ch04-solns-all_skuce_2e

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Instructors Solutions Manual - Chapter 4

Chapter 4 Solutions

Develop Your Skills 4.1

1a. Sample space:246 employees commute more than 40 km by car

350-246=104 commute 40 km by car

P(randomly selected employee commutes > 40 km by car) = 246/350 = 0.7029

1b. Sample space:150 employees arrange rides with others350-150=200 ride aloneP(randomly selected employee arranges rides with others) = 150/350 = 0.4286

Note that the sample space need not be more complicated than necessary. A fulldescription of the sample space, for both worker characteristics (commuting distance

and arranging rides) could look as follows.

Commuting Characteristics of Car Part Manufacturing Plant

Arrange Rides With Others Ride Alone Totals

Commute > 40 km 246-135=111 135 246

Commute 40 km 104-65=39 200-135=65 350-246=104Totals 150 350-150=200 350

2. Sample space:

We are interested only in managers, so the sample space is as follows:

Only High-School Education: 0Grad Degree Or Post-Grad Studies: 10up To 4 Years Of Post-Secondary Education: 37 10 0 = 27

P(randomly selected manager has up to 4 years of post-secondaryeducation)=27/37=0.7297

3. Sample space:Professional Employees: 372

Managers: 37(Note that "managerial" and "professional" are separate job classifications, somanagers are not included in the count of professional workers.)Clerical Employees: 520-372-37 = 111

P(randomly selected employee is professional) = 372/520 = 0.7154P(randomly selected employee is clerical) = 111/520 = 0.2135

Copyright 2011 Pearson Canada Inc. 65


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4. Sample space:

Shoppers Doing A Quick Trip: 62Shoppers Doing A Major Stock-Up: 13Shoppers Doing A Fill-In Shop: 100-62-13=25

P(randomly selected customer is doing a fill-in shop) = 25/100 = 0.25

5. Sample space:

Loved Previous Math Courses: 56Worked Very Hard In Previous Math Courses, But Did Not Enjoy Them, Or ThoughtPrevious Math Courses Were Far Too Difficult, Or Equated Previous Math CoursesWith Sticking Needles In The Eyes: 225-56=169

P(randomly selected student did not love his/her previous math

courses)=169/225=0.7511


6. These probabilities are given:P(pays with credit card) =P(CC) = 0.80P(buys something other than gas) P(OTG) = 0.25P(pays with credit card and buys something other than gas) =P(CC & OTG)= 0.20

Want to knowP(pays with credit card GIVEN buys something other than gas)

= P(CCOTG)

80.025.0

20.0

)OTG(P

)OTG&CC(P

Also want to knowP(buys something other than gas GIVEN pays with a credit card)

=P(OTG CC)

25.080.0

20.0

)CC(P

)OTG&CC(P

To check for independence, compare P(CCOTG) with P(CC).The two probabilities are equal. So, paying with a credit card and buying something

other than gas are independent (not related).



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7. P(honey-nut flavour GIVEN family size box)

5714.0315

180

)sizefamilyP(

)sizefamily&nutP(honey

To check for independence, compare this with

P(honey-nut flavour) 900

180315

= 0.55Since the two probabilities are NOT equal, the events are NOT independent, that is,the size of the box and the flavour are related, in terms of sales.

8. P(female given employed in agriculture industry)

2951.0327

5.96

)5.2305.96(

5.96

P(employed in public administration, given male)

0503.03.021,9

454

P(employed in public administration)

0541.08.125,17

7.925

)5.104,83.021,9(

7.471454

Since P(employed in public administration, given male) P(employed in publicadministration), we can say that sex of the employee and industry of employmentwere not independent in Canada in 2008. However, the difference in probabilities isnot that great (both are around 5%).

However, we can see thatP(employed in manufacturing)

1150.08.125,17

3.970,1

)5.81043.021,9(

9.5594.410,1

which is not equal to

P(employed in manufacturing given male)1563.0

3.021,9

4.410,1

This is more convincing evidence that sex of the employee was not independent of theindustry of employment in Canada in 2008.



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9. It is easiest to proceed if we first compute row and column totals for the table.

Accounts Receivable for a Roofing Company

amount

age $5,000 $5,000 -


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To test for independence, we could compare P(customer will not use the dry cleaningcompanys services again) with the conditional probability we just calculated above.

P(customer will not use the dry cleaning companys services again)=1 P(customers will use dry cleaning companys services again)

=1 0.620 = 0.38 0.0833

Since these probabilities are not equal, the events are not independent. There is arelationship between the rating of the service and the tendency to use it again.Overall, 38% of customers dont plan to use the service again. However, only about8% of those who rated the service positively dont plan to use it again.


11. We are told that the two employees live in different parts of the city, and sopresumably could not be held up by the same traffic problems. Assume that eachemployees lateness is independent of the others lateness. Then

P(both are late)=P(Jane is late andOscar is late)=0.02 0.04 = 0.0008The probability is low that both Jane and Oscar will be late for work.

12. We are told the friends are very different, and will assume that any one of themgetting a job in the financial services industry is independent of the others gettingsuch a job. Label the friends 1, 2 and 3.

a. P(all three of them succeed)=P(1 does and2 does and3 does)

=0.4 0.5 0.35 = 0.07

b. P(none of them succeeds)=P(1 does not and2 does not and3 does not)=(1-0.4)(1-0.5)(1-0.35)=0.60.50.65=0.195

c. P(at least one of them succeeds)=1-P(none of them succeeds)=1-0.195=0.805Using the complement rule here is a life-saver!



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13. P(game aimed at 15-25 year olds succeeding) = 0.34P(accounting program for small business succeeding) = 0.12P(payroll system for government organizations succeeding) = 0.10We are told to assume that the events are independent.P(all three succeed) = 0.34 0.12 0.10 = 0.00408

For calculation of at least two out of three succeeding, we need to think about whatthis means, in terms of the sample space. At least two out of three succeeding meansexactly two out of the three succeeding, or all three succeeding. A tree diagrammight be helpful to picture this. We need to calculate and add the probabilities forthe cases shown in bolded letters on the right-hand side of the tree diagram.

0.90

0.10 S

F

0.10 S

F

0.90

S

payroll

system

F

0.90

0.10

0.88

0.12

F

0.12

0.88

accounting

program

0.90

0.10

S

F

0.66

0.34

game

F

S

S

F

S

SSS

SSF

SFS

SFF

FSS

FSF

FFS

FFF

Once we have the cases identified, it is just a matter of arithmetic.

P(at least two out of three succeed)

=(0.34 0.12 0.10)+(0.34 0.12 0.90)+(0.34 0.88 0.10)+(0.66 0.12 0.10)=0.00408+0.03672+0.02992+0.00792=0.07864



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14. A fully-labelled tree diagram for the GeorgeConn customer data is shown below.

R

U

S

N

S

N

P(NR)=1/4

P(U)=6/10

P(NU)=2/6

P(SU)=4/6

P(R)=4/10

P(SR)=3/4

P(U and N)=0.2

P(U and S)=0.4

P(R and N)=0.1

P(R and S)=0.3

Another way to set up the tree diagram:

S

N

R

U

R

U

P(US)=4/7

P(N)=3/10

P(UN)=2/3

P(RN)=1/3

P(S)=7/10

P(RS)=3/7

P(N and U)=0.2

P(N and R)=0.1

P(S and U)=0.4

P(S and R)=0.3



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15. P(hourly worker oronly high school education)=P(hourly worker) + P(only high school education) P(hourly worker andonly highschool education)= (790+265+2)/1345 + (790+7+1+0)/1345 790/1345

= (790 + 265 + 2 + 7 + 1)/1345= 1065/1345= 0.7918

Chapter Review Exercises

1. P(account paid early)

750

1190.1587

P(account paid on time)

750320 0.4267

P(account paid late)

750

2000.2667

P(account uncollectible)

750

1110.1480

2. The probability calculations may seem easier if you organize the information into atable, as follows.

Business Diploma No Business Diploma Total

Men 30 30 60

Women 25 15 40

Total 55 45 100

a. P(employee is a man)

100

600.60

b. P(employee is a man with a Business diploma)



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100

300.30

c. P(employee is a woman)

100

400.40

d. P(employee is a woman with a Business diploma)

100

250.25

e. P(employee has a Business diploma)

100

550.55

f. P(employee is a man without a Business diploma)

100

300.30

g. P(employee is without a Business diploma)

100

450.45

h. P(employee is a woman without a Business diploma)

100

150.15

i. P(employee is a woman or employee with a Business diploma)

100

25

100

55

100

400.70

j. P(employee is a man or employee with a Business diploma)

100

30

100

55

100

600.85

k. P(employee is a woman or employee without a Business diploma)

10015

10045

10040 0.70

l. P(employee is a man or employee without a Business diploma)

100

30

100

45

100

600.75

3. P(employee has Business diploma given she is a woman)



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40

250.625

To test to see if gender and possession of a Business diploma are related for

GeorgeConn employees, we can compare the probability above to

P(employee has a Business diploma)

100

550.55

P(employee has Business diploma given she is a woman)) P(employee has aBusiness diploma), so gender and possession of a Business diploma are related forGeorgeConn employees. Female employees are more likely to have a Businessdiploma.

4. P(caller directly connected) = 0.80P(caller forced to wait) = 0.20

P(caller connected directly for three different calls)= P(caller connected on first day AND connected on second day AND connected on

third day)= 0.8 0.8 0.8= 0.512

P(caller forced to wait for three different calls)

= P(caller forced to wait on first day AND forced to wait on second day AND forcedto wait on third day)

= 0.2 0.2 0.2= 0.008

5. Start by totalling the rows and columns of the table. This will speed up theprobability calculations.

a. P(primary skill is bookkeeping)

100

300.30

b. P(employee has less than one year of experience)

100

500.50

c. P(primary skill is reception)



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100

250.25

d. P(employee has one to two years of experience)

100

230.23

e. P(primary skill is document management)

100

450.45

f. P(employee has more than two years of experience)

100

270.27

6. Begin by totalling rows and columns.

Survey of Restaurant Customers

Opinion About Food Satisfied with Service Not Satisfied with Service Totals

Excellent 0.36 0.06 0.42

Good 0.18 0.07 0.25

Fair 0.10 0.08 0.18

Poor 0.05 0.10 0.15

Totals 0.69 0.31

a. P(customer is satisfied with service and rates the food as poor)= 0.05

b. P(customer is not satisfied with service)= 0.31

c. P(not satisfied with service given food rated as poor)

15.0

10.00.6667

d. To test for independence, we could compare the probability in part c with

P(not satisfied with service)=0.31

The two probabilities are not equal, so the service rating and the food rating arerelated (that is, NOT independent). People who rate the food as poor are more likelyto be dissatisfied with the service.



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7. P(salesperson will exceed targets two years in a row)= P(salesperson exceeds target this year and exceeds target next year)

= P(exceeds target this year) P(exceeds target next year exceeds target this year)= 0.78 0.15 = 0.0117

8. Begin by summing rows and columns of the table.

Follow-up Survey of CustomersWho Bought Netbook Computers

Satisfied Not Satisfied Totals

1 GB of RAM or Less 0.30 0.05 0.35

More than 1 GB of RAM 0.50 0.15 0.65

Totals 0.80 0.20

a. P(satisfied with his/her purchase)

= 0.30 + 0.50 = 0.80

b. P(satisfied with his/her purchase more than 1 GB of RAM)= 0.50/(0.50 + 0.15) = 0.7692

c. The amount of RAM affects whether or not the purchaser was satisfied with his/her

purchase. P(satisfied more than 1 GB of RAM) = 0.7692 P(satisfied) = 0.80.Customers who bought netbooks with more than 1 GB of RAM were less likely to besatisfied with their purchases. This may seem odd, because generally more RAMmeans a better computer. However, it is possible that customers buying machineswith more RAM had higher performance expectations that could not be met with

slower processors.



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9. A tree diagram is helpful:

P(pass with no more than two attempts)= P(pass the first time) + P(fail the first time and pass the second time)= 0.75 + 0.225 = 0.975

10. Again, begin by summing rows and columns in the table.

Customers of an Insurance Company

Single Married Divorced TotalsMale 25 125 30 180

Female 50 50 20 120

Totals 75 175 50

P(P)=0.75P

F

P

F

P(F and P)=0.25 0.90=0.225

P(F and F)=0.25 0.10

=0.025

P(P)=0.75

P(PF)=0.90

P(FF)=0.10

P(F)=0.25

a. P(female or married)= P(female) + P(married) P( female andmarried)= (50+50+20)/300 + (125 + 50)/300 50/300= (50 + 50 + 20 + 125 + 50 - 50)/300 = 245/300 = 0.8167

b. P(married male)

= P(married and male)/P(male)= (125/300) / (25 + 125 + 30)/300= 125/ (180) =0.6944

c. P(married)

= (125 + 50)/300 = 0.5833 P(married male) = 0.6944Since the probabilities are not equal, gender and marital status are not independent.



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11. R: market risesR

C: market does not rise

P: newsletter predicts risePC: newsletter predicts market will not rise

R

RC

P

PC

P

PC

P(R and P)

=0.60 0.70

P(PCR)=0.30

P(RC)=0.40

P(PCRC)=0.70

P(PRC)=0.30

P(R)=0.60

P(PR)=0.70

=0.42

P(R and PC)

=0.60 0.30=0.18

P(RCand P)=0.40 0.30=0.12

P(RCand P

C)

=0.40 0.70

=0.28

P(correct prediction)

= P(market rises and newsletter predicts rise)+ P(market does not rise and newsletter predicts market will not rise)= 0.42 + 0.28 (from tree diagram)= 0.70

12. P(all of the students selected by the company are female)= 6/10 5/9 4/8= 120/720 = 0.1667

P(all of the students selected by the company are male)= 4/10 3/9 2/8

= 24/720 = 0.0333



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13. If we can identify one situation where gender and tendency to use the health facilitiesare related, we can say that gender and the tendency to use health facilities arerelated.

Compare P(used the facilities) with P(used the facilities male)

P(used the facilities) = 210/350 = 0.6P(used the facilities male) = 65/170 = 0.3824Since these two probabilities are not equal, gender and tendency to use the health andfitness facilities are not independent (that is, they are related).

14.

U

D

M

F

M

F

P(U and M)=65/350

P(FU)=145/210

P(D)= 140/350

P(FD)=35/140

P(MD)=105/140

P(U)= 210/350

P(MU)=65/210

=0.1857

P(U and F)=145/350=0.4143

P(D and M)=105/350=0.3

P(D and F)=35/350=0.1

The joint probabilities are the same, as we would expect them to be.

15. One of the ways to test for independence (or lack of it) is as follows.

P(purchased the product)

300228 0.76

P(purchased the product given saw the TV ad)

188

152

36152

152

= 0.8085



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These two probabilities are not equal, so purchasing behaviour is related to seeingthe TV ad. Those who saw the ad were more likely to purchase the product.

16. A: Brenda moves to AlbertaA

C: Brenda does not move to Alberta

B: Brenda is offered the job at Canadas largest bankBC: Brenda is not offered the job at Canadas largest bank

B

BC

A

AC

A

P(BCand AC)=0.75 0.65

=0.4875

P(ACB)=0.3

P(BC)=0.75

P(ACBC)=0.65

P(ABC)=0.35

P(B)=0.25

P(AB)=0.7

AC

P(B and A)=0.25 0.7=0.175

P(B and AC)=0.25 0.3=0.075

P(BCand A)=0.75 0.35=0.2625

From the tree diagram:P(Brenda will be offered the job and not move to Alberta) = 0.075



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17. I: Canadian adult has taken instruction in canoeingIC: Canadian adult has not taken instruction in canoeing

CT: Canadian adult is going on a canoe trip this summerCTC: Canadian adult is not going on a canoe trip this summer

P(CTC IC)=0.80

P(CT IC)=0.20

P(CTC I)=0.54

P(CT I)=0.46

P(IC

)=0.97

P(I)=0.03

CT

CTC

CTC

CT

IC

I

P(I and CT)=0.03 0.46=0.0138P(I and CT

C)

=0.03 0.54=0.0162

P(ICand CT)=0.97 0.20=0.194

P(ICand CT

C)

=0.97 0.80=0.776

18. P(a randomly-selected Canadian adult is going on a canoe trip this summer, and hastaken some canoeing instruction) = 0.0138

P(a randomly-selected Canadian adult is going on a canoe trip this summer, and hasnot taken any canoeing instruction) = 0.194

19. P(a randomly-selected customer from one of these stores uses a cash/debit card or acredit card for payment) = (150 + 180)/500 = 0.66

20. If we can identify one situation where payment method and store location are related,we can say that they are related in general.

One approach is to compare P(cheque) with P(cheque Store A)P(cheque) = 170/500 = 0.34

P(cheque Store A) = 30/100 = 0.30Since these two probabilities are not equal, payment method and store location arenot independent.



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21. For people who visit the facility:P(buy a membership) = 0.40P(buy a membership and sign up for fitness classes) = 0.30

P(fitness classes bought a membership)=0.30/0.40 = 0.75

22. Buying a membership and signing up for fitness classes are NOT mutually exclusive.We are told that P(buy a membership and sign up for fitness classes) = 0.30 0. Aperson can do both, so the events are not mutually exclusive.

We cannot assess independence without more information. For example if we knew

P(fitness classes), we could compare that with P(fitness classes bought amembership).

23. If we can identify one situation where gender and type of alcoholic drink are related,we can say that they are related in general.

However, there is no case where gender and type of alcoholic drink are related.

P(wine) = (36 + 54)/(42 + 63 + 36 + 54 + 22 + 33)=90/250=0.36

P(wine female) = 54/(63 + 54 + 33) = 0.36P(wine male) = 36/(42 + 36 + 22) = 0.36So we can see that P(wine) = P(wine female) = P(wine male).

Similarly, P(beer) = P(beer female) = P(beer male).

As well, P(other alcoholic drinks) = P(other alcoholic drinks female)= P(other alcoholic drinks male).

We cannot identify a situation where gender and type of alcoholic drink are notindependent, so we conclude that gender and type of alcoholic drink ordered areindependent in this sample.

24. 500 circuit boards, of which 30 are defective, 470 are not defectiveP(all three are defective)=30/500 29/499 28/498=24,360/124,251,000 = 0.000196054

P(defective board on 1stselection) = 30/500 = 0.06

P(all 3 boards defective, assuming independence) = 0.06 0.06 0.06 = 0.000216The probabilities agree, to four decimal places.

25. In this case, we are stuck. We have only one probability (25%), but we do not haveindependent events. Once the first randomly-selected Canadian is asked about RRSPplans, he/she is removed from further consideration. Depending whether this personplans to make an RRSP contribution over the next year, this will affect the 25%probability of making a contribution. However, it will not affect it very much,



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because there are many millions of Canadians. So, although the events are not reallyindependent, we can still use the probability as if they were.

P(all four intend to contribute to their RRSPs over the next year)= 0.25 0.25 0.25 0.25 = 0.0039

26. This looks like a long and complicated question, but it isn't, as long as theinformation is organized properly. You can use the Sort tool (under the Data tab) inExcel to help organize the data as you require it (the use of the Sort tool wasdescribed in Chapter 1). You might also explore the use of the Filter tool (use Excel'sHelp function if you can't see how it works.) There are 24 employees in total.

a. P(employee has low experience)

24

130.5417

P(employee has high experience)

24

110.4583

P(employee has a specialty in spreadsheet software)

24

110.4583

P(employee has a specialty in presentation software)

24

40.1667

b. There are 11 employees who specialize in spreadsheet software.

P(first employee selected has high experience given specialization in spreadsheetsoftware)

11

40.3636

P(second employee selected has high experience given specialization in spreadsheetsoftware given first employee had high experience given specialization in

spreadsheet software)

10

30.30

P(both employees selected have high experience, given specialization in spreadsheetsoftware)

110

12

10

3

11

4 =0.1091



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P(at least one of the two employees has high experience)= 1 P(none of the employees has high experience)

6182.0110

68

110

421

10

6

11

71

c. The joint probability table is shown below.

DatabaseSoftware

PresentationSoftware

SpreadsheetSoftware

Word ProcessingSoftware

Totals

Low 0 4 7 2 13

High 1 0 4 6 11

Totals 1 4 11 8 24

P(high experience, given specialty is word processing)

8

60.75

P(low experience, given specialty is spreadsheet)

11

70.6364

d. The joint probability table is shown below.

DatabaseSoftware

PresentationSoftware

SpreadsheetSoftware

Word ProcessingSoftware

Totals

F 0 3 5 4 12

M 1 1 6 4 12

Totals 1 4 11 8 24

P(word processing specialization given female)

12

4

0.3333



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P(male given spreadsheet specialization)

11

60.5455

e. Because the tree diagram has three stages, it takes up an entire page (see the nextpage).

Notice that the end-stage probabilities could have been calculated directly from thetable of information about the employees. It is useful to explore the structure of thesample space with the tree diagram.

M: MaleF: Female

D: Specializes in Database Software

P: Specializes in Presentation SoftwareS: Specializes in Spreadsheet SoftwareW: Specializes in Word Processing Software

L: Low ExperienceH: High Experience



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L 00/1

D 1/11/24H

1/121/24L

1/1

P 0/11/12

H 0

M 3/24L6/12 3/6

S 3/6

3/24H12/244/12

1/241/4 L

W 3/4

3/24H

00/0 L

0/0D

0H

0/12

3/2412/24 3/3 L

3/12 P0/3

0H

F4/245/12 4/5 L

S1/5

1/24H

4/12

1/241/4 L

W3/4

H 3/24



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f. This tree diagram is just another way of representing the sample space. The end- stage probabilities match those in part e.

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D

P

S

W

D

P

S

W

D

P

S

W

D

M

1/7

0/7

3/7

3/7

F

M

3/8

0/8

1/8

4/8F

H

7/11

L

8/13

1/5

3/5

1/5

0/5

3/4

1/4

0/4

0/4

5/13

4/11

11/24

13/24

W

S

P 0/24

3/24

3/24

0/24

0/24

1/24

3/24

0/24

1/24

3/24

1/24

0/24

3/24

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