13/07/54
CHS316StatisticsforChemicalEngineering,Semester1,AY2011 CHS316StatisticsforChemicalEngineering,Semester1,AY2011
Dr.Wanwipa Siriwatwechakul Dr.Wanwipa Siriwatwechakul
Example332:CalculationsforWireFlaws1 Example332:CalculationsforWireFlaws2
Forthecaseofthethincopperwire,supposethatthe Determinetheprobabilityof10flawsin5mmofwire.
numberofflawsfollowsaPoissondistributionof2.3 NowletXdenotethenumberofflawsin5mmof
flawspermm.LetX denotethenumberofflawsin1 wire.
mmofwire.Findtheprobabilityofexactly2flawsin1 Answer: E ( X ) = = 5 mm 2.3 flaws/mm =11.5 flaws
mmofwire.
11.510
e 2.3 2.32 P ( X = 10 ) = e 11.5 = 0.113
Answer: P ( X = 2) = = 0.265 10!
2!
InExcel
0.1129 =POISSON(10,11.5,FALSE)
Sec 23-9 Poisson Distribution 76 Sec 23-9 Poisson Distribution 77
CHS316StatisticsforChemicalEngineering,Semester1,AY2011 CHS316StatisticsforChemicalEngineering,Semester1,AY2011
Dr.Wanwipa Siriwatwechakul Dr.Wanwipa Siriwatwechakul
Example332:CalculationsforWireFlaws3 PoissonMean&Variance
Determinetheprobabilityofatleast1flawin2mmof IfXisaPoissonrandomvariablewithparameter,then:
wire.NowletXdenotethenumberofflawsin2mm
ofwire.NotethatP(X1)requiresinfiniteterms./ =E(X)= and2=V(X)= (317)
Answer: E ( X ) = = 2 mm 2.3 flaws/mm =4.6 flaws
ThemeanandvarianceofthePoissonmodelarethesame.
4.60
P ( X 1) = 1 P ( X = 0 ) = 1 e 4.6
= 0.9899 Ifthemeanandvarianceofadatasetarenotaboutthe
0! same,thenthePoissonmodelwouldnotbeagood
representation of that set
representationofthatset.
InExcel
0.989948 =1POISSON(0,4.6,FALSE)
Thederivationofthemeanandvarianceisshowninthetext.
Sec 23-9 Poisson Distribution 79 Sec 2- 80
1
13/07/54
CHS316StatisticsforChemicalEngineering,Semester1,AY2011 CHS316StatisticsforChemicalEngineering,Semester1,AY2011
Dr.Wanwipa Siriwatwechakul Dr.Wanwipa Siriwatwechakul
Exercise1 Exercise2
Supposethatthenumberofcustomerswho Thenumberoftelephonecallsthatarrivesataphoneswitchboard
ismodeledasaPoissonrandomvariables.Assumethatonthe
enterabankinanhourisaPoissonrandom average, there are 20 phone calls per hours.
average,thereare20phonecallsperhours.
variableandthatP(X=0) =0.04.Determinethe a) What is the probability that a) Let X denote the number of calls in one hour. Then, X
is a Poisson random variable with = 20.
meanandvarianceofX there are exactly 18 calls in 1
hour? e20 2018
P(X = 18) = = 0.0844
18!
P(X = 0) = exp(-). b) What is the probability that b) = 10 for a thirty minute period.
there are 3 or fewer calls in e10101 e10102 e10103
P(X 3) = e 20 + + + = 0.0103
Therefore, = ln(0.04) = 3.219. 30 minutes? 1! 2! 3!
c) Let Y denote the number of calls in two hours.
c) What is the probability that Then, Y is a Poisson random variable
are exactly 30 calls in 2 hrs? e40 4030
P(Y = 30) = = 0.0185
Consequently, E(X) = V(X) = 3.219. 30!
d) What is the probability that d) Let W denote the number of calls in 30 minutes. Then W
there are exactly 10 calls in is a Poisson random variable with = 10.
30 minutes?
e101010
P(W = 10) = = 0.1251
82 10! 84
