﻿
231
P(x>2)=(base)(height)=(4 2)(0.4)=0.8
Example4Figure1
probabilitythattherepairtimexisgreaterthan2
Problem2
Findtheprobabilitythatarandomlyselectedfurnacerepairrequireslessthan3hours.Describe
howthegraphdiffersfromthegraphintheﬁrstpartofthisexample.
Solution
P(x<3)=(base)(height)=(3 1.5)(0.4)=0.6
Thegraphoftherectangleshowingtheentiredistributionwouldremainthesame.Howeverthe
thanatx=0;sinceXU(1.5,4),xcannotbelessthan1.5.
Example4Figure2
probabilitythattherepairtimexislessthan3
Problem3
Findthe30thpercentileoffurnacerepairtimes.
Solution
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232
CHAPTER5. CONTINUOUSRANDOMVARIABLES
Example4Figure3
shortest30%ofrepairtimes.
P(x<k)=0.30
P(x<k)=(base)(height)=(1.5)(0.4)
0.3=(k 1.5)(0.4);Solvetoﬁndk:
0.75=k 1.5,obtainedbydividingbothsidesby0.4
The30thpercentileofrepairtimesis2.25hours.30%ofrepairtimesare2.5hoursorless.
Problem4
The longest25% offurnacerepairtimestakeatleast howlong? ? (Findtheminimumtimefor
thelongest25%ofrepairs.)
Solution
Example4Figure4
thelongest25%ofrepairtimes.
P(x>k)=0.25
P(x>k)=(base)(height)=(k)(0.4)
0.25=(4 k)(0.4);Solvefork:
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233
0.625=4 k,obtainedbydividingbothsidesby0.4
3.375= k,obtainedbysubtracting4frombothsides
k=3.375
Thelongest25%offurnacerepairstakeatleast3.375hours(3.375hoursorlonger).
Note: Since25%ofrepairtimesare3.375hoursorlonger,thatmeansthat75%ofrepairtimesare
3.375hoursorless.3.375hoursisthe75thpercentileoffurnacerepairtimes.
Problem5
Findthemeanandstandarddeviation
Solution
m=
a+b
2
ands=
q
(b a)
2
12
m=
1.5+4
2
=2.75hoursands=
q
(4 1.5)
2
12
=0.7217hours
NOTE
:See"SummaryoftheUniformandExponentialProbabilityDistributions(Section5.5)"for
afullsummary.
**Example5contributedbyRobertaBloom
5.4TheExponentialDistribution
4
Theexponentialdistributionisoftenconcernedwiththeamountoftimeuntilsomespeciﬁceventoccurs.
Forexample, theamountoftime(beginningnow)untilanearthquakeoccurshasanexponentialdistri-
amountoftime,inmonths,acarbatterylasts. Itcanbeshown,too,thatthevalueofthechangethatyou
haveinyourpocketorpurseapproximatelyfollowsanexponentialdistribution.
Valuesforanexponentialrandomvariableoccurinthefollowingway. Therearefewerlargevaluesand
followsanexponentialdistribution. Therearemorepeoplethatspendlessmoneyandfewerpeoplethat
spendlargeamountsofmoney.
Theexponentialdistributioniswidelyusedintheﬁeldofreliability. Reliabilitydealswiththeamountof
timeaproductlasts.
Example5.7
Illustratestheexponentialdistribution: Let X=amountoftime (inminutes)apostalclerk
spendswithhis/hercustomer. Thetimeisknowntohaveanexponentialdistributionwiththe
averageamountoftimeequalto4minutes.
Xisacontinuousrandomvariablesincetimeismeasured. Itisgiventhatm=4minutes. Todo
anycalculations,youmustknowm,thedecayparameter.
m=
1
m
.Therefore,m=
1
4
=0.25
4
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234
CHAPTER5. CONTINUOUSRANDOMVARIABLES
Thestandarddeviation,s,isthesameasthemean.m=s
ThedistributionnotationisX~Exp(m).Therefore,X~Exp(0.25).
Theprobabilitydensityfunctionisf(x)=me
mx
Thenumbere=2.71828182846...Itisanumber
thatisusedofteninmathematics.Scientiﬁccalculatorshavethekey"e
x
."Ifyouenter1forx,the
calculatorwilldisplaythevaluee.
Thecurveis:
f(x)=0.25e
0.25x
wherexisatleast0andm=0.25.
Forexample,f(5)=0.25e
0.255
=0.072
Thegraphisasfollows:
f(x)=0.25e
0.250
=0.251=0.25=m
Example5.8
Problem1
Findtheprobabilitythataclerkspendsfourtoﬁveminuteswitharandomlyselectedcustomer.
Solution
FindP(4<x<5).
Thecumulativedistributionfunction(CDF)givestheareatotheleft.
P(x<x)=e
mx
P(x<5)=e
0.255
=0.7135andP(x<4)=1 e
0.254
=0.6321
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235
NOTE
:Youcandothesecalculationseasilyonacalculator.
Theprobabilitythatapostalclerkspendsfourtoﬁveminuteswitharandomlyselectedcustomer
is
P(4<x<5)=P(x<5P(x<4)=0.7135 0.6321=0.0814
NOTE
:TI-83+andTI-84:Onthehomescreen,enter(1-e^(-.25*5))-(1-e^(-.25*4))orentere^(-.25*4)-
e^(-.25*5).
Problem2
Halfofallcustomersareﬁnishedwithinhowlong?(Findthe50thpercentile)
Solution
Findthe50thpercentile.
P(x<k)=0.50,k=2.8minutes(calculatororcomputer)
Halfofallcustomersareﬁnishedwithin2.8minutes.
Youcanalsodothecalculationasfollows:
P(x<k)=0.50andP(x<k)=e
0.25k
Therefore,0.50=e
0.25k
ande
0.25k
=1 0.50=0.5
Takenaturallogs:ln
e
0.25k
=ln
(0.50).So,
0.25k=ln
(0.50)
Solvefork:k=
ln(.50)
0.25
=2.8minutes
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236
CHAPTER5. CONTINUOUSRANDOMVARIABLES
NOTE
:Aformulaforthepercentilekisk=
LN(AreaToTheLeft)
m
whereLNisthenaturallog.
NOTE
:TI-83+andTI-84:Onthehomescreen,enterLN(1-.50)/-.25.Pressthe(-)forthenegative.
Problem3
Whichislarger,themeanorthemedian?
Solution
Isthemeanormedianlarger?
Frompartb,themedianor50thpercentileis2.8minutes.Thetheoreticalmeanis4minutes.The
meanislarger.
5.4.1OptionalCollaborativeClassroomActivity
Haveeachclassmembercountthechangehe/she hasinhis/herpocketorpurse. . Yourinstructorwill
recordtheamountsindollarsandcents. Constructahistogramofthe e datatakenbythe class. . Use5
intervals.Drawasmoothcurvethroughthebars.Thegraphshouldlookapproximatelyexponential.Then
calculatethemean.
LetX=theamountofmoneyastudentinyourclasshasinhis/herpocketorpurse.
ThedistributionforXisapproximatelyexponentialwithmean,m=_______andm=_______.Thestandard
deviation,s=________.
Drawtheappropriateexponentialgraph.Youshouldlabelthexandyaxes,thedecayrate,andthemean.
Example5.9
Ontheaverage,acertaincomputerpartlasts10years.Thelengthoftimethecomputerpartlasts
isexponentiallydistributed.
Problem1
Whatistheprobabilitythatacomputerpartlastsmorethan7years?
Solution
Letx=theamountoftime(inyears)acomputerpartlasts.
m=10som=
1
m
=
1
10
=0.1
FindP(x>7).Drawagraph.
P(x>7)=P(x<7).
SinceP(X<x)=e
mx
thenP(X>x)=(e
mx
)=e
mx
P(x>7)=e
0.17
=0.4966.Theprobabilitythatacomputerpartlastsmorethan7yearsis0.4966.
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237
NOTE
:TI-83+andTI-84:Onthehomescreen,entere^(-.1*7).
Problem2
Ontheaverage,howlongwould5computerpartslastiftheyareusedoneafteranother?
Solution
Ontheaverage,1computerpartlasts10years.Therefore,5computerparts,iftheyareusedone
rightaftertheotherwouldlast,ontheaverage,
(5)(10)=50years.
Problem3
Eightypercentofcomputerpartslastatmosthowlong?
Solution
Findthe80thpercentile.Drawagraph.Letk=the80thpercentile.
Solvefork:k=
ln(.80)
0.1
=16.1years
Eightypercentofthecomputerpartslastatmost16.1years.
NOTE
:TI-83+andTI-84:Onthehomescreen,enterLN(1-.80)/-.1
Problem4
Whatistheprobabilitythatacomputerpartlastsbetween9and11years?
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238
CHAPTER5. CONTINUOUSRANDOMVARIABLES
Solution
FindP(9<x<11).Drawagraph.
P(9<x<11) = P(x<11P(x<9) =
e
0.111
e
0.19
0.6671 0.5934 =
0.0737.(calculatororcomputer)
Theprobabilitythatacomputerpartlastsbetween9and11yearsis0.0737.
NOTE
:TI-83+andTI-84:Onthehomescreen,entere^(-.1*9)-e^(-.1*11).
Example5.10
Supposethatthelengthofaphonecall,inminutes,isanexponentialrandomvariablewithdecay
parameter=
1
12
.Ifanotherpersonarrivesatapublictelephonejustbeforeyou,ﬁndtheprobability
thatyouwillhavetowaitmorethan5minutes.LetX=thelengthofaphonecall,inminutes.
Problem
(Solutiononp.257.)
Whatism,m,ands?Theprobabilitythatyoumustwaitmorethan5minutesis_______.
NOTE
: Asummaryfor r exponentialdistributionisavailable in"SummaryofTheUniformand
ExponentialProbabilityDistributions(Section5.5)".
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239
5.5SummaryoftheUniformandExponentialProbabilityDistributions
5
Formula5.1:Uniform
X=arealnumberbetweenaandb(insomeinstances, Xcantakeonthevaluesaandb). a=
smallestX;b=largestX
XU(a,b)
Themeanism=
a+b
2
Thestandarddeviationiss=
q
(b a)
2
12
Probabilitydensityfunction: f(X)=
1
b a
foraXb
AreatotheLeftofx:P(X<x)=(base)(height)
AreatotheRightofx:P(X>x)=(base)(height)
AreaBetweencandd:P(c<X<d)=(base)(height)=(c)(height).
Formula5.2:Exponential
XExp(m)
X=arealnumber,0orlarger.m=theparameterthatcontrolstherateofdecayordecline
Themeanandstandarddeviationarethesame.
m=s=
1
m
andm=
1
m
=
1
s
Theprobabilitydensityfunction: f(X)=me
mX
,X0
AreatotheLeftofx:P(X<x)=e
mx
AreatotheRightofx:P(X>x)=e
mx
AreaBetweencandd: P(c<X<d)=P(X<dP(X<c)=
e
md
(1 e
mc
)=
e
mc
e
md
Percentile,k:k=
LN(1-AreaToTheLeft)
m
5
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