The exponential distribution is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) ...
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IntroductiontoStatistics
Module5:ContinuousRandomVariables
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TheExponentialDistribution
LearningOutcomes
Recognizetheexponentialprobabilitydistributionandapplyitappropriately
The exponentialdistributionisoftenconcernedwiththeamountoftimeuntilsomespecificeventoccurs.Forexample,theamountoftime(beginningnow)untilanearthquakeoccurshasanexponentialdistribution.Otherexamplesincludethelength,inminutes,oflongdistancebusinesstelephonecalls,andtheamountoftime,inmonths,acarbatterylasts.Itcanbeshown,too,thatthevalueofthechangethatyouhaveinyourpocketorpurseapproximatelyfollowsanexponentialdistribution.
Valuesforanexponentialrandomvariableoccurinthefollowingway.Therearefewerlargevaluesandmoresmallvalues.Forexample,theamountofmoneycustomersspendinonetriptothesupermarketfollowsanexponentialdistribution.Therearemorepeoplewhospendsmallamountsofmoneyandfewerpeoplewhospendlargeamountsofmoney.
Theexponentialdistributioniswidelyusedinthefieldofreliability.Reliabilitydealswiththeamountoftimeaproductlasts.
Example
Let X=amountoftime(inminutes)apostalclerkspendswithhisorhercustomer.Thetimeisknowntohaveanexponentialdistributionwiththeaverageamountoftimeequaltofourminutes.
Xisacontinuousrandomvariablesincetimeismeasured.Itisgiventhatμ=4minutes.Todoanycalculations,youmustknowm,thedecayparameter.
[latex]{m}=\frac{1}{\mu}[/latex].Therefore, [latex]{m}=\frac{1}{4}={0.25}[/latex]
Thestandarddeviation,σ,isthesameasthemean.μ=σ
ThedistributionnotationisX~Exp(m).Therefore,X~Exp(0.25).
Theprobabilitydensityfunctionisf(x)=me–mx.Thenumbere=2.71828182846…Itisanumberthatisusedofteninmathematics.Scientificcalculatorshavethekey“ex.”Ifyouenteroneforx,thecalculatorwilldisplaythevaluee.
Thecurveis:
f(x)=0.25e–0.25xwherexisatleastzeroandm=0.25.
Forexample,f(5)=0.25e−(0.25)(5)=0.072.Thepostalclerkspendsfiveminuteswiththecustomers. Thegraphisasfollows:
Noticethegraphisadecliningcurve.Whenx=0,
f(x)=0.25e(−0.25)(0)=(0.25)(1)=0.25=m.Themaximumvalueonthey-axisism.
Example
Usingtheinformationinexample1,findtheprobabilitythataclerkspendsfourtofiveminuteswitharandomlyselectedcustomer.
Thecurveis:
X~Exp(0.125);f(x)=0.125e–0.125x
a)FindP(47).Drawthegraph.
P(x>7)=1–P(x<7).
SinceP(Xx)=1–(1–e–mx)=e-mx
P(x>7)=e(–0.1)(7)=0.4966.Theprobabilitythatacomputerpartlastsmorethansevenyearsis0.4966.
Onthehomescreen,entere^(-.1*7).
b) Ontheaverage,howlongwouldfivecomputerpartslastiftheyareusedoneafteranother?
Solution:
Ontheaverage,onecomputerpartlaststenyears.Therefore,fivecomputerparts,iftheyareusedonerightaftertheotherwouldlast,ontheaverage,(5)(10)=50years.
c) Eightypercentofcomputerpartslastatmosthowlong?
Solution:
Findthe80thpercentile.Drawthegraph.Letk=the80thpercentile.
Solvefork: [latex]{k}=\frac{ln(1-0.80)}{-0.1}={16.1}[/latex]
Eightypercentofthecomputerpartslastatmost16.1years.
Solution:
FindP(95)=0.6592
Example
Thetimespentwaitingbetweeneventsisoftenmodeledusingtheexponentialdistribution.Forexample,supposethatanaverageof30customersperhourarriveatastoreandthetimebetweenarrivalsisexponentiallydistributed.
Onaverage,howmanyminuteselapsebetweentwosuccessivearrivals?
Whenthestorefirstopens,howlongonaveragedoesittakeforthreecustomerstoarrive?
Afteracustomerarrives,findtheprobabilitythatittakeslessthanoneminuteforthenextcustomertoarrive.
Afteracustomerarrives,findtheprobabilitythatittakesmorethanfiveminutesforthenextcustomertoarrive.
Seventypercentofthecustomersarrivewithinhowmanyminutesofthepreviouscustomer?
Isanexponentialdistributionreasonableforthissituation?
Solutions:
Sinceweexpect30customerstoarriveperhour(60minutes),weexpectonaverageonecustomertoarriveeverytwominutesonaverage.
Sinceonecustomerarriveseverytwominutesonaverage,itwilltakesixminutesonaverageforthreecustomerstoarrive.
LetX=thetimebetweenarrivals,inminutes.Byparta,μ=2,som=12=0.5.
Therefore,X∼Exp(0.5).ThecumulativedistributionfunctionisP(X5)=1–P(X<5)=1–(1–e(–5)(0.5))=e–2.5≈0.0821.
Wewanttosolve0.70=P(Xr+t|X>r)=P(X>t)forallr≥0andt≥0
Forexample,iffiveminuteshaselapsedsincethelastcustomerarrived,thentheprobabilitythatmorethanoneminutewillelapsebeforethenextcustomerarrivesiscomputedbyusingr=5andt=1intheforegoingequation.
P(X>5+1|X>5)=P(X>1)=e(–0.5)(1)≈0.6065.
Thisisthesameprobabilityasthatofwaitingmorethanoneminuteforacustomertoarriveafterthepreviousarrival.
Theexponentialdistributionisoftenusedtomodelthelongevityofanelectricalormechanicaldevice.Inexample1,thelifetimeofacertaincomputerparthastheexponentialdistributionwithameanoftenyears(X~Exp(0.1)).Thememorylesspropertysaysthatknowledgeofwhathasoccurredinthepasthasnoeffectonfutureprobabilities.Inthiscaseitmeansthatanoldpartisnotanymorelikelytobreakdownatanyparticulartimethanabrandnewpart.Inotherwords,thepartstaysasgoodasnewuntilitsuddenlybreaks.Forexample,iftheparthasalreadylastedtenyears,thentheprobabilitythatitlastsanothersevenyearsisP(X>17|X>10)=P(X>7)=0.4966.
Example
Refertoexample1,wherethetimeapostalclerkspendswithhisorhercustomerhasanexponentialdistributionwithameanoffourminutes.Supposeacustomerhasspentfourminuteswithapostalclerk.Whatistheprobabilitythatheorshewillspendatleastanadditionalthreeminuteswiththepostalclerk?
ThedecayparameterofXism=14=0.25,soX∼Exp(0.25).
ThecumulativedistributionfunctionisP(X7|X>4).ThememorylesspropertysaysthatP(X>7|X>4)=P(X>3),sowejustneedtofindtheprobabilitythatacustomerspendsmorethanthreeminuteswithapostalclerk.
ThisisP(X>3)=1–P(X<3)=1–(1–e–0.25⋅3)=e–0.75≈0.4724.
RelationshipbetweenthePoissonandtheExponentialDistribution
ThereisaninterestingrelationshipbetweentheexponentialdistributionandthePoissondistribution.Supposethatthetimethatelapsesbetweentwosuccessiveeventsfollowstheexponentialdistributionwithameanofμunitsoftime.Alsoassumethatthesetimesareindependent,meaningthatthetimebetweeneventsisnotaffectedbythetimesbetweenpreviousevents.Iftheseassumptionshold,thenthenumberofeventsperunittimefollowsaPoissondistributionwithmeanλ=1/μ.Recall thatifXhasthePoissondistributionwithmeanλ,then[latex]P(X=k)=\frac{{\lambda}^{k}{e}^{-\lambda}}{k!}[/latex].Conversely,ifthenumberofeventsperunittimefollowsaPoissondistribution,thentheamountoftimebetweeneventsfollowstheexponentialdistribution.(k!=k*(k-1*)(k–2)*(k-3)…3*2*1)
Example
Atapolicestationinalargecity,callscomeinatanaveragerateoffourcallsperminute.Assumethatthetimethatelapsesfromonecalltothenexthastheexponentialdistribution.Takenotethatweareconcernedonlywiththerateatwhichcallscomein,andweareignoringthetimespentonthephone.Wemustalsoassumethatthetimesspentbetweencallsareindependent.Thismeansthataparticularlylongdelaybetweentwocallsdoesnotmeanthattherewillbeashorterwaitingperiodforthenextcall.WemaythendeducethatthetotalnumberofcallsreceivedduringatimeperiodhasthePoissondistribution.
Findtheaveragetimebetweentwosuccessivecalls.
Findtheprobabilitythatafteracallisreceived,thenextcalloccursinlessthantenseconds.
Findtheprobabilitythatexactlyfivecallsoccurwithinaminute.
Findtheprobabilitythatlessthanfivecallsoccurwithinaminute.
Findtheprobabilitythatmorethan40callsoccurinaneight-minuteperiod.
Solutions:
Onaveragetherearefourcallsoccurperminute,so15seconds,or[latex]\frac{15}{60}[/latex]=0.25minutes occurbetweensuccessivecallsonaverage.
LetT=timeelapsedbetweencalls. Fromparta,[latex]\mu={0.25}[/latex],som=[latex]\frac{1}{0.25}[/latex]=4.Thus,T~Exp(4). ThecumulativedistributionfunctionisP(T40)=1–P(Y≤40)=1–0.9294=0.0707.
ConceptReview
IfXhasanexponentialdistributionwithmean[latex]\mu[/latex]thenthedecayparameteris[latex]m=\frac{1}{\mu}[/latex], andwewriteX∼Exp(m)wherex≥0andm>0.TheprobabilitydensityfunctionofXisf(x)= me-mx(orequivalently[latex]f(x)=\frac{1}{\mu}{e}^{\frac{-x}{\mu}}[/latex].ThecumulativedistributionfunctionofXisP(X≤x)=1–e–mx.
Theexponentialdistributionhasthememorylessproperty,whichsaysthatfutureprobabilitiesdonotdependonanypastinformation.Mathematically,itsaysthatP(X>x+k|X>x)=P(X>k).
IfTrepresentsthewaitingtimebetweenevents,andifT∼Exp(λ),thenthenumberofeventsXperunittimefollowsthePoissondistributionwithmeanλ.Theprobabilitydensityfunctionof[latex]P\left(X=k\right)=\frac{\lambda^{k}}{e^{-\lambda}}k![/latex]. ThismaybecomputedusingaTI-83,83+,84,84+calculatorwiththecommandpoissonpdf(λ,k).ThecumulativedistributionfunctionP(X≤k)maybecomputedusingtheTI-83,83+,84,84+calculatorwiththecommandpoissoncdf(λ,k).
FormulaReview
Exponential:X~Exp(m)wherem=thedecayparameter
pdf:f(x)=m[latex]{e}^{-mx}[/latex] wherex≥0andm>0
cdf:P(X≤x)=1–[latex]{e}^{-mx}[/latex]
mean[latex]\mu=\frac{1}{m}[/latex]
standarddeviationσ=µ
percentile,k:k=[latex]\frac{ln(\text{AreaToTheLeftOfK})}{-m}[/latex]
Additionally
P(X>x)=e(–mx)
P(ax+k|X>x)=P(X>k)
Poissonprobability: P(X=k)=[latex]\frac{{\lambda}^{k}{e}^{-\lambda}}{k!}[/latex]withmean[latex]\lambda[/latex]
k!=k*(k-1)*(k-2)*(k-3)…3*2*1
References
DatafromtheUnitedStatesCensusBureau.
DatafromWorldEarthquakes,2013.Availableonlineathttp://www.world-earthquakes.com/(accessedJune11,2013).
“No-hitter.”Baseball-Reference.com,2013.Availableonlineathttp://www.baseball-reference.com/bullpen/No-hitter(accessedJune11,2013).
Zhou,Rick.“ExponentialDistributionlectureslides.”Availableonlineatwww.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf(accessedJune11,2013).
LicensesandAttributions
CClicensedcontent,SharedpreviouslyOpenStax,TheExponentialDistribution.Providedby:OpenStax.Locatedat:http://cnx.org/contents/[email protected]:37/Introductory_Statistics.License:CCBY:AttributionIntroductoryStatistics.Authoredby:BarbaraIllowski,SusanDean.Providedby:OpenStax.Locatedat:http://cnx.org/contents/[email protected]:CCBY:Attribution.LicenseTerms:Downloadforfreeathttp://cnx.org/contents/[email protected]
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