WEEKEND OF DECEMBER 9 th & 10 th, 10AM-2PM … · as well as past exam questions with solutions will be provided during the upcoming: qms 102 final exam crash course december 9th

ITM 102 MKT 300

WEEKEND OF DECEMBER 9th & 10th, 10AM-2PM

2 DAY REVIEW TO PREPARE YOU FOR YOUR FINAL EXAM

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!!! REMINDER !!!

FULL CHAPTER NOTES, CALCULATOR STEPS AS WELL AS PAST EXAM QUESTIONS WITH SOLUTIONS WILL

BE PROVIDED DURING THE UPCOMING:

QMS 102 FINAL EXAM CRASH COURSE DECEMBER 9th & 10th

(10:00am-2:00pm)

THE FINAL EXAM CRASH COURSE WILL BE 2 DAYS OF COMPREHENSIVE REVIEW THAT WILL GO OVER EVERYTHING

YOU NEED TO KNOW TO EXCEL ON YOUR QMS 102 FINAL EXAM.

IF YOU WOULD LIKE TO ATTEND THIS CRASH COURSE / PREP

SESSION, PLEASE REGISTER ONLINE:

http://easygradetutorials.com/qms-102



COURSE DEVELOPMENT BY:

www.EasyGradeTutorials.com UNIVERSITY EXAM PREP

QMS 102

SAMPLE TEST-3 COURSE

PACK (F2017)



CH5BasicProbability

Thereare3typesofprobabilities:(MCQ)

• Priori-theprobabilityofsuccessisbasedonpriorknowledgeoftheprocess.• Empirical-basedonobserveddata,notonpriorknowledgeofaprocess.• Subjective-theprobabilitythatdiffersfrompersontoperson

Probabilityofoccurrence=X/T

Where,X=#ofwaysinwhichaneventoccurs

Y=total#ofpossibleoutcomes

Moredefinitions:

Event:eachpossibleoutcomeofavariable.AsimpleeventisasinglecharacteristicJointevent:aneventthathas2ormorecharacteristicsComplement:complementofeventA(representedasA’)includesalltheeventsthatareNOTpartofeventA.

Samplespace:collectionofallpossibleevents

Formulas

P(A&B)=!&$%&'()

P(AorB)=P(A)+P(B)–P(AandB)

P(A|B)=*(!&$)*($)

P(B|A)=*(!&$)*(!)



ContingencyTables

ActuallyPurchased

PlannedtoPurchase

Yes No Total

Yes

200 50 250

No 100 650 750

Total 300 700 1000

a) WhatistheprobabilityoffamiliesplanningtopurchaseabigscreenTV?

P(Plannedtopurchase)=-./01234&5)(6617'&5.284(91

%&'()6./012&:4&.914&)79 = ;<=>====0.25

b) Whatistheprobabilityoffamilieswhoplannedtopurchaseandactuallypurchased?

P(Plannedtopurchaseandpurchased)=

5)(6617'&5.284(91&(8'(.))?5.284(917%&'()6./012 =

;==>,====0.20

c) Whatistheprobabilityoffamilieswhoplannedtopurchaseoractuallypurchased?

P(Plannedtopurchaseoractuallypurchased)=

P(plannedtopurchase)+P(actuallypurchased)–P(plannedtopurchaseand

actuallypurchased)

;<=>=== +

B==>=== −

;==>====0.35

d) WhatistheprobabilityoffamilieswhoplannedtopurchaseaTV,giventhattheyactuallypurchasedit?

P(Plannedtopurchase|actuallypurchased)=5)(6617'&5.284(91&5.284(917

(8'.())?5.284(917

P(Plannedtopurchase|actuallypurchased)=;==B===0.67



TreeMethod

Let“plannedtopurchase”=ALet“actuallypurchased”=B

P(A)=0.25

P(AandB)=0.2

P(AandB’)=0.05

P(A’andB)=0.1

P(A’andB’)=0.65

Entire

setof

house-

holds

0.25

0.75

0.20

0.05

0.1

0.65



CH6DiscreteProbabilityDistributions

Continuousvariables-fromameasuringprocessDiscretevariables-fromacountingprocess

FormulasforDiscreteprobabilities:

(Youdon’treallyneedthoseformulasasyoucanfindalltheaboveusingyour

calculator,stepsareshownbelow)



Example1

Acompanyplanstoinstallanewunitinoneoftheir2locations.Theprobabilityof

theunitbeingsuccessfulinlocationAis¾andtheannualprofitinthiscaseis

$150,000.However,ifitisn’tsuccessfultherewillbealossof$80,000.Atthesecond

location,locationB,thesuccessprobabilityis½andtheprofitis$240,000,and

losses$48,000.

Whereshouldthecompanylocateinordertomaximizeitsmaximumprofit?

X=profit

LocationA:Profit Prob

$150,000 0.75

-$80,000 0.25

Calculatorsteps:

ForlocationA&Bseparately:

• Entertheprofitcolumnintolist1

• Entertheprobabilitycolumnintolist2

• PressCALC

• SET-XLIST:LIST1&YLIST:LIST2

• Exit

• 1VARtoobtainresults

LocationA:

E(X)=92,500

Sigmax=115,000

LocationB:

E(X)=96,000

Sigmax=144,000

LocationB:Profit Prob

$240,000 0.5

-$48,000 0.5

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a)Usingthemeantodecidewherethecompanyshouldlocatetomaximizeexpectedprofit:Answer:LocationB(higherexpectedprofit)b)Whichofthe2locationsislessrisky(haslowestrelativevariability)?USETHEBELOWFORMULA:

LocationA:CV=($99,593/$92,500)*100=108%LocationB:CV=($144,000/$96000)*100=150%SincelocationBhasahigherCV(150%)thanlocationA(108%),locationAislessriskythanlocationB.


EXAMPLE2Aninvestmentadvisorsendsinformationtovariousclientsviaexpressdeliverycompanies.Inallcasesitisimportantthattheclientreceivetheinformationonthedayitissent.Threecompaniesarecapableofprovidingthedeliveryservice.Thefollowinginformationisgiven:

Company Cost($) Probabilityofsamedayarrival

Can-Express 22 0.99

Can-Parcel 14 0.97

CanadaMail 6 0.89

Theaveragepayoffis$200whentheinformationarrivesontime,butthereisnopayoffifthedeliveryislate.Basedonmaximizingexpectedprofit,whichdeliveryserviceshouldbeused?SolutionUseyourcalculatortofindtheexpectedprofitforeachofthedeliveryservices:Can-Express

Can-ParcelX P(X)

186 0.97-14 0.03CanadaMailX P(X)

194 0.89-6 0.11Therefore,thecompanyshouldchooseCan-Parcelasithasthehighestexpectedprofit.

X P(X)178 0.99-22 0.01

E(X)=$176

E(X)=$180

E(X)=$172


Summaryofprobabilities(Binomial,PoissonandNormalProbabilities)Atleastà≥Atmostà≤Morethat,greaterthanà>Lessthanà<Binomial&Poisson

1) P(X=#)àUsebpd,ppd(number(x)asis)

2) P(X≤#)àUsebcd,pcd(number(x)asis)

3) P(X≥#)

Ex:P(X≥3)à1-P(X≤ 2)àbcd/pcdwherex=2,then1-answer

4) P(X<#)

Ex:P(X<10)àP(X≤9)àbcd/pcdwherex=9

5) P(X>#)

Ex:P(X>40)à1-P(X≤40)àbcd/pcdwherex=40

Meanofbinomial:n*p

Standarddeviationofbinomial: %& 1 − &

NormalDistribution

Onlyuse:Ncd&InvN

1) P(3≤X≤5)àNcd,lower:3,upper:52) P(X<40)àNcd,upper:40,lower:-10000(sameifP(X≤ #))3) P(X>40)àNcd,upper:10000,lower:40(sameifP(X≥ #))


BinomialDistributionTouseabinomialdistribution,theremustbe:

• Sampleconsistingofafixednumberofobservations“n”• Eachobservationclassifiedintooneoftwomutuallyexclusiveand

collectivelyexhaustivecategories.Eachtrialwillresultineitherofthe2outcomes:successorfailure.Thesuccess(+),andthereforefailureis(1-+).Example:Tossacoin6times,andfindtheprobabilityofgetting2tails.Find:P(X=2)Usingyourcalculator:Stat-DIST-BINM-BpdData:F2(Var)X:2Numtrial:6P:0.5PressexeBinomialP.DP(x=2)=0.23437


***IMPORTANT***

Yourcalculatorhas2binomialprobabilityfunctions:Bpd:thiscalculatestheprobabilityoftheformofP(X=#)Bcd:thiscalculatestheprobabilityoftheformofP(X≤#)EXAMPLE:P(X=2)wheren=6&+ = 0.5Calculatorsteps:STAT-BINM-BpdData:F2X:2Numtrial:6P:0.5Resultafterexe:p=0.234Example2:P(X≤2),wheren=18&+ = 0.43Calculatorsteps:STAT-BINM-BcdX:2Numtrial:18P:0.43Resultafterexe:p=0.0041


PoissonProbabilityDistributionCharacteristicsofPoisson:

• Consistsofobservingasituationforaperiodoftimeoramountofspace(length,area,volumeorweight)

• Successmustoccurrandomly,andthesuccessesareindependentofeachother

• Theaveragerateofsuccessisdenotedusing3(lambda)

***IMPORTANT***Yourcalculatorhas2settingsforPoissondistribution:Pbd:calculatestheprobabilityoftheformP(X=#)Pcd:calculatestheprobabilityoftheformP(X≤#)Example:Recordshavebeenkeptforthepastseveralmonths,andtheyshowthatcustomersarrivetouseacertainmachineatanaveragerateof15perhour.

a. Whatistheprobabilitythat12customerswillusethemachinein1hour?

X=numberofcustomersusingthemachinein1hour3 = 15/ℎ6P(X=12)=Ppd(12,15)=0.0829

b. Whatistheprobabilitythattherewillbelessthan3customersinthenext10minutes?

X=numberofcustomersusingthemachinein10minutes3 = 15/ℎ6=2.5/10mins(crossmultiplication)P(X<3)=P(7 ≤ 2)Pcd(2,2.5)=0.5438

c. Whatistheprobabilitythattherewillbemorethan40customersinthenext2hours?

X=#ofcustomersusingmachinein2hours3 = 15/ℎ6,30/2hrs(crossmultiplication)P(X>40)=1-P(≤ 40)=1-Pcd(40,30)=1-0.9677=0.0323


Practicequestions

1. Astudentistakingamultiple-choiceexaminwhicheachquestionhas4

choices(A,B,C,&D).Assumingshehasnoknowledgeofthecorrectanswers

toanyofthequestions,shedecidestoplacefourballsmarkedA,B,C,&D

intoabox,andsherandomlyselectsoneballforeachquestionandreplaces

theballbackinthebox.Themarkingontheballwilldetermineheranswer

tothequestion.Thereare5multiple-choicequestionsontheexam.

a) Whatistheprobabilitythatshewillgetfivequestionscorrect?

b) Whatistheprobabilitythatshewillgetatleastfourquestionscorrect?

c) Whatistheprobabilitythatshewillgetnoquestionscorrect?

d) Whatistheprobabilitythatshewillgetnomorethantwoquestions

correct?


2. Theincreaseordecreaseinthepriceofastockbetweenthebeginningand

theendofatradingdayisassumedtobeanequallyrandomevent.Whatis

theprobabilitythatastockwillshowadecreaseinitsclosingpriceonten

consecutivedays?

3. Amanufacturingcompanyregularlyconductsqualitycontrolchecksat

specificperiodsontheproductsitmanufactures.Historically,thefailurerate

forLEDlightbulbsthatthecompanymanufacturesis13%.Supposea

randomsampleof10LEDlightbulbsisselected:

a) WhatistheprobabilitythatnoneoftheLEDlightbulbsaredefective?

b) WhatistheprobabilitythatexactlyoneoftheLEDlightbulbsisdefective?

c) WhatistheprobabilitythattwoorfeweroftheLEDlightbulbsaredefective?

d) WhatistheprobabilitythatthreeormoreoftheLEDlightbulbsare

defective?


4. Pastrecordsindicatethattheprobabilityofonlineretailordersthatturnout

tobefraudulentis0.05.Supposethat,onagivenday,21onlineretailstore

ordersareplaced.Assumethatthenumberofonlineretailordersthatturn

outtobefraudulentisdistributedasabinomialrandomvariable.

a) Whatisthemeanandstandarddeviationofthenumberofonlineretail

ordersthatturnouttobefraudulent?

Mean:

Standarddeviation:

b) Whatistheprobabilitythattwoormoreonlineretailordersturnouttobe

fraudulent?

5. Supposethatyouandtwofriendsgotoarestaurant,whichlastmonthfilledapproximately83%oftheirorderscorrectly.a) Whatistheprobabilitythatatleasttwoofthethreeorderswillbefilled

correctly?

b) Whatarethemeanandstandarddeviationofthebinomialdistribution?Mean:Standarddeviation:


GOODLUCK J

PLEASE NOTE THIS IS ONLY A SAMPLE COURSE PACKAGE FOR TEST-3!!

THE ACTUAL TEST-3 PACKAGE INCLUDES CHAPTER 7 NOTES,

FULL PRACTICE PROBLEMS FROM PREVIOUS SEMESTER’S TESTS AND ANSWERS TO ALL THE PROBLEMS.

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FOR THE QMS 102 FINAL EXAM CRASH COURSE!

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-----------

THE FINAL EXAM CRASH COURSE INCLUDES:

- FULL CHAPTER NOTES - CALCULATOR STEPS

- PAST EXAM QUESTIONS WITH SOLUTIONS - 2 DAYS OF COMPREHENSIVE REVIEW

- GO OVER EVERYTHING YOU NEED TO KNOW TO EXCEL ON YOUR QMS 102 FINAL EXAM.

QMS 102 FINAL EXAM CRASH COURSE DECEMBER 9th & 10th

REGISTER ONLINE: http://easygradetutorials.com/qms-102


QMS102TEST-3CRIBSHEETFALL2017(FRONT)www.EasyGradeTutorials.com

Ch5

P(A&B)=!&$%&'()

P(AorB)=P(A)+P(B)–P(AandB)

P(A|B)=*(!&$)*($)

P(B|A)=*(!&$)*(!)

CH6Tofindexpectedmean/profit:

• Entertheprofitcolumnintolist1

• Entertheprobabilitycolumnintolist2

• PressCALC

• SET-XLIST:LIST1&YLIST:LIST2

• Exit

• 1VARtoobtainresults

CoefficientofVariation:

BinomialDistributionTouseabinomialdistribution,theremustbe:

• Sampleconsistingofafixednumberofobservations“n”

• Eachobservationclassifiedintooneoftwomutuallyexclusiveand

collectivelyexhaustivecategories.

Eachtrialwillresultineitherofthe2outcomes:successorfailure.

Thesuccess(-),andthereforefailureis(1--).

PoissonProbabilityDistributionCharacteristicsofPoisson:

• Consistsofobservingasituationforaperiodoftimeoramountofspace

(length,area,volumeorweight)

• Successmustoccurrandomly,andthesuccessesareindependentofeach

other

• Theaveragerateofsuccessisdenotedusing.(lambda)

QMS102TEST-3CRIBSHEETFALL2017(BACK)www.EasyGradeTutorials.com

NormaldistributionPropertiesofnormaldistribution:

• Itissymmetrical,therefore,themeanandmedianareequal

• ItsIQR=1.33standarddeviations

• Ithasaninfiniterange

Summaryofprobabilities

Atleastà≥Atmostà≤Morethat,greaterthanà>

Lessthanà<

Binomial&Poisson

1) P(X=#)àUsebpd,ppd(number(x)asis)

2) P(X≤#)àUsebcd,pcd(number(x)asis)

3) P(X≥#)

Ex:P(X≥3)à1-P(X≤ 2)àbcd/pcdwherex=2,then1-answer

4) P(X<#)

Ex:P(X<10)àP(X≤9)àbcd/pcdwherex=9

5) P(X>#)

Ex:P(X>40)à1-P(X≤40)àbcd/pcdwherex=40

Meanofbinomial:n*p

Standarddeviationofbinomial: 23 1 − 3 NormalDistribution

Onlyuse:Ncd&InvN

1) P(3≤X≤5)àNcd,lower:3,upper:5

2) P(X<40)àNcd,upper:40,lower:-10000(sameifP(X≤ #))3) P(X>40)àNcd,upper:10000,lower:40(sameifP(X≥ #))

Documents

WEEKEND OF DECEMBER 9 th & 10 th, 10AM-2PM … · as well as past exam questions with solutions will be provided during the upcoming: qms 102 final exam crash course december 9th