) 0.1005

2!

4.7 2.71878(3) 0.1574

3!0.0091 0.0427 0.1005 0.1574( 3) 0.3097

P

P

P

P

P x

−

−

−

−

= =

= =

= =

= =

+ + +≤ = =

5.24

a) ( )( )4 2.02.0 2.71828

(4) 0.09024!

P−

= =

b) ( )( )4 3.03.0 2.71828

(4) 0.16804!

P−

= =

5-14 Chapter 5


c) ( )( )4 4.04.0 2.71828

(4) 0.19544!

P−

= =

d) As the value of λ approached 4.0, the probability that 4x = increases.

5.25 . 8.5 2.928 5; μ λ σ == = =

5.26

a) ( )( )0 2.62.6 2.71828

(0) 0.07430!

P−

= =

b)

( )( )

( )( )

( )( )

( )( )

( )

0 2.6

1 2.6

2 2.6

3 2.6

2.6 2.71828(0) 0.0743

0!

2.6 2.71828(1) 0.1931

1!

2.6 2.71828(2) 0.2510

2!

2.6 2.71828(3) 0.2176

3!4 0.0743 0.1931 0.2510 0.2176 0.7360

P

P

P

P

P x

−

−

−

−

= =

= =

= =

= =

< = + + + =

c)

( )( )

( )( )

( )

0 2.6

1 2.6

2.6 2.71828(0) 0.0743

0!

2.6 2.71828(1) 0.1931

1!( 1) 1 1 1 0.0743 0.1931 0.7326

P

P

P x P x

−

−

= =

= =

> = − ≤ = − − =

5.27

a) ( )( )5 2.12.1 2.71828

(5) 0.04175!

P−

= =

b) ( )( )0 2.12.1 2.71828

(0) 0.12250!

P−

= =



c)

( )

( ) ( )( )

( ) ( )( )

( ) ( )( )

0 2.1

1 2.1

2 2.1

( 2) 1 2

2.1 2.718280 0.1225

0!

2.1 2.718281 0.2572

1!

2.1 2.718282 0.2700

2!( 2) 1 0.1225 0.2572 0.2700 0.3503

P x P x

P

P

P

P x

−

−

−

> = − ≤

= =

= =

= =

> = − − − =

d) 2.1 1.45σ ==

5.28

a) ( )( )7 4.54.5 2.71828

(7) 0.08247!

P−

= =

b)

( )( )

( )( )

( )( )

( )

0 4.5

1 4.5

2 4.5

4.5 2.71828(0) 0.0111

0!

4.5 2.71828(1) 0.0500

1!

4.5 2.71828(2) 0.1125

2!3 0.0111 0.0500 0.1125 0.1736

P

P

P

P x

−

−

−

= =

= =

= =

< = + + =

c)

( )( )

( )( )

( )( )

( )( )

( )( )

( )

0 4.5

1 4.5

2 4.5

3 4.5

4 4.5

4.5 2.71828(0) 0.0111

0!

4.5 2.71828(1) 0.0500

1!

4.5 2.71828(2) 0.1125

2!

4.5 2.71828(3) 0.1687

3!

4.5 2.71828(4) 0.1898

4!( 5) 1 5

( 5) 1 0.0111 0.0500 0.1125 0.1687 0.189

P

P

P

P

P

P x P xP x

−

−

−

−

−

= =

= =

= =

= =

= =

≥ = − <≥ = − − − − − 8 0.4679=

5-16 Chapter 5


5.29

a) ( )( )0 1.51.5 2.71828

(0) 0.22310!

P−

= =

b)

( ) ( )( )

( ) ( )( )

( ) ( )( )

( )

0 1.5

1 1.5

2 1.5

1.5 2.718280 0.2231

0!

1.5 2.718281 0.3347

1!

1.5 2.718282 0.2510

2!( 3) 1 3 1 0.2231 0.3347 0.2510 0.1912

P

P

P

P x P x

−

−

−

= =

= =

= =

≥ = − < = − − − =

c) ( )( )3 3.03.0 2.71828

(3) 0.22403!

P−

= =

d)

( )( )

( )( )

( )( )

( )

0 3.0

1 3.0

2 3.0

3.0 2.71828(0) 0.0498

0!

3.0 2.71828(1) 0.1494

1!

3.0 2.71828(2) 0.2240

2!2 0.0498 0.1494 0.2240 0.4232

P

P

P

P x

−

−

−

= =

= =

= =

≤ = + + =

5.30

a) 2 2022!(2,22) (0.015) (0.985) 0.0384

(22 2)!2!P = =

−

b) 2 (22)(0.015)((22)(0.015)) 2.71828

(2) 0.03912!

P−

= =

c) The probabilities are close enough, but fewer calculations were needed for the result when using the Poisson distribution



5.31 a)

( )

0 25

1 24

2 23

25!(0,25) (0.016) (0.984) 0.6682

(25 0)!0!

25!(1, 25) (0.016) (0.984) 0.2716

(25 1)!1!

25!(2,25) (0.016) (0.984) 0.0530

(25 2)!2!

3 0.6682 0.2716 0.0530 0.9928

P

P

P

P x

= =−

= =−

= =−

< = + + =

b)

( )( )

( )

0 (0.40)

1 (0.40)

2 (0.40)

25 0.016 0.40

(0.40) 2.71828(0) 0.6703

0!

(0.40) 2.71828(1) 0.2681

1!

(0.40) 2.71828(2) 0.0536

2!3 0.6703 0.2681 0.0536 0.9920

np

P

P

P

P x

−

−

−

= =

= =

= =

= =

< = + + =

5.32

a) ( ) 12 6 5 3 6 3

12 5

(3) 0.3788C CP x P

C− −= = =

b) 12 6 5 2 6 2

12 5

( ) (2) 0.3788C CP x P

C− −= = =

c) 12 6 5 0 6 0 12 6 5 1 6 1

12 5 12 5

( ) (0) (1) 0.1212C C C CP x P P

C C− − − −= + = + =

d) ( )( )2

12 5(5)(6) (5)(6)(12 6)2.5; 0.89

12 12 12 1μ σ −−= =

−= =

5.33

a) 9 5 3 0 5 0

9 3

( ) (0) 0.0476C CP x P

C− −= = =

b) ( ) 9 5 3 0 5 0 9 5 3 1 5 1

9 3 9 3

( ) 1 (0) (1) 1 0.5952C C C CP x P P

C C− − − −

= − + = − + =

c) 9 5 3 0 5 0 9 5 3 1 5 1 9 5 3 2 5 2

9 3 9 3 9 3

( ) (0) (1) (2) 0.8810C C C C C CP x P P P

C C C− − − − − −= + + = + + =

5-18 Chapter 5


d) ( )( )2

9 3(3)(5) (3)(5)(9 5)1.67; 0.7453

9 9 9 1μ σ −−= =

−= =

5.34 14, 6, 4, 3;N R n x= = = = 14 6 4 3 6 3

14 4

( ) (3) 0.1598C CP x P

C− −= = =

5.35 18, 10, 6, 6;N R n x= = = = 18 10 6 6 10 6

18 6

( ) (6) 0.0113C CP x P

C− −= = =

5.36

a) 9, 2, 2, 0;N R n x= = = = 9 2 2 0 2 0

9 2

( ) (0) 0.5833C CP x P

C− −= = =

b) 9, 2, 2, 1;N R n x= = = = 9 2 2 1 2 1

9 2

( ) (1) 0.3889C CP x P

C− −= = =

c) 9, 2, 2, 2;N R n x= = = = 9 2 2 2 2 2

9 2

( ) (2) 0.0278C CP x P

C− −= = =

d) ( )( )2

9 2(2)(2) (2)(2)(9 2)0.444; 0.55

9 9 9 1μ σ −−= =

−= =

e) HYPGEOM.DIST(0, 2, 2, 9, FALSE) = 0.5833 HYPGEOM.DIST(1, 2, 2, 9, FALSE) = 0.3889 HYPGEOM.DIST(2, 2, 2, 9, FALSE) = 0.0278

5.37

a) 12, 5, 6, 4;N R n x= = = = 12 5 6 4 5 4

12 6

( ) (4) 0.1136C CP x P

C− −= = =

b) 12, 5, 6, 2;N R n x= = = = 12 5 6 2 5 2

12 6

( ) (2) 0.3788C CP x P

C− −= = =

c) 12, 5, 6, 3;N R n x= = = = 12 5 6 3 5 3

12 6

( ) (3) 0.3788C CP x P

C− −= = =

d) ( )( )2

12 6(6)(5) (6)(5)(12 5)2.50; 0.8919

12 12 12 1μ σ −−= =

−= =




5.38

a) 20, 8, 3, 3;N R n x= = = = 20 8 3 3 8 3

20 3

( ) (3) 0.0491C CP x P

C− −= = =

b) 20, 8, 3, 1;N R n x= = = = 20 8 3 1 8 1

20 3

( ) (1) 0.4632C CP x P

C− −= = =

c) 20, 8, 3, 2;N R n x= = = = 20 8 3 2 8 2

20 3

( ) (2) 0.2947C CP x P

C− −= = =

d) 20, 8, 3, 0;N R n x= = = = 20 8 3 0 8 0

20 3

( ) (0) 0.1930C CP x P

C− −= = =

e) ( )( )2

20 3(3)(8) (3)(8)(20 8)1.2; 0.80

20 20 20 1μ σ −−= =

−= =

f) HYPGEOM.DIST(3, 3, 8, 20, FALSE) = 0.0491 HYPGEOM.DIST(1, 3, 8, 20, FALSE) = 0.4632 HYPGEOM.DIST(2, 3, 8, 20, FALSE) = 0.2947 HYPGEOM.DIST(0, 3, 8, 20, FALSE) = 0.1930

5.39 1 0.10 45 2 0.20 90 3 0.50 225 4 0.20 90 5.40 a) µ = (0)(0.23) + (1)(0.32) + (2)(0.22) + (3)(0.15) + (4)(0.08) = 1.53

b) 23.83 (1.53)σ = − = 1.22

5.41 a) µ = (0)(0.37) + (1)(0.28) + (2)(0.21) + (3)(0.10) + (4)(0.04) = 1.16

b) 2 1.152.66 (1.16)σ == −

5.42 a) 8 AM class: µ = 2.58; 10 AM class: µ = 1.79 b) 8 AM class: 1.39σ = ; 10 AM class: 1.33σ = c) These results were expected 5.43 EMV = ($10,000)(0.03) + ($0)(0.97) = $300. The cost of insurance is $100 greater than EMV, so it should not be purchased.

5-20 Chapter 5


5.44 Portfolio A: EMV = (-$1,500)(0.25) + ($2,500)(0.60) + ($4,000)(0.15) = $1,725 Portfolio B: EMV = (-$4,000)(0.25) + ($3,000)(0.60) + ($6,000)(0.15) = $1,700 Investor should choose the portfolio A. 5.45 Profit = $16 revenue + $5 revenue – cost to purchase shirts The $16 revenue is based on the number of shirts sold at $16, which is the minimum of the supply of shirts (3000) and the eventual demand. Demand Profit (3,000 shirts) Probability 1,000 (1,000)($16) + (2,000)($5) – (3,000)($10) = -$4,000 0.15 2,000 (2,000)($16) + (1,000)($5) – (3,000)($10) = $7,000 0.25 3,000 (3,000)($16) + (0)($5) – (3,000)($10) = $18,000 0.30 4,000 (3,000)($16) + (0)($5) – (3,000)($10) = $18,000 0.30 EMV = (-$4,000)(0.15) + ($7,000)(0.25) + ($18,000)(0.30) + ($18,000)(0.30) = $11,950 5.46 Profit = $14 revenue + $2 revenue – cost to purchase calendars The $14 revenue is based on the number of calendars sold at $14, which is the minimum of the supply of calendars and the eventual demand. Demand Profit (Order 100 calendars) Probability 100 (100)($14) + (0)($2) – (100)($6) = $800 0.25 200 (100)($14) + (0)($2) – (100)($6) = $800 0.40 300 (100)($14) + (0)($2) – (100)($6) = $800 0.35 EMV = ($800)(0.25) + ($800)(0.40) + ($800)(0.35) = $800 Demand Profit (Order 200 calendars) Probability 100 (100)($14) + (100)($2) – (200)($6) = $400 0.25 200 (200)($14) + (0)($2) – (200)($6) = $1,600 0.40 300 (200)($14) + (0)($2) – (200)($6) = $1,600 0.35 EMV = ($400)(0.25) + ($1,600)(0.40) + ($1,600)(0.35) = $1,300 Demand Profit (Order 300 calendars) Probability 100 (100)($14) + (200)($2) – (300)($6) = $0 0.25 200 (200)($14) + (100)($2) – (300)($6) = $1,200 0.40 300 (300)($14) + (0)($2) – (300)($6) = $2,400 0.35 EMV = ($0)(0.25) + ($1,200)(0.40) + ($2,400)(0.35) = $1,320



Bob’s Bookstore should order 300 calendars.

5.47 14 014!(14,14) (0.5) (0.5) 0.000061

(14 14)!14!P = =

−

5.48

a) 6 915!(6,15) (0.45) (0.55) 0.1914

(15 6)!6!P = =

−

b) 9 615!(9,15) (0.45) (0.55) 0.1048

(15 9)!9!P = =

− c)

( )

4 11

5 10

15!(4,15) (0.45) (0.55) 0.0780

(15 4)!4!

15!(5,15) (0.45) (0.55) 0.1404

(15 5)!5!

4 or 5 0.0780 0.1404 0.2184

P

P

P x x

= =−

= =−

= = = + =

d) (15)(0.45) 6.75; (15)(0.45)(0.55) 1.926μ σ= = = =

e)

5-22 Chapter 5


5.49

a) 3 25!(3,5) (0.24) (0.76) 0.0798

(5 3)!3!P = =

−

b) 0 55!(0,5) (0.24) (0.76) 0.2536

(5 0)!0!P = =

−

c)

0 5

1 4

2 3

5!(0,5) (0.24) (0.76) 0.2536

(5 0)!0!

5!(1,5) (0.24) (0.76) 0.4003

(5 1)!1!

5!(2,5) (0.24) (0.76) 0.2529

(5 2)!2!

( 2) 0.2536 0.4003 0.2529 0.9068

P

P

P

P x

= =−

= =−

= =−

≤ = + + =

d)

5.50

a) 4 37!(4,7) (0.37) (0.63) 0.1640

(7 4)!4!P = =

−



b)

0 7

1 6

2 5

7!(0,7) (0.37) (0.63) 0.0394

(7 0)!0!

7!(1,7) (0.37) (0.63) 0.1619

(7 1)!1!

7!(2,7) (0.37) (0.63) 0.2853

(7 2)!2!

( 2) 1.0 0.0394 0.1619 0.2853 0.5134

P

P

P

P x

= =−

= =−

= =−

> = − − − =

c) (7)(0.37) 2.59; (7)(0.37)(0.63) 1.28μ σ= = = = d)

5.51

a) 10 010!(10,10) (0.80) (0.20) 0.1074

(10 10)!10!P = =

−

b)

8 2

9 1 10 0

10!(8,10) (9,10) (10,10) (0.80) (0.20)

(10 8)!8!

10! 10!(0.80) (0.20) (0.80) (0.20) 0.6778

(10 9)!9! (10 10)!10!

P P P+ + = +−

+ =− −

c) 6 4 7 310! 10!(6,10) (7,10) (0.80) (0.20) (0.80) (0.20) 0.2894

(10 6)!6! (10 7)!7!P P+ = + =

− −

5-24 Chapter 5


d)

e) 5 510!(5,10) (0.8) (0.2) 0.0264

(10 5)!5!P = =

−

The probability of 5 out of 10 rounds being below 90 is very low assuming that 80% of the rounds are below 90. Based on this sample, it is not likely that 80% of the rounds are below 90. 5.52

a) 12 012!(12,12) (0.79) (0.21) 0.0591

(12 12)!12!P = =

−

b)

( )

10 2

11 1

12 0

12!(10,12) (0.79) (0.21) 0.2756

(12 10)!10!

12!(11,12) (0.79) (0.21) 0.1885

(12 11)!11!

12!(12,12) (0.79) (0.21) 0.0591

(12 12)!12!

9 0.2756 0.1185 0.0591 0.5232

P

P

P

P x

= =−

= =−

= =−

> = + + =



c)

8 4

9 3

10 2

11 1

12!(8,12) (0.79) (0.21) 0.1460

(12 8)!8!

12!(9,12) (0.79) (0.21) 0.2442

(12 9)!9!

12!(10,12) (0.79) (0.21) 0.2756

(12 10)!10!

12!(11,12) (0.79) (0.21) 0.1885

(12 11)!11!

12!(12,12) (0.79

(12 12)!12!

P

P

P

P

P

= =−

= =−

= =−

= =−

=−

( )

12 0) (0.21) 0.0591

7 1 0.1460 0.2442 0.2756 0.1885 0.0591 0.0866P x

=

≤ = − − − − − =

d) (12)(0.79) 9.48; (12)(0.79)(0.21) 1.411μ σ= = = = e)

5.53

a) 12 113!(12,13) (0.70) (0.30) 0.0540

(13 12)!12!P = =

−

5-26 Chapter 5


b)

11 2

12 1 13 0

13!(11,13) (12,13) (13,13) (0.70) (0.30)

(13 11)!11!

13! 13!(0.70) (0.30) (0.70) (0.30) 0.2025

(13 12)!12! (13 13)!13!

P P P+ + = +−

+ =− −

c)

( )12 1 13 0

( ) 1 (12,13) (13,13)

13! 13!1 (0.70) (0.30) (0.70) (0.30) 0.9363

(13 12)!12! (13 13)!13!

P x P P= − + =

− + = − −

d)

e) 5 813!(5,13) (0.7) (0.3) 0.0142

(13 5)!5!P = =

−

The probability that 5 out of 13 men did not use their shoes on the basketball court is very low assuming that 70% of men do not wear them on courts. Based on this sample, it is not likely that 70% of men do not wear their basketball shoes on the basketball court. 5.54

a) 6 06!(6,6) (0.86) (0.14) 0.4046

(6 6)!6!P = =

−



b)

4 2

5 1

6 0

6!(4,6) (0.86) (0.14) 0.1608

(6 4)!4!

6!(5,6) (0.86) (0.14) 0.3952

(6 5)!5!

6!(6,6) (0.86) (0.14) 0.4046

(6 6)!6!

( 4) 0.1608 0.3952 0.4046 0.9606

P

P

P

P x

= =−

= =−

= =−

≥ = + + =

c)

5 1

6 0

6!(5,6) (0.86) (0.14) 0.3952

(6 5)!5!

6!(6,6) (0.86) (0.14) 0.4046

(6 6)!6!

( 5) 1 0.3952 0.4046 0.2002

P

P

P x

= =−

= =−

< = − − =

d) (6)(0.86) 5.16; (6)(0.86)(0.14) 0.850μ σ= = = = e)

5.55

a) 9 110!(9,10) (0.79) (0.21) 0.2517

(10 9)!9!P = =

−

5-28 Chapter 5


b)

8 2

9 1

10 0

10!(8,10) (0.79) (0.21) 0.3011

(10 8)!8!

10!(9,10) (0.79) (0.21) 0.2517

(10 9)!9!

10!(10,10) (0.79) (0.21) 0.0947

(10 10)!10!

( 7) 0.3011 0.2517 0.0947 0.6475

P

P

P

P x

= =−

= =−

= =−

> = + + =

c)

10 010!

(10,10) (0.79) (0.21) 0.0947(10 10)!10!

( 10) 1 0.0947 0.9053

P

P x

= =−

< = − =

d) (10)(0.79) 7.9; (10)(0.79)(0.21) 1.29μ σ= = = =

e

5.56 a) (12,20) 0.0727P =

b)

(13, 20) (14, 20) (15, 20) (16, 20)

(17, 20) (18, 20) (19, 20) (20, 20) 0.0580

P P P PP P P P

+ + + ++ + + + =



c)

(0, 20) (1, 20) (2, 20) (3, 20) (4, 20)

(5, 20) (6, 20) (7, 20) (8, 20) (9, 20) 0.5914

P P P P PP P P P P

+ + + + ++ + + + + =

d) (20)(0.45) 9; (20)(0.45)(0.55) 2.2μ σ= = = =

e)

5.57

a) 2 79!(2,9) (0.23) (0.77) 0.3056

(9 2)!2!P = =

−

b)

0 9

1 8

2 7

3 6

9!(0,9) (0.23) (0.77) 0.0952

(9 0)!0!

9!(1,9) (0.23) (0.77) 0.2558

(9 1)!1!

9!(2,9) (0.23) (0.77) 0.3056

(9 2)!2!

9!(3,9) (0.23) (0.77) 0.2130

(9 3)!3!

( 4) 0.0952 0.2558 0.3056 0.2130 0.8696

P

P

P

P

P x

= =−

= =−

= =−

= =−

< = + + + =

5-30 Chapter 5


c)

0 9

1 8

2 7

3 6

4 5

9!(0,9) (0.23) (0.77) 0.0952

(9 0)!0!

9!(1,9) (0.23) (0.77) 0.2558

(9 1)!1!

9!(2,9) (0.23) (0.77) 0.3056

(9 2)!2!

9!(3,9) (0.23) (0.77) 0.2130

(9 3)!3!

9!(4,9) (0.23) (0.77) 0.0954

(9 4)!4!

( 5) 1 0

P

P

P

P

P

P x

= =−

= =−

= =−

= =−

= =−

≥ = − .0952 0.2558 0.3056 0.2130 0.0954 0.0350− − − − =

d) (9)(0.23) 2.07; (9)(0.23)(0.77) 1.26μ σ= = = =

e)

5.58

a) 2 68!(2,8) (0.62) (0.38) 0.0324

(8 2)!2!P = =

−



b)

( )

6 2

7 1

8 0

8!(6,8) (0.62) (0.38) 0.2297

(8 6)!6!

8!(7,8) (0.62) (0.38) 0.1071

(8 7)!7!

8!(8,8) (0.62) (0.38) 0.0218

(8 8)!8!

5 0.2297 0.1071 0.0218 0.3586

P

P

P

P x

= =−

= =−

= =−

> = + + =

c)

0 8

1 7

2 6

3 5

8!(0,8) (0.62) (0.38) 0.0004

(8 0)!0!

8!(1,8) (0.62) (0.38) 0.0057

(8 1)!1!

8!(2,8) (0.62) (0.38) 0.0324

(8 2)!2!

8!(3,8) (0.62) (0.38) 0.1058

(8 3)!3!

( 3) 0.0004 0.0057 0.0324 0.1058 0.1443

P

P

P

P

P x

= =−

= =−

= =−

= =−

≤ = + + + =

d) (8)(0.62) 4.96; (8)(0.62)(0.38) 1.37μ σ= = = = e)

5-32 Chapter 5


5.59

a) ( )36 = 10 0.9

400λ =

;

( )( )1 0.90.9 2.71828( ) (1) 0.3659

1!P x P

−

= = =

b) ( )36 = 20 1.8

400λ =

;

( )( )1 1.81.8 2.71828( ) (1) 0.2975

1!P x P

−

= = =

c) ( )36 = 50 4.5

400λ =

;

( )( )4 4.54.5 2.71828( ) (4) 0.1898

4!P x P

−

= = =

d) ( )36 = 110 9.9

400λ =

;

( )( )12 9.99.9 2.71828( ) (12) 0.0928

12!P x P

−

= = =

5.60

a) ( )( )6 11.011.0 2.71828

(6) 0.04116!

P−

= =

b)

( )( )3 5.55.5 2.71828(3) 0.1133

3!P

−

= =

c)

11.0

( 7) 1.0 ( 7) 1.0 0.0786 0.9214P x P xλ =

≥ = − < = − =

d)

11

( 10) 0.3405P xλ =

< =

5.61

a) ( )( )8 1313 2.71828

(8) 0.04578!

P−

= =

b) (0) (1) (2) (3) (4) (5) (6) (7) (8) (9) 0.1658P P P P P P P P P P+ + + + + + + + + =

c)

()

( ) 1 (0) (1) (2) (3) (4) (5) (6) (7) (8) (9)

(10) (11) (12) (13) (14) (15) 0.2364

P x P P P P P P P P P P

P P P P P P

= − + + + + + + + + + +

+ + + + + + =

d) ( )( )4 6.56.5 2.71828

(0.5)(13) 6.5; ( ) (4) 0.11184!

P x Pλ−

= = = = =



5.62

a) ( )( )2 44 2.71828

(0.25)(16) 4; ( ) (2) 0.14652!

P x Pλ−

= = = = =

b) ( )( )4 44 2.71828

(0.25)(16) 4; ( ) (4) 0.19544!

P x Pλ−

= = = = =

c) ( )( )6 88 2.71828

(0.5)(16) 8; ( ) (6) 0.12216!

P x Pλ−

= = = = =

d) ( )( )6 1212 2.71828

(0.75)(16) 12; ( ) (6) 0.02556!

P x Pλ−

= = = = =

5.63

a) ( )( )0 3.873.87 2.71828

(0) 0.02090!

P−

= =

b)

( )( )

( )( )

( )( )

0 3.87

1 3.87

2 3.87

3.87 2.71828(0) 0.0209

0!

3.87 2.71828(1) 0.0807

1!

3.87 2.71828(2) 0.1562

2!0.0209 0.0807 0.1562 0.2578( 3)

P

P

P

P x

−

−

−

= =

= =

= =

+ + =< =

c) ( )( )0 1.9351.935 2.71828

(0) 0.14440!

P−

= =

d)

( )( )

( )( )

( )( )

0 1.935

1 1.935

2 1.935

1.935 2.71828(0) 0.1444

0!

1.935 2.71828(1) 0.2795

1!

1.935 2.71828(2) 0.2704

2!0.1444 0.2795 0.2704 0.6943( 3)

P

P

P

P x

−

−

−

= =

= =

= =

+ + =< =

5.64

a) ( )( )1 6.56.5 2.71828

(1) 0.00981!

P−

= =

5-34 Chapter 5


b)

( )( )

( )( )

( )( )

0 6.5

1 6.5

2 6.5

6.5 2.71828(0) 0.0015

0!

6.5 2.71828(1) 0.0098

1!

6.5 2.71828(2) 0.0318

2!( 3) 0.0015 0.0098 0.0318 0.0431

P

P

P

P x

−

−

−

= =

= =

= =

< = + + =

c)

( )( )

( )( )

( )( )

( )( )

( )( )

0 6.5

1 6.5

2 6.5

3 6.5

4 6.5

( 4) 1.0 0.0015 0.0098 0.0318 0.0688 0.1118 0.7763

6.5 2.71828(0) 0.0015

0!

6.5 2.71828(1) 0.0098

1!

6.5 2.71828(2) 0.0318

2!

6.5 2.71828(3) 0.0688

3!

6.5 2.71828(4) 0.1118

4!P x

P

P

P

P

P

−

−

−

−

−

> = − − − − − =

= =

= =

= =

= =

= =

d) 6.5 2.55σ = =

5.65

a) ( )( )1 5.35.3 2.71828

(3) 0.12393!

P−

= =



b)

( )( )

( )( )

( )( )

( )( )

0 5.3

1 5.3

2 5.3

3 5.3

5.3

5.3

5.3

5.3

2.71828(0) 0.0050

0!

2.71828(1) 0.0265

1!

2.71828(2) 0.0701

2!

2.71828(3) 0.1239

3!( 4) 0.0050 0.0265 0.0701 0.1239 0.2255

P

P

P

P

P x

−

−

−

−

= =

= =

= =

= =

< = + + + =

c)

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

0 5.3

1 5.3

2 5.3

3 5.3

4 5.3

5 5.3

5.3

5.3

5.3

5.3

5.3

5.3

2.71828(0) 0.0050

0!

2.71828(1) 0.0265

1!

2.71828(2) 0.0701

2!

2.71828(3) 0.1239

3!

2.71828(4) 0.1641

4!

2.71828(5) 0.1740

5!( 6) 1 0.0050 0.0265 0.0

P

P

P

P

P

P

P x

−

−

−

−

−

−

= =

= =

= =

= =

= =

= =

> = − − − 701 0.1239 0.1641 0.1740 0.4364− − − =

d) ( )( )1 10.610.6 2.71828

(6) 0.04916!

P−

= =

5.66

a) 1 2324!(1, 24) (0.025) (0.975) 0.3352

(24 1)!1!P = =

−

b) 4 2024!(4, 24) (0.025) (0.975) 0.0025

(24 4)!4!P = =

−

5-36 Chapter 5


c)

( )( )( )( )1 0.60

24 0.025 0.60

0.60 2.71828(1) 0.3293

1!

np

P−

= =

= =

d)

( )( )( )( )4 0.60

24 0.025 0.60

0.60 2.71828(4) 0.0030

4!

np

P−

= =

= =

5.67

a) 0 2020!(0, 20) (0.034) (0.966) 0.5007

(20 0)!0!P = =

−

b)

( )

0 20

1 19

20!(0, 20) (0.034) (0.966) 0.5007

(20 0)!0!

20!(1, 20) (0.034) (0.966) 0.3524

(20 1)!1!

2 0.5007 0.3524 0.8531

P

P

P x

= =−

= =−

< = + =

c)

( )( )( )( )0 0.68

20 0.034 0.68

0.68 2.71828(0) 0.5066

0!

np

P−

= =

= =

d)

( )( )( )( )

( )( )

( )

0 0.68

1 0.68

20 0.034 0.68

0.68

0.68

2 0.5066 0.3445 0.8511

2.71828(0) 0.5066

0!

2.71828(1) 0.3445

1!

np

P x

P

P

−

−

= =

< = + =

= =

= =

5.68

a) 3 2124!( ) (3,24) (0.035) (0.965) 0.0411

(24 3)!3!P x P= = =

−

b) 3 (24)(0.035)((24)(0.035)) 2.71828

( ) (3) 0.04263!

P x P−

= = =



c) The probabilities are very close. 5.69

a) 18 4 3 0 4 0

18 3

( ) (0) 0.4461C CP x P

C− −= = =

b) ( ) 18 4 3 0 4 0 18 4 3 1 4 1

18 3 18 3

( ) 1 (0) (1) 1 0.1078C C C CP x P P

C C− − − −

= − + = − + =

c) ( )( )2

18 3(3)(4) (3)(4)(18 4)0.667; 0.6763

18 18 18 1μ σ −−= =

−= =

d) HYPGEOM.DIST(0, 3, 4, 18, FALSE) = 0.4461 1 – (HYPGEOM.DIST(0, 3, 4, 18, FALSE) + HYPGEOM.DIST(1, 3, 4, 18, FALSE) ) = 1 - 0.8922 = 0.1078

5.70

a) 19 14 8 8 14 8

19 8

( ) (8) 0.0397C CP x P

C− −= = =

b) 19 14 8 6 14 6 19 14 8 7 14 7 19 14 8 8 14 8

19 8 19 8 19 8

( ) (6) (7) (8) 0.6640C C C C C CP x P P P

C C C− − − − − −= + + = + + =

c) There are only five returns that are not using the short form. If eight returns are selected, at least three returns must be short forms.

d) ( )( )2

19 8(8)(14) (8)(14)(19 14)5.895; 0.9737

19 19 19 1μ σ −−= =

−= =

e) HYPGEOM.DIST(8, 8, 14, 19, FALSE) = 0.0397 HYPGEOM.DIST(6, 8, 14, 19, FALSE) + HYPGEOM.DIST(7, 8, 14, 19, FALSE) + HYPGEOM.DIST(8, 8, 14, 19, FALSE) = 0.3973 + 0.2270 + 0.0397 = 0.6640

5.71

a) 16, 10, 6, 2;N R n x= = = = 16 10 6 2 10 2

16 6

( ) (2) 0.0843C CP x P

C− −= = =

b) 16, 10, 6, 4;N R n x= = = = 16 10 6 4 10 4

16 6

( ) (4) 0.3934C CP x P

C− −= = =

c) 16, 10, 6, 6;N R n x= = = = 16 10 6 6 10 6

16 6

( ) (6) 0.0262C CP x P

C− −= = =

d) ( )( )2

16 6(6)(10) (6)(10)(16 10)3.75; 0.968

16 16 16 1μ σ −−= =

−= =

5-38 Chapter 5



5.72

a) 22, 8, 10, 4;N R n x= = = = 22 8 10 4 8 4

22 10

( ) (4) 0.3251C CP x P

C− −= = =

b) 22, 8, 10, 6;N R n x= = = = 22 8 10 6 8 6

22 10

( ) (6) 0.0433C CP x P

C− −= = =

c) The proportion of cars (0.636) is much higher than the proportion of trucks (0.364) on the lot. Four trucks out of 10 vehicles have a higher probability than 4 cars out of 10 vehicles because the 0.40 proportion is very close to the proportion of trucks on the lot.

d) ( )( )2

22 10(10)(8) (10)(8)(22 8)3.64; 1.150

22 22 22 1μ σ −−= =

−= =

e) HYPGEOM.DIST(4, 10, 8, 22, FALSE) = 0.3251 HYPGEOM.DIST(6, 10, 8, 22, FALSE) = 0.0433

5.73

a) 53, 22, 3, 0;N R n x= = = = 53 22 3 0 22 0

53 3

( ) (0) 0.1919C CP x P

C− −= = =

b) 53, 22, 3, 3;N R n x= = = = 53 22 3 3 22 3

53 3

( ) (3) 0.0657C CP x P

C− −= = =

c) 53, 22, 3, 2;N R n x= = = = 53 22 3 2 22 2

53 3

( ) (2) 0.3057C CP x P

C− −= = =

d) ( )( )2

53 3(3)(22) (3)(22)(53 22)1.245; 0.8368

53 53 53 1μ σ −−= =

−= =


5.74

a) 52, 12, 5, 0;N R n x= = = = 52 12 5 0 12 0

52 5

( ) (0) 0.2532C CP x P

C− −= = =

b) 52, 12, 5, 1;N R n x= = = = 52 12 5 1 12 1

52 5

( ) (1) 0.4220C CP x P

C− −= = =

c) 52, 12, 5, 2;N R n x= = = = 52 12 5 2 12 2

52 5

( ) (2) 0.2509C CP x P

C− −= = =



d) 52, 12, 5, 4;N R n x= = = = 52 12 5 4 12 4

52 5

( ) (4) 0.0076C CP x P

C− −= = =

e) ( )( )2

52 5(5)(12) (5)(12)(52 12)1.154; 0.9044

52 52 52 1μ σ −−= =

−= =

f) HYPGEOM.DIST(0, 5, 12, 52, FALSE) = 0.2532 HYPGEOM.DIST(1, 5, 12, 52, FALSE) = 0.4220 HYPGEOM.DIST(2, 5, 12, 52, FALSE) = 0.2509 HYPGEOM.DIST(4, 5, 12, 52, FALSE) = 0.0076

CHAPTER 5 Discrete Probability Distributions€¦ · CHAPTER 5 Discrete Probability Distributions...

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Transcript of CHAPTER 5 Discrete Probability Distributions€¦ · CHAPTER 5 Discrete Probability Distributions...