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Lectures BIT2405QuantitativeMethodsI
Chapter5b
DiscreteProbabilityDistributions
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Chapter5
DiscreteProbabilityDistributions
KEY
Lecture #
Chapter
TEXT
Quantitative Methods I,
Anderson, Sweeney & Williams
Hawkes Learning Systems:
Statistics
HLS
LastTime
5.2
Expected Value and
Variance
5.4
Today
Discrete Probability
Distributions
5.3
09
Ch05b.ppt
Sectio
n
Topic
5.1
Discrete Random
Variables
5.2
08
Ch05a.ppt
Sectio
n
Topic
5.1
Random Variables
Binomial Probability
Distribution
The Binomial
Distribution
Chapter5
DiscreteProbabilityDistributions
s
s
s
s
RandomVariables
DiscreteProbabilityDistributions
ExpectedValueandVariance
BinomialProbabilityDistribution
.40
.30
.20
.10
01234
RandomVariables
Wetypicallylabeltherandomvariableasx
Arandomvariableisanumericaldescriptionofthe
outcomeofanexperiment.
Adiscreterandomvariablemayassumeeithera
finitenumberofvaluesoraninfinitesequenceof
values.
asequencetoolargetospecifyeachvalue
Acontinuousrandomvariablemayassumeany
numericalvalueinanintervalorcollectionof
intervals.
DiscreteandContinuousVariables.
0
1
2
WhetherBoundorUnbound
WhetherBound
3
4
DISCRETEandContinuousVariables.
0
1
2
3
4
DISCRETE:YouareCOUNTINGWHOLE
OBJECTS.
Youcoulddeviseascalewherebythetick
marksrepresentallpossiblevaluesofthe
outcomes.
Nopossibleoutcomeinrealitycouldfallbetween
thevaluesofyourscale.
EXAMPLES:People,votes,cars,lighteningstrikes,hurricanes...
DiscreteandCONTINUOUSVariables.
0
1
2
3
4
CONTINUOUS:YouareMEASURINGA
CHARACTERISTICofanOBJECT.
NomatterhowFINEASCALEyoudeviseto
measurethecharacteristic,itwouldbepossible
foravaluetoexistthatfallsBETWEENtwoof
thetickmarksonyourscale.
EXAMPLES:Height,weight,temperature,energy...
DiscreteProbabilityDistributions
Theprobabilitydistributionforarandomvariable
describeshowprobabilitiesaredistributedover
thevaluesoftherandomvariable.
Wecandescribeadiscreteprobabilitydistribution
withatable,graph,orequation.
DiscreteProbabilityDistributions
Theprobabilitydistributionisdefinedbya
probabilityfunction,denotedbyf(xi),whichprovides
theprobabilityforeachvalueoftherandomvariable.
Therequiredconditionsforadiscreteprobability
functionare:
0 f (x i ) 1
f (x ) = 1
i
NOTE: Hawkes uses P(x) instead of f(xi)
asthesymbolfortheDiscreteProbabilityFunction.
DiscreteProbabilityDistributions
s
Example:JSLAppliances
Probability
UsingpastdataonTVsales,
Random Variable
atabularrepresentationoftheprobability
distributionforTVsaleswasdeveloped.
Function
Number Relative
UnitsSoldofDays Frequency
0 80
.40
80/200
1 50
.25
2 40
.20
3 10 .05
4 20
.10
200
1.00
x f(x)
0
.40
1
.25
2
.20
3
.05
4
.10
1.00
DiscreteUniformProbabilityDistribution
Thediscreteuniformprobabilitydistributionisthe
simplestexampleofadiscreteprobability
distributiongivenbyaformula.
Thediscreteuniformprobabilityfunctionis
f(xi)=1/n
thevaluesofthe
randomvariable
areequallylikely
where:
n=thenumberofvaluestherandom
variablemayassume
ExpectedValueandVariance
Theexpectedvalue,ormean,ofarandomvariable
isameasureofitscentrallocation.
E(x)==xif(xi)
Thevariancesummarizesthevariabilityinthe
valuesofarandomvariable.
2=(xi)2f(xi)
Equivalentform(Hawkes)
2=[xi2f(xi)]2
Thestandarddeviation,,isdefinedasthepositive
squarerootofthevariance.
BinomialProbabilityDistribution
s
SpecialFORMofaDISCRETEPROBABILITY
DISTRIBUTION.
BinomialProbabilityDistribution
s
FourPropertiesofaBinomialExperiment
1.Theexperimentconsistsofasequenceofn
identicaltrials.
2.Twooutcomes,successandfailure,arepossible
oneachtrial.
3.Theprobabilityofasuccess,denotedbyp,does
notchangefromtrialtotrial.
stationarity
assumption
4.Thetrialsareindependent.
These are IMPORTANT for you to Understand and to Remember so You
Can RECOGINZE Binomial Experiments.
BinomialProbabilityDistribution
Ourinterestisinthenumberofsuccesses
occurringinthentrials.
Weletxdenotethenumberofsuccesses
occurringinthentrials.
BinomialProbabilityDistribution
s
BinomialProbabilityFunction
xi
n!
f (x i ) =
p (1 p )( n x i )
x i !( n x i )!
where:
f(xi)=theprobabilityofxisuccessesinntrials
n=thenumberoftrials
p=theprobabilityofsuccessonanyonetrial
BinomialProbabilityDistribution
s
BinomialProbabilityFunction
xi
n!
f (x i ) =
p (1 p)( n x i )
x i !(n x i )!
Numberofexperimental
outcomesprovidingexactly
xsuccessesinntrials
Probabilityofaparticular
sequenceoftrialoutcomes
withxsuccessesinntrials
BinomialProbabilityDistribution
s
BinomialProbabilityFunction
xi
n!
f (x i ) =
p (1 p)( n x i )
x i !(n x i )!
Numberofexperimental
outcomesprovidingexactly
xsuccessesinntrials
Experiment
Combination
Probabilityofaparticular
sequenceoftrialoutcomes
withxsuccessesinntrials
CountingRulefor
NumberofSamplePoints
C=
N
n
()
N
n
N!
=
n!( N n )!
WeUsetheCombinationNotationinthef(xi)
Equation:nchoosexi
s
BinomialProbabilityFunction
n xi
( n xi )
f ( x i ) = p (1 p)
xi
Experiment
Combination
CountingRulefor
NumberofSamplePoints
C=
N
n
()
N
n
N!
=
n!( N n )!
BinomialProbabilityDistribution
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
BinomialProbabilityDistribution
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
IsthisaDiscreteBinomialProbabilityExperiment?
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
BinomialProbabilityDistribution
s
FourPropertiesofaBinomialExperiment
1.Theexperimentconsistsofasequenceofn
identicaltrials.
2.Twooutcomes,successandfailure,arepossible
oneachtrial.
3.Theprobabilityofasuccess,denotedbyp,does
notchangefromtrialtotrial.
4.Thetrialsareindependent.
1.Theexperimentconsistsofasequenceofn
IsthisaDiscreteBinomialProbabilityExperiment?
identicaltrials.
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
Each employee chosen at RANDOM is a TRIAL.
2.Twooutcomes,successandfailure,arepossible
IsthisaDiscreteBinomialProbabilityExperiment?
oneachtrial.
Notice that success for the experimentisnotnecessarilywhatweinsociety
s
Example:EvansElectronics
wouldcallaSUCCESSFULOUTCOME.
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
Each employee will either LEAVE or NOT LEAVE.
thisyear?
success or
failure
3.Theprobabilityofasuccess,denotedbyp,does
IsthisaDiscreteBinomialProbabilityExperiment?
notchangefromtrialtotrial.
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
For each employee chosen at Random (each TRIAL),
the probability that person leaves within the year is
0.1.
IsthisaDiscreteBinomialProbabilityExperiment?
4.Thetrialsareindependent.
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
Since each employee is chosen at RANDOM, we can
assume they are INDEPENDENT.
IsthisaDiscreteBinomialProbabilityExperiment?
YES,WecanASSUMEitis.
s
Example:EvansElectronics
Evansisconcernedaboutalowretentionratefor
employees.Inrecentyears,managementhasseena
turnoverof10%ofthehourlyemployeesannually.
Thus,foranyhourlyemployeechosenatrandom,
managementestimatesaprobabilityof0.1thatthe
personwillnotbewiththecompanynextyear.
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
BinomialProbabilityDistribution
s
Example:EvansElectronics
Choosing3hourlyemployeesatrandom,whatis
theprobabilitythat1ofthemwillleavethecompany
thisyear?
Let:p=.10,n=3,xi=1
Usingthe
binomial
probability
function
xi
n!
p (1 p) n xi
f ( xi ) =
x !( n x )!
i
i
3!
f (1) =
(0.1)1 (0.9)2 = 3(.1)(.81) = .243
1!(3 1)!
BinomialProbabilityDistribution
s
Usingatreediagram
Example:EvansElectronics
Leaves
(.1)
3
f(x)
.0010
2
.0090
L(.1)
2
.0090
1
.0810
L(.1)
2
.0090
S(.9)
Leaves(.1)
x
S(.9)
2ndWorker
3rdWorker
L(.1)
S(.9)
1stWorker
1
.0810
1
.0810
0
.7290
Stays(.9)
Leaves(.1)
Stays
(.9)
L(.1)
Stays(.9)
S(.9)
0.0810 + 0.0810 + 0.0810 = 0.243
BinomialProbabilityDistribution
1stWorker
Leaves(.1)
Leaves
(.1)
x
3
f(x)
.0010
2
.0090
1
.0810
L(.1)
2
.0090
S(.9)
2ndWorker
1
.0810
1
.0810
0
.7290
3rdWorker
L(.1)
(p)(1p)(1p)
L(.1)
.0090
2
S(.9)
Stays(.9)
S(.9)
Leaves(.1)
Stays
(.9)
L(.1)
Stays(.9)
S(.9)
BinomialProbabilityDistribution
1stWorker
Leaves(.1)
Leaves
(.1)
x
3
f(x)
.0010
2
.0090
1
.0810
L(.1)
2
.0090
S(.9)
2ndWorker
1
.0810
1
.0810
0
.7290
3rdWorker
L(.1)
(p)(1p)(1p)
(p)1(1p)2.0090
L(.1)
2
S(.9)
Stays(.9)
S(.9)
Leaves(.1)
Stays
(.9)
L(.1)
Stays(.9)
S(.9)
BinomialProbabilityDistribution
1stWorker
Notice there are 3 2ndWorker with 3rdWorker
outcomes
L(.1)
1 success
Leaves(.1)
AND
S(.9)
the probability of each is the same.
Leaves
(.1)
Stays(.9)
3Stays (p)1(1-p)2
x
(.9)
3
f(x)
.0010
(p)1(1p)2.0090
L(.1)
2
2
.0090
1
.0810
L(.1)
2
.0090
S(.9)
1
.0810
1
.0810
0
.7290
Therefore,theprobabilityof1success
S(.9)
wouldbe
Leaves(.1)
x
L(.1)
Stays(.9)
S(.9)
BinomialProbabilityDistribution
xi
n!
( n x i )
f (x i ) =
p (1 p )
x i !(n x i )!
Notice there are 3 outcomes with
1 success
AND
the probability of each is the same.
Therefore,theprobabilityof1success
wouldbe
3 x (p)1(1-p)2
BinomialProbabilityDistribution
xi
n!
( n x i )
f (x i ) =
p (1 p )
x i !(n x i )!
3 x (p)1(1-p)2
Let:p=.10,n=3,x=1
n!
f ( x) =
p x (1 p ) (n x )
x !( n x ) !
3!
f (1) =
(0.1)1 (0.9)2 = 3(.1)(.81) = .243
1!(3 1)!
BinomialProbabilityDistribution
s
ExpectedValue
E(x)==np
s
Variance
2=np(1p)
s
StandardDeviation
= np(1 p)
BinomialProbabilityDistribution
Binomial
s
ExpectedValue
E(x)==np
s
E(x)==xif(xi)
Variance
2=np(1p)
s
Discrete in General
StandardDeviation
= np(1 p)
2=(xi)2f(xi)
Binomial is a Special Case of
Discrete so the formulas can be
simplified.
BinomialProbabilityDistribution
s
ExpectedValue
E(x)==np
s
Example:EvansElectronics
E(x)==3(.1)=.3employeesoutof3
Variance
2=np(1p)
2=3(.1)(.9)=.27
s
StandardDeviation
= np(1 p)
Questions?
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