Chapter 16 – Probability Models - Weeblycschneidermath.weebly.com/uploads/2/5/4/6/25460506/chapter_16_hw... · Chapter 16 Probability Models 165 Copyright © 2012 Pearson Education,

Chapter 16 Probability Models 163

Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 16 – Probability Models

1. Expected value.

a) ( )E Yµ = = 10(0.3) + 20(0.5) + 30(0.2) = 19

b) ( )E Yµ = = 2(0.3) + 4(0.4) + 6(0.2) + 8(0.1) = 4.2

2. Expected value.

a) ( )E Yµ = = 0(0.2) + 1(0.4) + 2(0.4) = 1.2

b) ( )E Yµ = = 100(0.1) + 200(0.2) + 300(0.5) + 400(0.2) = 280

3. Variation 1.

a)

b)

4. Variation 2.

a)

b)

5. Repairs.

a) (Number of Repair Calls) 0(0.1) 1(0.3) 2(0.4) 3(0.2) 1.7 callsEµ = = + + + =

b)

6. Red lights.

a)

b)

2 2 2 2( ) (10 19) (0.3) (20 19) (0.5) (30 19) (0.2) 49

( ) ( ) 49 7

Var Y

SD Y Var Y

σ

σ

= = − + − + − =

= = = =2 2 2 2 2( ) (2 4.2) (0.3) (4 4.2) (0.4) (6 4.2) (0.2) (8 4.2) (0.1) 3.56

( ) ( ) 3.56 1.89

Var Y

SD Y Var Y

σ

σ

= = − + − + − + − =

= = = ≈

2 2 2 2( ) (0 1.2) (0.2) (1 1.2) (0.4) (2 1.2) (0.4) 0.56

( ) ( ) 0.56 0.75

Var Y

SD Y Var Y

σ

σ

= = − + − + − =

= = = ≈2

2 2 2 2

( )

(100 280) (0.1) (200 280) (0.2) (300 280) (0.5) (400 280) (0.2) 7600

( ) ( ) 7600 87.18

Var Y

SD Y Var Y

σ

σ

=

= − + − + − + − =

= = = ≈

2

2 2 2 2

(Calls)

(0 1.7) (0.1) (1 1.7) (0.3) (2 1.7) (0.4) (3 1.7) (0.2) 0.81

(Calls) (Calls) 0.81 0.9 calls

Var

SD Var

σ

σ

=

= − + − + − + − =

= = = =

(Red lights)

0(0.05) 1(0.25) 2(0.35) 3(0.15) 4(0.15) 5(0.05) 2.25 red lights

Eµ == + + + + + =

2 2 2 2

2 2 2

(Red lights) (0 2.25) (0.05) (1 2.25) (0.25) (2 2.25) (0.35)

(3 2.25) (0.15) (4 2.25) (0.15) (5 2.25) (0.05) 1.5875

(Red lights) (Red lights) 1.5875 1.26

Var

SD Var

σ

σ

= = − + − + −

+ − + − + − =

= = = ≈ red lights

164 Part IV Randomness and Probability


7. Defects.

a) The percentage of cars with no defects is 61%.

b) (Defects) 0(0.61) 1(0.21) 2(0.11) 3(0.07) 0.64 defectsEµ = = + + + =

c)

8. Insurance.

a)

b) (Profit) 100(0.9975) 9900(0.0005) 2900(0.002) $89Eµ = = − − =

c)

9. Bernoulli.

a) These are not Bernoulli trials. The possible outcomes are 1, 2, 3, 4, 5, and 6. There are more than two possible outcomes.

b) These may be considered Bernoulli trials. There are only two possible outcomes, Type A and not Type A. Assuming the 120 donors are representative of the population, the probability of having Type A blood is 43%. The trials are not independent, because the population is finite, but the 120 donors represent less than 10% of all possible donors.

c) These are not Bernoulli trials. The probability of getting a heart changes as cards are dealt without replacement. The trials are not independent.

d) These are not Bernoulli trials. We are sampling without replacement, so the trials are not independent. Samples without replacement may be considered Bernoulli trials if the sample size is less than 10% of the population, but 500 is more than 10% of 3000.

e) These may be considered Bernoulli trials. There are only two possible outcomes, sealed properly and not sealed properly. The probability that a package is unsealed is constant, at about 10%, as long as the packages checked are a representative sample of all packages. Finally, the trials are not independent, since the total number of packages is finite, but the 24 packages checked probably represent less than 10% of the packages.

Profit $100 – $9900 – $2900P(Profit) 0.9975 0.0005 0.002

2 2 2

2 2

(Defects) (0 0.64) (0.61) (1 0.64) (0.21)

(2 0.64) (0.11) (3 0.64) (0.07) 0.8704

(Defects) (Defects) 0.8704 0.93 defects

Var

SD Var

σ

σ

= = − + −

+ − + − ≈

= = ≈ ≈

X = # of defects 0 1 2 3 P(X = x) 0.61 0.21 0.11 0.07

2 2 2

2

(Profit) (100 89) (0.9975) ( 9900 89) (0.0005)

( 2900 89) (0.002) 67,879

(Profit) (Profit) 67,879 $260.54

Var

SD Var

σ

σ

= = − + − −

+ − − =

= = ≈ ≈



10. Bernoulli 2.

a) These may be considered Bernoulli trials. There are only two possible outcomes, getting a 6 and not getting a 6. The probability of getting a 6 is constant at 1/6. The rolls are independent of one another, since the outcome of one die roll doesn’t affect the other rolls.

b) These are not Bernoulli trials. There are more than two possible outcomes for eye color.

c) These can be considered Bernoulli trials. There are only two possible outcomes, properly attached buttons and improperly attached buttons. As long as the button problem occurs randomly, the probability of a doll having improperly attached buttons is constant at about 3%. The trails are not independent, since the total number of dolls is finite, but 37 dolls is probably less than 10% of all dolls.

d) These are not Bernoulli trials. The trials are not independent, since the probability of picking a council member with a particular political affiliation changes depending on who has already been picked. The 10% condition is not met, since the sample of size 4 is more than 10% of the population of 19 people.

e) These may be considered Bernoulli trials. There are only two possible outcomes, cheating and not cheating. Assuming that cheating patterns in this school are similar to the patterns in the nation, the probability that a student has cheated is constant, at 74%. The trials are not independent, since the population of all students is finite, but 481 is less than 10% of all students.

11. Hoops.

These may be considered Bernoulli trials. There are only two possible outcomes, making the foul shot and missing the foul shot. The probability of making the foul shot is given, p = 0.70. The shots are assumed to be independent.

Let X = the number of foul shots made in n = 5 shots.

a) Use (5,0.70)Binom .

3 25 3

(exactly 3 foul shots made) ( 3)

(0.70) (0.30)0.31

P P X

C

= =

=≈

b) Use (5,0.70)Binom .

4 15 4

(exactly 4 foul shots made) ( 4)

(0.70) (0.30)0.36

P P X

C

= =

=≈



c) Use (5,0.70)Binom .

5 05 5

(all 5 foul shots made) ( 5)

(0.70) (0.30)0.17

P P X

C

= =

=≈

d) Use (5,0.70)Binom . = = + = + =

= + +≈

3 2 4 1 5 05 3 5 4 5 5

(at least 3 shots made) ( 3) ( 4) ( 5)

(0.70) (0.30) (0.70) (0.30) (0.70) (0.30)0.84

P P X P X P X

C C C

12. Arrows.

These may be considered Bernoulli trials. There are only two possible outcomes, hitting the bull’s-eye and not hitting the bull’s-eye. The probability of hitting the bull’s-eye is given, p = 0.80. The shots are assumed to be independent.

Let X = the number of bull’s-eyes in n = 6 shots.


4 26 4

(exactly four hits) ( 4)

(0.80) (0.20)0.246

P P X

C

= =

=≈


5 16 5

(exactly five hits) ( 5)

(0.80) (0.20)0.393

P P X

C

= =

=≈


6 06 6

(all six hits) ( 6)

(0.80) (0.20)0.262

P P X

C

= =

=≈

d) Use (6,0.80)Binom .

4 2 5 1 6 06 4 6 5 6 6

(at least four hits) ( 4) ( 5) ( 6)

(0.80) (0.20) (0.80) (0.20) (0.80) (0.20)0.901

P P X P X P X

C C C

= = + = + =

= + +≈

13. More hoops.

a) In a previous exercise, we determined that the shots could be considered Bernoulli trials. Since the player is shooting 5 shots, use (5,0.70)Binom .

Let X = the number of shots made in n = 5 foul shots.

( ) 5(0.70) 3.5E X npµ = = = = foul shots made.



b) ( ) 5(0.70)(0.30) 1.025SD X npqσ = = = ≈ foul shots made.

14. More arrows.

a) In a previous exercise, we determined that the shots could be considered Bernoulli trials. Since the archer is shooting 6 arrows, use (6,0.80)Binom .

Let X = the number of bull’s-eyes in n = 6 shots.

( ) 6(0.80) 4.8E X npµ = = = = bull’s-eyes.

b) ( ) 6(0.80)(0.20) 0.98SD X npqσ = = = ≈ bull’s-eyes.

15. Chips.

These may be considered Bernoulli trials. There are only two possible outcomes, working properly, and failing to work properly. The probability of failing to work properly is given, p = 0.02. The defects in the chips are assumed to be independent.

Let X = the number of chips that fail in n = 10 chips.

a) Use (10,0.02)Binom .

0 1010 0

(no failures) ( 0)

(0.02) (0.98)0.817

P P X

C

= =

=≈

b) Use (10,0.02)Binom .

1 910 1

(exactly 1 failure) ( 1)

(0.02) (0.98)0.167

P P X

C

= =

=≈

c) Use (10,0.02)Binom .

2 810 2

(exactly 2 failures) ( 2)

(0.02) (0.98)0.015

P P X

C

= =

=≈

d) Use (10,0.02)Binom .

3 710 3

(exactly 3 failures) ( 3)

(0.02) (0.98)0.0008

P P X

C

= =

=≈

e) Use (10,0.02)Binom . (no more than 3 failures) ( 0) ( 1) ( 2) ( 3)

0.817 0.167 0.015 0.00080.9998

P P X P X P X P X= = + = + = + == + + +≈



16. Colorblindness.

These may be considered Bernoulli trials. There are only two possible outcomes, being colorblind and not being colorblind. The probability of being colorblind is given, p = 0.08. Colorblindness is assumed to be independent. For example, the volunteers can’t be related, as colorblindness may be hereditary.

Let X = the number of colorblind men in n = 12 men.

a) Use (12,0.08)Binom .

0 1212 0

(no colorblind) ( 0)

(0.08) (0.92)0.368

P P X

C

= =

=≈

b) Use (12,0.08)Binom .

1 1112 1

(exactly 1 colorblind) ( 1)

(0.08) (0.92)0.384

P P X

C

= =

=≈

c) Use (12,0.08)Binom .

2 1012 2


(0.08) (0.92)0.183

P P X

C

= =

=≈

d) Use (12,0.08)Binom .

3 912 3


(0.08) (0.92)0.053

P P X

C

= =

=≈

b) Use (12,0.08)Binom . (at most 3 colorblind) ( 0) ( 1) ( 2) ( 3)

0.368 0.384 0.183 0.0530.988

P P X P X P X P X= = + = + = + == + + +≈

17. Many chips.

In a previous exercise, we determined that the production of the chips could be considered Bernoulli trials. Since the company is producing 1000 chips per day, use (1000,0.02)Binom .

Let X = the number of failed chips in n = 1000 chips.

( ) 1000(0.02) 20E X npµ = = = = failed chips.

( ) 1000(0.02)(0.98) 4.43SD X npqσ = = = ≈ failed chips.



18. Many men.

In a previous exercise, we determined that the selection of men could be considered Bernoulli trials. Since the researcher is testing 450 men, use

(450,0.08)Binom .


( ) 450(0.08) 36E X npµ = = = = colorblind men.

( ) 450(0.08)(0.92) 5.75SD X npqσ = = = ≈ colorblind men.

19. Pick a card, any card.

a)

b) 26 13 12 1

(amount won) $0 $5 $10 $30 $4.1352 52 52 52

Eµ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + + ≈⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

c) Answers may vary. In the long run, the expected payoff of this game is $4.13 per play. Any amount less than $4.13 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less.

20. You bet!

a)

b) µ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + ≈⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠1 5 25

(amount won) $100 $50 $0 $23.616 36 36

E

c) Answers may vary. In the long run, the expected payoff of this game is $23.61 per play. Any amount less than $23.61 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less.

Amount Won $0 $5 $10 $30

P(amount won) 2652

1352

1252

152

Amount Won $100 $50 $0

P(amount won) 16

1 5 56 6 36⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5 5 256 6 36⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



21. Pick another card.

Answers may vary slightly (due to rounding of the mean)

σ

σ

⎛ ⎞ ⎛ ⎞= = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + − ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= = = ≈

2 2 2

2 2

26 13(Won) (0 4.13) (5 4.13)

52 5212 1

(10 4.13) (30 4.13) 29.539652 52

(Won) (Won) 29.5396 $5.44

Var

SD Var

22. The die.

Answers may vary slightly (due to rounding of the mean)

2 2

2 2

1(Won) (100 23.61)

65 25

(50 23.61) (0 23.61) 1456.404336 36

(Won) (Won) 1456.4043 $38.16

Var

SD Var

σ

σ

⎛ ⎞= = − ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞+ − + − ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= = ≈ ≈

23. Cancelled flights.

a) ( ( )( . ) ( . ) $gain) 150 0 20 100 0 80 50µ = = − + =E

b)

24. Day trading.

a) ( ( . ) ( )( . ) ( . ) $gain) 800 0 20 200 0 30 0 0 50 100µ = = + − + =E

b)

25. Batteries.

a)

b) 6 42 42

(number good) 0 1 2 1.4 batteries90 90 90

Eµ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( ) ( ) ( . ) ( ) ( . ) ,

( ) ( ) , $

2 2 2gain 150 50 0 20 100 50 0 80 10 000

gain gain 10 000 100

σ

σ

= = − − + − =

= = ≈ =

Var

SD Var

( ) ( ) ( . ) ( ) ( . )

( ) ( . ) ,

( ) ( ) , $ .

2 2 2

2

gain 800 100 0 20 200 100 0 30

0 100 0 50 130 000

gain gain 130 000 360 56

σ

σ

= = − + − −

+ − =

= = ≈ ≈

Var

SD Var

Number good 0 1 2

P(number good) 3 2 6

10 9 90⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 3 7 7 3 42

10 9 10 9 90⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 7 6 42

10 9 90⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



c)

26. Kittens.

a)

b) 6 24 12

(number of males) 0 1 2 1.14 males42 42 42

Eµ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + ≈⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

c) Answers may vary slightly (due to rounding of the mean)

( ) ( ) ( )2 2 22 6 24 12(number of males) 0 1.14 1 1.14 2 1.14

42 42 420.4082

(number of males) (number of males) 0.4082 0.64 males

Var

SD Var

σ

σ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠≈

= = ≈ ≈

27. On time.

These departures cannot be considered Bernoulli trials. Departures from the same airport during a 2-hour period may not be independent. They all might be affected by weather and delays.

28. Lost luggage.

The fate of these bags cannot be considered Bernoulli trials. What happens to 22 pieces of luggage, all checked on the same flight probably won’t be independent.

29. Still more hoops.

In a previous exercise, we determined that the shots could be considered Bernoulli trials.

Let X = the number of shots made.


8 08 8

(8 made shots in a row) ( 8)

(0.70) (0.30)0.058

P P X

C

= =

=≈

( ) ( ) ( )2 2 22 6 42 42(number good) 0 1.4 1 1.4 2 1.4

90 90 900.3733

(number good) (number good) 0.3733 0.61 batteries.

Var

SD Var

σ

σ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠≈

= = ≈ ≈

Number of males 0 1 2 P(number of males)

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠3 2 67 6 42

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠4 3 3 4 247 6 7 6 42

⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠4 3 127 6 42




6 28 6

(exactly 6 of 8) ( 6)

(0.70) (0.30)0.296

P P X

C

= =

=≈


0 8 1 78 0 8 1

6 28 6

(no more than 6 of 8) ( 0) ( 1) ... ( 6)

(0.70) (0.30) (0.70) (0.30) ...

... (0.70) (0.30)0.745

P P X P X P X

C C

C

= = + = + + =

= + +

+≈

d) Use (20,0.70)Binom .

10 1020 10

(exactly 10 of 20) ( 10)

(0.70) (0.30)0.031

P P X

C

= =

=≈

e) Use (20,0.70)Binom .

10 10 11 920 10 20 11

20 020 20

(at least 10 of 20) ( 10) ( 11) ... ( 20)

(0.70) (0.30) (0.70) (0.30) ...

... (0.70) (0.30)0.983

P P X P X P X

C C

C

= = + = + + =

= + +

+≈

30. Still more arrows.

In a previous exercise, we determined that the archer’s shots could be considered Bernoulli trials.

Let X = the number of bull’s-eyes hit.

a) Use (10,0.80)Binom .

10 010 10

(no misses out of 10 shots) (all hits out of 10 shots)

( 10)

(0.80) (0.20)0.107

P P

P X

C

== =

=≈

b) Use (10,0.80)Binom .

8 210 8

(exactly 8 out of 10 shots) ( 8)

(0.80) (0.20)0.302

P P X

C

= =

=≈



c) Use (10,0.80)Binom

0 10 1 910 0 10 1

8 210 8

(no more than 8 hits) ( 8)( 0) ( 1) ( 2) ( 8)

(0.80) (0.20) (0.80) (0.20) ...

... (0.80) (0.20)0.624

P P X

P X P X P X P X

C C

C

= ≤= = + = + = + + =

= + +

+≈

…

d) Use (20,0.80)Binom = =

=≈

12 820 12

(no more than 12 hits) ( 12)

(0.80) (0.20)0.022

P P X

C

e) Use (20,0.80)Binom

12 8 13 720 12 20 13

20 020 20

(at least 12 hits) ( 12)( 12) ( 13) ( 14) ( 20)

(0.80) (0.20) (0.80) (0.20) ...

... (0.80) (0.20)0.99

P P X

P X P X P X P X

C C

C

= ≥= = + = + = + + =

= + +

+≈

…

31. Tennis, anyone?

The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0.70. Finally, we are assuming that each serve is independent of the others. Since she is serving 6 times, use Binom(6,0.70).

Let X = the number of successful serves in n = 6 first serves.

a)

b)

c)

6 06 6

(all six serves in ) ( 6)

(0.70) (0.30)0.118

P P X

C

= =

=≈

4 26 4

(exactly four serves in) ( 4)

(0.70) (0.30)0.324

P P X

C

= =

=≈

4 2 5 1 6 06 4 6 5 6 6

(at least four serves in) ( 4) ( 5) ( 6)

(0.70) (0.30) (0.70) (0.30) (0.70) (0.30)0.744

P P X P X P X

C C C

= = + = + =

= + +≈



d)

32. Frogs.

The frog examinations can be considered Bernoulli trials. There are only two possible outcomes, having the trait and not having the trait. If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 0.125. The trials are not independent since the population of frogs is finite, but 12 frogs is fewer than 10% of all frogs. Since the biologist is collecting 12 frogs, use

Binom(12,0.125).

Let X = the number of frogs with the trait, from n = 12 frogs.

a)

b)

c)

d)

0 6 1 5 2 46 0 6 1 6 2

3 3 4 26 3 6 4

(no more than 4 in) ( 0) ( 1) ( 2) ( 3) ( 4)

(0.70) (0.30) (0.70) (0.30) (0.70) (0.30)

(0.70) (0.30) (0.70) (0.30)0.580

P P X P X P X P X P X

C C C

C C

= = + = + = + = + =

= + +

+ +≈

0 1212 0

(no frogs have the trait) ( 0)

(0.125) (0.875)0.201

P P X

C

= =

=≈

2 10 3 912 2 12 3

12 012 12

(at least two frogs) ( 2)( 2) ( 3) ( 12)

(0.125) (0.875) (0.125) (0.875) ...

... (0.125) (0.875)0.453

P P X

P X P X P X

C C

C

= ≥= = + = + + =

= + +

+≈

…

3 9 4 812 3 12 4

(three or four frogs have trait) ( 3) ( 4)

(0.125) (0.875) (0.125) (0.875)0.171

P P X P X

C C

= = + =

= +≈

0 12 1 1112 0 12 1

4 812 4

(no more than four) ( 4) ( 0) ( 1) ( 4)

(0.125) (0.875) (0.125) (0.875) ...

... (0.125) (0.875)0.989

P P X P X P X P X

C C

C

= ≤ = = + = + + =

= + +

+≈

…



33. And more tennis.

The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0.70. Finally, we are assuming that each serve is independent of the others. Since she is serving 80 times, use Binom(80,0.70).


a) ( ) 80(0.70) 56E X np= = = first serves in.

( ) 80(0.70)(0.30) 4.10SD X npq= = ≈ first serves in.

b) Since np = 56 and nq = 24 are both greater than 10, Binom(80,0.70) may be approximated by the Normal model, N(56, 4.10).

c) Using Binom(80, 0.70):

65 15 80 080 65 80 80

(at least 65 first serves) ( 65)

( 65) ( 66) ( 80)

(0.70) (0.30) (0.70) (0.30)0.0161

P P X

P X P X P X

C C

= ≥= = + = + + =

= + +≈

……

According to the Binomial model, the probability that she makes at least 65 first serves out of 80 is approximately 0.0161.

Using N(56, 4.10):

( 65) ( 2.195) 0.0141P X P z≥ ≈ > ≈

According to the Normal model, the probability that she makes at least 65 first serves out of 80 is approximately 0.0141.

34. More arrows.

These may be considered Bernoulli trials. There are only two possible outcomes, hitting the bull’s-eye and not hitting the bull’s-eye. The probability of hitting the bull’s-eye is given, p = 0.80. The shots are assumed to be independent. Since she will be shooting 200 arrows, use Binom(200, 0.80).

Let Y = the number of bull’s-eyes in n = 200 shots.

65 564.10

2.195

xz

z

z

µσ−

=

−=

≈



a) ( ) 200(0.80) 160E Y np= = = bull’s-eyes.

( ) 200(0.80)(0.20) 5.66SD Y npq= = ≈ bull’s-eyes.

b) Since np = 160 and nq = 40 are both greater than 10, (200,0.80)Binom may be approximated by the Normal model, N(160, 5.66).


151 49 200 0

200 151 200 200

(over 150 hits) ( 150)

( 151) ( 152) ( 200)

(0.80) (0.20) (0.80) (0.70)0.951

P P Y

P Y P Y P Y

C C

= >= = + = + + =

= + +≈

……

According to the Binomial model, the probability that she makes over 150 bull’s-eyes out of 200 is approximately 0.951.

Using N(160, 5.66):

( 150) ( 1.590) 0.944P Y P z> ≈ > − ≈ .

According to the Normal model, the probability that she hits over 150 bull’s-eyes out of 200 is approximately 0.944.

35. Seatbelts II.

These stops may be considered Bernoulli trials. There are only two possible outcomes, belted or not belted. Police estimate that the probability that a driver is buckled is 80%. (The probability of not being buckled is therefore 20%.) Provided the drivers stopped are representative of all drivers, we can consider the probability constant. The trials are not independent, since the population of drivers is finite, but the police will not stop more than 10% of all drivers.

a) Let X = the number of drivers wearing their seatbelts in 120 cars. Use Binom(120, 0.8).

( ) 120(0.8) 96E X np= = = drivers.

( ) 120(0.8)(0.2) 4.38SD X npq= = ≈ drivers.

151 1605.66

1.590

yz

z

z

µσ−

=

−=

≈ −



b) Let Y = the number of drivers NOT wearing their seat belts in 120 cars.

Using Binom(120, 0.2):

20 100 120 0

120 20 120 120

(at least 20) ( 20)

( 20) ( 120)

(0.2) (0.8) (0.2) (0.8)0.848

P P Y

P Y P Y

C C

= ≥= = + + =

= + +≈

……

According to the Binomial model, the probability that at least 20 out of 120 drivers are not wearing their seatbelts is approximately 0.848.

Using N(24, 4.38):

( ) 120(0.2) 24E Y np= = = drivers.

( ) 120(0.2)(0.8) 4.38SD Y npq= = ≈ drivers.

Since np = 24 and nq = 96 are both greater than 10, (120,0.2)Binom may be approximated by the Normal model, N(24, 4.38).

( 120) ( 0.913) 0.8194P Y P z≥ ≈ > − ≈

According to the Normal model, the probability that at least 20 out of 120 drivers stopped are not wearing their seatbelts is approximately 0.8194.

36. Vitamin D.

The selection of these children may be considered Bernoulli trials. There are only two possible outcomes, vitamin D deficient or not vitamin D deficient. Recent research indicates that 20% of British children are vitamin D deficient. (The probability of not being vitamin D deficient is therefore 80%.) Provided the students at this school are representative of all British children, we can consider the probability constant. The trials are not independent, since the population of British children is finite, but the children at this school represent fewer than 10% of all British children.

20 244.380.913

yz

z

z

µσ−

=

−=

≈ −



a) Let Y = the number of children who are vitamin D deficient out of 320 children. Use Binom(320, 0.2).

= = =( ) 320(0.2) 64E Y np children.

= = ≈( ) 320(0.2)(0.8) 7.16SD Y npq children.

b) Using Binom(320, 0.2):

0 320 50 270

320 0 320 50

(no more than 50 are deficient) ( 50)

( 0) ( 50)

(0.2) (0.8) (0.2) (0.8)0.027

P P Y

P Y P Y

C C

= ≤= = + + =

= + +≈

……

According to the Binomial model, the probability that no more than 50 of the 320 children have the vitamin D deficiency is approximately 0.027.

Using N(64, 7.16):


( 50) ( 1.955) 0.0253P Y P z≤ ≈ < − ≈

According to the Normal model, the probability that no more than 50 out of 320 children have the vitamin D deficiency is approximately 0.0253.

37. Apples.

a) Let X = the number of cider apples found in the n = 300 apples from the tree.

The quality of the apples may be considered Bernoulli trials. There are only two possible outcomes, cider apple or not a cider apple. The probability that an apple must be used for a cider apple is constant, given as p = 0.06. The trials are not independent, since the population of apples is finite, but the apples on the tree are undoubtedly less than 10% of all the apples that the farmer has ever produced, so model with Binom(300, 0.06).

µσ−

=

−=

≈ −

50 647.161.955

yz

z

z



( ) 300(0.06) 18E X np= = = cider apples.

( ) 300(0.06)(0.94) 4.11SD X npq= = ≈ cider apples.

b) Since np = 18 and nq = 282 are both greater than 10, (300,0.06)Binom may be approximated by the Normal model, N(18, 4.11).

c) It is extremely unlikely that the tree will bear more than 50 cider apples. Using the Normal model, N(18, 4.11), 50 cider apples is about 7.8 standard deviations above the mean.

38. Frogs, part II.

The frog examinations can be considered Bernoulli trials. There are only two possible outcomes, having the trait and not having the trait. If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 0.125. The trials are not independent since the population of frogs is finite, but 150 frogs is fewer than 10% of all frogs. Since the biologist is collecting 150 frogs, use

(150,0.125).Binom

Let =X the number of frogs with the trait, from n = 150 frogs.

a) ( ) 150(0.125) 18.75E X np= = = frogs.

( ) 150(0.125)(0.875) 4.05SD X npq= = ≈ frogs.

b) Since np = 18.75 and nq = 131.25 are both greater than 10, (150,0.125)Binom may be approximated by the Normal model, N(18.75, 4.05).


= ≥= = + + =

= + +≈

……22 128 150 0

150 22 150 150

(at least 22 frogs) ( 22)

( 22) ( 150)

(0.125) (0.875) (0.125) (0.875)0.2433

P P X

P X P X

C C

According to the Binomial model, the probability that at least 22 frogs out of 150 have the trait is approximately 0.2433.

Using N(18.75, 4.05):

µσ−

=

−=

≈

22 18.754.05

0.802

xz

z

z



( 22) ( 0.802) 0.2111P X P z≥ ≈ > ≈

According to the Normal model, the probability that at least 22 frogs out of 150 have the trait is approximately 0.2111.

Using either model, the probability that the biologist discovers 22 of 150 frogs with the trait simply as a result of natural variability is quite high. This doesn’t prove that the trait has become more common.

39. Annoying phone calls.

Let X = the number of sales made after making n = 200 calls.

Using Binom(200, 0.12): These may be considered Bernoulli trials. There are only two possible outcomes, making a sale and not making a sale. The telemarketer was told that the probability of making a sale is constant at about p = 0.12. The trials are not independent, since the population is finite, but 200 calls is fewer than 10% of all calls. Therefore, the number of sales made after making 200 calls may be modeled by Binom(200, 0.12).

0 200 10 190

200 0 200 10

(at most 10) ( 10)

( 0) ( 10)

(0.12) (0.88) (0.12) (0.88)0.0006

P P X

P X P X

C C

= ≤= = + + =

= + +≈

……

According to the Binomial model, the probability that the telemarketer would make at most 10 sales is approximately 0.0006.

Using N(24, 4.60):

( ) 200(0.12) 24E X np= = = sales.

( ) 200(0.12)(0.88) 4.60SD X npq= = ≈ sales.

Since np = 24 and nq = 176 are both greater than 10, Binom(200,0.12) may be approximated by the Normal model, N(24, 4.60).

( 10) ( 3.043) 0.0012P X P z≤ ≈ < − ≈ According to the Normal model, the probability that the telemarketer would make at most 10 sales is approximately 0.0012.

µσ−

=

−=

≈ −

10 244.603.043

xz

z

z



Since the probability that the telemarketer made 10 sales, given that the 12% of calls result in sales is so low, it is likely that he was misled about the true success rate.

40. The euro.

Let X = the number of heads after spinning a Belgian euro n = 250 times.

Using Binom(250, 0.5): These may be considered Bernoulli trials. There are only two possible outcomes, heads and tails. The probability that a fair Belgian euro lands heads is p = 0.5. The trials are independent, since the outcome of a spin does not affect other spins. Therefore, Binom(250, 0.5) may be used to model the number of heads after spinning a Belgian euro 250 times.

140 110 250 0

250 140 250 250

(at least 140) ( 140)

( 140) ( 250)

(0.5) (0.5) (0.5) (0.5)0.0332

P P X

P X P X

C C

= ≥= = + + =

= + +≈

……

According to the Binomial model, the probability that a fair Belgian euro comes up heads at least 140 times is 0.0332.

Using N(125, 7.91):

= = =( ) 250(0.5) 125E X np heads.

( ) 250(0.5)(0.5) 7.91SD X npq= = ≈ heads.

Since np = 125 and nq = 125 are both greater than 10, Binom(250,0.5) may be approximated by the Normal model, N(125, 7.91).

( 140) ( 1.896) 0.0290P X P z≥ ≈ > ≈

According to the Normal model, the probability that a fair Belgian euro lands heads at least 140 out of 250 spins is approximately 0.0290.

Since the probability that a fair Belgian euro lands heads at least 140 out of 250 spins is low, it is unlikely that the euro spins fairly. However, the probability is not extremely low, and we aren’t sure of the source of the data, so it might be a good idea to spin it some more.

µσ−

=

−=

≈

140 1257.91

1.896

xz

z

z



41. Kids.

a)

b) Eµ = (Kids) = 1(0.5) + 2(0.25) + 3(0.25) = 1.75 kids

c)

42. Carnival.

a)

b) Eµ = (winnings)

= $95(0.1) + $90(0.09) + $85(0.081) + $80(0.0729) – $20(0.6561) ≈ $17.20

c)

43. Contest.

a) The two games are not independent. The probability that you win the second depends on whether or not you win the first.

b)

c)

d)

Kids 1 2 3 P(Kids) 0.5 0.25 0.25

Net winnings $95 $90 $85 $80 -$20

number of darts 1 dart 2 darts 3 darts 4 darts (win) 4 darts (lose)

P(Amount won)⎛ ⎞⎜ ⎟⎝ ⎠=

1100.1

9 1

10 100.09

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

29 110 100.081

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

39 110 100.0729

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠=

49100.6561

⎛ ⎞⎜ ⎟⎝ ⎠=

X 0 1 2

( )P X x= 0.42 0.50 0.08

σ

σ

= = − + − + − =

= = = ≈

2 2 2 2(Kids) (1 1.75) (0.5) (2 1.75) (0.25) (3 1.75) (0.25) 0.6875

(Kids) (Kids) 0.6875 0.83 kids

Var

SD Var

2 2 2 2

2 2

(Winnings) (95 17.20) (0.1) (90 17.20) (0.09) (85 17.20) (0.081)

(80 17.20) (0.0729) ( 20 17.20) (0.6561) 2650.057

(Winnings) (Winnings) 2650.057 $51.48

Var

SD Var

σ

σ

= = − + − + −

+ − + − − ≈

= = ≈ ≈

(losing both games) (losing the first) ( losing the second first was lost)

(0.6)(0.7) 0.42

P P P== =

(win both games) (winning the first) ( winning the second first was won)

(0.4)(0.2) 0.08

P P P== =



e) ( ) 0(0.42) 1(0.50) 2(0.08) 0.66 gamesE Xµ = = + + =

2 2 2 2( ) (0 0.66) (0.42) (1 0.66) (0.50) (2 0.66) (0.08) 0.3844

( ) ( ) 0.3844 0.62 games

Var X

SD X Var X

σ

σ

= = − + − + − =

= = = =

44. Contracts.

a) The contracts are not independent. The probability that your company wins the second contract depends on whether or not your company wins the first contract.

b)

c)

d)

e) ( ) 0(0.14) 1(0.70) 2(0.16) 1.02 contractsE Xµ = = + + =

2 2 2 2( ) (0 1.02) (0.14) (1 1.02) (0.70) (2 1.02) (0.16) 0.2996

( ) ( ) 0.2996 0.55 contracts

Var X

SD X Var X

σ

σ

= = − + − + − =

= = = ≈

45. LeBron again.

a) Answers will vary. A component is the simulation of the picture in one box of cereal. One possible way to model this component is to generate random digits 0-9. Let 0 and 1 represent LeBron James and 2-9 a picture of another sports star. Each run will consist of generating five random numbers. The response variable will be the number of 0s and 1s in the five random numbers.

b) Answers will vary.

c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that you get no pictures of LeBron in the five boxes. This is the number of trials in which neither 0 nor 1 were generated divided by the total number of trials. Perform similar calculations for the simulated probability that you would get one picture, 2 pictures, etc.

X 0 1 2

( )P X x= 0.14 0.70 0.16

(getting both contracts) (getting #1) (getting #2 got #1)

(0.8)(0.2) 0.16

P P P===

getting no contract) (not getting #1) ( not getting #2 didn't get #1)

0 2 0 7 0.14

=

==

(

( . )( . )

P P P



d) Let X = the number of LeBron James pictures in 5 boxes.

e) Answers will vary.

46. Seatbelts.

a) Answers will vary. A component is the simulation of one driver in a car. One possible way to model this component is to generate pairs of random digits 00-99. Let 01-75 represent a driver wearing his or her seatbelt and let 76-99 and 00 represent a driver not wearing his or her seatbelt. Each run will consist of generating five pairs of random digits. The response variable will be the number of pairs of digits that are 01-75.

b) Answers will vary.

c) Answers will vary. To construct your simulated probability model, start by calculating the simulated probability that none of the five drivers are wearing seatbelts. This is the number of trials in which no pairs of digits were 00-75, divided by the total number of trials. Perform similar calculations for the simulated probability that one driver is wearing his or her seatbelt, two drivers, etc.

d) Let X = the number of drivers wearing seatbelts in 5 cars.

e) Answers will vary.

47. Lefties again.

Let X = the number of righties among a class of n = 188 students.

Using Binom(188, 0.87):

These may be considered Bernoulli trials. There are only two possible outcomes, right-handed and not right-handed. The probability of being right-handed is assumed to be constant at about 87%. The trials are not independent, since the population is finite, but a sample of 188 students is certainly fewer than 10% of all people. Therefore, the number of righties in a class of 188 students may be modeled by Binom(188, 0.87).

X 0 1 2 3 4 5

P(X)

0 5(0.20) (0.80)0.33≈

1 45 1(0.20) (0.80)

0.41C

≈

2 35 2 (0.20) (0.80)

0.20C≈

3 25 3(0.20) (0.80)

0.05C≈

4 15 4(0.20) (0.80)

0.01C≈

5 0(0.20) (0.80)0.0≈

X 0 1 2 3 4 5

P(X)

0 5(0.75) (0.25)0.0≈

1 45 1(0.75) (0.25)

0.01C

≈

2 35 2 (0.75) (0.25)

0.09C

≈

3 25 3(0.75) (0.25)

0.26C

≈

4 15 4(0.75) (0.25)

0.40C

≈

5 0(0.75) (0.25)0.24≈



If there are 171 or more righties in the class, some righties have to use a left-handed desk.

= ≥= = + + =

= + +≈

……171 17 188 0

188 171 188 188

(at least 171 righties) ( 171)

( 171) ( 188)

(0.87) (0.13) (0.87) (0.13)0.061

P P X

P X P X

C C

According to the binomial model, the probability that a right-handed student has to use a left-handed desk is approximately 0.061.

Using N(163.56, 4.61):

( ) 188(0.87) 163.56E X np= = = righties.

( ) 188(0.87)(0.13) 4.61SD X npq= = ≈ righties.

Since np = 163.56 and nq = 24.44 are both greater than 10, Binom(188,0.87) may be approximated by the Normal model, N(163.56, 4.61).

( 171) ( 1.614) 0.053P X P z≥ ≈ > ≈

According to the Normal model, the probability that there are at least 171 righties in the class of 188 is approximately 0.0533.

48. No-shows.

Let X = the number of passengers that show up for the flight of n = 275 passengers.

Using Binom(275, 0.95): These may be considered Bernoulli trials. There are only two possible outcomes, showing up and not showing up. The airlines believe the probability of showing up is constant at about 95%. The trials are not independent, since the population is finite, but a sample of 275 passengers is certainly fewer than 10% of all passengers. Therefore, the number of passengers who show up for a flight of 275 may be modeled by Binom(275, 0.95).

µσ−=

−=

≈

171 163.564.61

1.614

xz

z

z



If 266 or more passengers show up, someone has to get bumped off the flight.

266 9 275 0

275 266 275 275

(at least 266 passengers) ( 266)

( 266) ( 275)

(0.95) (0.05) (0.95) (0.05)0.116

P P X

P X P X

C C

= ≥= = + + =

= + +≈

……

According to the binomial model, the probability someone on the flight must be bumped is approximately 0.116.

Using N(261.25, 3.61):

( ) 275(0.95) 261.25E X np= = = passengers.

( ) 275(0.95)(0.05) 3.61SD X npq= = ≈ passengers.

Since np = 261.25 and nq = 13.75 are both greater than 10, (275,0.95)Binom may be approximated by the Normal model, N(261.25, 3.61).

( 266) ( 1.316) 0.0941P X P z≥ ≈ > ≈

According to the Normal model, the probability that at least 266 passengers show up and someone gets bumped is approximately 0.0941.

49. Tennis model.

a) The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0.70. Finally, we are assuming that each serve is independent of the others. Since she is serving 80 times, use (80,0.70).Binom


( ) 80(0.70) 56E X npµ = = = = first serves in.

( ) 80(0.70)(0.30) 4.10SD X npqσ = = = ≈ first serves in.


µσ−=

−=

≈

266 261.253.61

1.316

xz

z

z



b)

c) According to the Normal model, in matches with 80 serves, she is expected to make between 51.9 and 60.1 first serves approximately 68% of the time, between 47.8 and 64.2 first serves approximately 95% of the time, and between 43.7 and 68.3 first serves approximately 99.7% of the time.

50. Colorblindness model.

a) These may be considered Bernoulli trials. There are only two possible outcomes, being colorblind and not being colorblind. The probability of being colorblind is given, p = 0.08. Colorblindness is assumed to be independent. For example, the volunteers can’t be related, as colorblindness may be hereditary. Use (450,0.08)Binom .


( ) 450(0.08) 36E X npµ = = = = colorblind men.

( ) 450(0.08)(0.92) 5.75SD X npqσ = = = ≈ colorblind men.


b)

c) According to the Normal model, the researcher would expect to find between 30.25 and 41.75 colorblind men in 68% of groups of 450 men, between 24.5 and 47.5 in 95% of the groups, and between 18.75 and 53.25 colorblind men in 99.7% of the groups.



51. Chips model.

a) These may be considered Bernoulli trials. There are only two possible outcomes, working properly, and failing to work properly. The probability of failing to work properly is given, p = 0.02. The defects in the chips are assumed to be independent. Use (1000,0.02)Binom .

Let X = the number of chips that fail in n = 1000 chips.

( ) 1000(0.02) 20E X npµ = = = = defective chips.

( ) 1000(0.02)(0.08) 4.43SD X npqσ = = = = chips.


b)

c) According to the Normal model, on 68% of days, the company can expect to produce between 15.57 and 24.43 defective chips. They should have between 11.14 and 28.86 defective chips on 95% of days, and between 6.71 and 33.29 defective chips on 99.7% of days.

52. Frog Model.

a) The frog examinations can be considered Bernoulli trials. There are only two possible outcomes, having the trait and not having the trait. If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 0.125. The trials are not independent since the population of frogs is finite, but 120 frogs is fewer than 10% of all frogs. Since the biologist is collecting 120 frogs, use

(120,0.125).Binom

Let X = the number of frogs with the trait, from n = 120 frogs.

( ) 120(0.125) 15E X npµ = = = = frogs.

( ) 120(0.125)(0.875) 3.6SD X npqσ = = = ≈ frogs.




b)

c) According to the Normal model, the biologist would expect to find between 11.4 and 18.6 frogs with the trait in 68% of all samples of 120 frogs, between 7.8 and 22.2 frogs with the trait in 95% of samples, and between 4.2 and 25.8 frogs with the trait in 99.75 of the samples.

53. ESP.

Choosing symbols may be considered Bernoulli trials. There are only two possible outcomes, correct or incorrect. Assuming that ESP does not exist, the probability of a correct identification from a randomized deck is constant, at p = 0.20. The trials are independent, as long as the deck is shuffled after each attempt. Since 100 trials will be performed, use Binom(100, 0.2).

Let X = the number of symbols identified correctly out of 100 cards.

( ) 100(0.2) 20E X npµ = = = = correct identifications.

( ) 100(0.2)(0.8) 4SD X npqσ = = = = correct identifications.

Answers may vary. In order be convincing, the “mind reader” would have to identify at least 32 out of 100 cards correctly, since 32 is three standard deviations above the mean. Identifying fewer cards than 32 could happen too often, simply due to chance.

54. True-False.

Guessing at answers may be considered Bernoulli trials. There are only two possible outcomes, correct or incorrect. If the student was guessing, the probability of a correct response is constant, at p = 0.50. The trials are independent, since the answer to one question should not have any bearing on the answer to the next. Since 50 questions are on the test use Binom(50, 0.5).

Let X = the number of questions answered correctly out of 50 questions.

( ) 50(0.5) 25E X npµ = = = = correct answers.

( ) 50(0.5)(0.5) 3.54SD X npqσ = = = ≈ correct answers.

Answers may vary. In order be convincing, the student would have to answer at least 36 out of 50 questions correctly, since 36 is approximately three standard



deviations above the mean. Answering fewer than 36 questions correctly could happen too often, simply due to chance.

55. Hot hand.

The shots may be considered Bernoulli trials. There are only two possible outcomes, make or miss. The probability of success is constant at 55%, and the shots are independent of one another. Therefore, we can model this situation with Binom(40, 0.55).

Let X = the number of free throws made out of 40.

( ) 40(0.55) 22E X npµ = = = = free throws made.

( ) 40(0.55)(0.45) 3.15SD X npqσ = = = ≈ free throws.

Answers may vary. The player’s performance seems to have increased. 32 made free throws is (32 − 22) / 3.15 ≈ 3.17 standard deviations above the mean, an extraordinary feat, unless his free throw percentage has increased. This does NOT mean that the sneakers are responsible for the increase in free throw percentage. Some other variable may account for the increase. The player would need to set up a controlled experiment in order to determine what effect, if any, the sneakers had on his free throw percentage.

56. New bow.

The shots may be considered Bernoulli trials. There are only two possible outcomes, hit or miss the bulls-eye. The probability of success is constant at 80%, and the shots are independent of one another. Therefore, we can model this situation with Binom(50, 0.8).

Let X = the number of bulls-eyes hit out of 50.

( ) 50(0.8) 40E X npµ = = = = bulls-eyes hit.

( ) 50(0.8)(0.2) 2.83SD X npqσ = = = ≈ bulls-eyes.

Answers may vary. The archer’s performance doesn’t seem to have increased. 45 bulls-eyes is (45 − 40) / 2.83 ≈ 1.77 standard deviations above the mean. This isn’t unusual for an archer of her skill level.

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