weekend of december 9 th & 10 th, 10am-2pm … · as well as past exam questions with solutions...
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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
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COURSE DEVELOPMENT BY:
www.EasyGradeTutorials.com UNIVERSITY EXAM PREP
QMS 102
SAMPLE TEST-3 COURSE
PACK (F2017)
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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)=!&$%&'()
P(AorB)=P(A)+P(B)–P(AandB)
P(A|B)=*(!&$)*($)
P(B|A)=*(!&$)*(!)
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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
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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
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CH6DiscreteProbabilityDistributions
Continuousvariables-fromameasuringprocessDiscretevariables-fromacountingprocess
FormulasforDiscreteprobabilities:
(Youdon’treallyneedthoseformulasasyoucanfindalltheaboveusingyour
calculator,stepsareshownbelow)
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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/$96000)*100=150%SincelocationBhasahigherCV(150%)thanlocationA(108%),locationAislessriskythanlocationB.
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EXAMPLE2Aninvestmentadvisorsendsinformationtovariousclientsviaexpressdeliverycompanies.Inallcasesitisimportantthattheclientreceivetheinformationonthedayitissent.Threecompaniesarecapableofprovidingthedeliveryservice.Thefollowinginformationisgiven:
Company Cost($) 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
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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≥ #))
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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
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***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
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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
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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?
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2. Theincreaseordecreaseinthepriceofastockbetweenthebeginningand
theendofatradingdayisassumedtobeanequallyrandomevent.Whatis
theprobabilitythatastockwillshowadecreaseinitsclosingpriceonten
consecutivedays?
3. Amanufacturingcompanyregularlyconductsqualitycontrolchecksat
specificperiodsontheproductsitmanufactures.Historically,thefailurerate
forLEDlightbulbsthatthecompanymanufacturesis13%.Supposea
randomsampleof10LEDlightbulbsisselected:
a) WhatistheprobabilitythatnoneoftheLEDlightbulbsaredefective?
b) WhatistheprobabilitythatexactlyoneoftheLEDlightbulbsisdefective?
c) WhatistheprobabilitythattwoorfeweroftheLEDlightbulbsaredefective?
d) WhatistheprobabilitythatthreeormoreoftheLEDlightbulbsare
defective?
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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:
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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.
IF YOU WOULD LIKE THE FULL TEST-3 COURSE PACKAGE, IT IS AVAILABLE FOR DOWNLOAD TO EVERYONE THAT REGISTERS
FOR THE QMS 102 FINAL EXAM CRASH COURSE!
REGISTER ONLINE: http://easygradetutorials.com/qms-102
-----------
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
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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≥ #))