Download CONTINUOUS RANDOM VARIABLES Continuous Probability

ContinuousProbabilityDistributions
y Recall that a random variable is discrete if the range of values
that it takes on is finite. This refers to the number of values X
can take on,, not the size of the values. On the contrary,
y, a
continuous random variable can assume any value in an interval
or in a collection of intervals. Typically an uncountably infinite
range can result from a random variable that makes a physical
measurement — the position,
position size,
size time,
time age,
age flow,
flow volume,
volume or
area of something.
y For a continuous random variable, it is not possible to talk
about the probability of the random variable assuming a
particular value.
y Instead, we talk about the probability of the random variable
assuming a value within a given interval.
CONTINUOUSRANDOMVARIABLES
y Thus the probability of the random variable assuming a value
within some given interval from a to b is defined to be the area
under the ggraph
p of the p
probabilityy densityy function between a
and b.
2
Continuousprobabilitydistribution
Continuousprobabilitydistributions
Twocharacteristics
y Theprobabilityoftherandomvariableassumingavalue
Theprobabilitythat
Th
b bilit th t x assumesavalueinany
l i
intervalliesintherangeoffrom0to1.
2 Thetotalprobabilityofalltheintervals
2.
The total probability of all the intervals
withinwhichx canassumeavalueof1.
1
1.
f (x)
withinsomegivenintervalfromx
i hi
i
i
lf
i d fi d b
1 tox2 isdefinedtobe
theareaunderthegraphoftheprobabilitydensity
functionbetweenx1 andx2.
f (x) Exponential
Uniform
f (x )
x1 x2
Normal
x
x1 x2
x1
3
x2
x
x
4
Someexamples
CumulativeDistributionFunction
y ThecumulativedistributionfunctionF ofacontinuous
a
P(x<a)
random variable has the same definition as that for a
randomvariablehasthesamedefinitionasthatfora
discreterandomvariable.Thatis,
F ( x) P ( X d x)
y InpracticethismeansthatFisessentiallyaparticular
h
h
ll
l
“antiderivative”ofthepdf since
b
F ( x)
P(x>b)
³
x
f
f ( t ) dt
y ThusatthepointswherefiscontinuousF’(x)=f(x).
y Knowingthecdf
Knowing the cdf ofarandomvariablegreatly
of a random variable greatly
Noticethatforacontinuousrandomvariablex,
(
)
P(X=a)=0
foranyspecificvalueabecausethe“areaaboveapoint”underthecurve
isalinesegmentandhencehas0area.Specificallythismeans
P(X<a)=P(Xa)
P(a<X<b)=P(a X<b)=P(a<X b)=P(aX b)
5
P ( X d x)
facilitatescomputationofprobabilitiesinvolvingthat
randomvariablesince
6
P ( a d X d b)
F (b) F ( a )
MethodofProbabilityCalculation
MeanandVarianceofacontinuousrandom
variable
y Theprobabilitythatacontinuousrandomvariable
y TheexpectedvalueofacontinuousrandomvariableXisdefined
by
xliesbetweenalowerlimitaandanupperlimitb
is
P(a<x<b)=(cumulativeareatotheleftofb)–
(
) (
)
(cumulativeareatotheleftofa)
=P(x<b)– P(x<a)
E( X )
a
-
b
xf ( x)dx
E (( X P ) 2 )
Onceagainthestandarddeviationisthesquarerootofvariance.
Variance and standard deviation do not exist if the expected
Varianceandstandarddeviationdonotexistiftheexpected
valuebywhichtheyaredefineddoesnotconverge.
a
b
f
f
y
Notethesimilaritytothedefinitionfordiscreterandomvariables.
Onceagainweoftendenoteitbyμ.Asinthediscretecasethis
integralmaynotconverge,inwhichcasetheexpectationifXis
undefined.
undefined.
y Asinthediscretecasewedefinethevariance by
Var ( X )
=
³
7
8
Areaunderaprobabilitydistributioncurve
Theprobabilityofarandomvariabletakingon
aparticularvalue
Shaded area is
between a and b; in other
words
d what
h t we have
h
i
is
P (a x b)
x=a
P(x=34)=0
x
x=b
Shaded area is 1.0 or 100%
Shadedareais1(or
(
100%)
9
x
10
FamiliesofContinuousprobability
distributions
DefinitionandBasicProperties
Theprobabilitydistributionfunctionofacontinuous
randomvariableshouldsatisfythefollowing
properties:
i
1.
2.
³
f
3.
11
y UniformProbabilityDistribution
Itisanonnegativefunction(butunlikeinthediscrete
caseitmaytakeonvaluesexceeding1).
y
g )
Itsdefiniteintegraloverthewholereallineequalsone.
Thatis:
f
f ( x ) dx
d
y NormalProbabilityDistribution
1
y ExponentialProbabilityDistribution
Andfinally,followingfromthedefinitionofacontinuous
y,
g
randomvariableandtheformulaforcomputingthearea
underacurvewehave:
³
b
a
f ( x)dx P(a d X d b)
X=34
Inotherwords,sincetheprobabilityofacontinuousrandom
In
other words since the probability of a continuous random
variableassumingavaluewithinsomegivenintervalisdefined
tobethearea undertheprobabilitydistributioncurvewithin
thatinterval,theareaunderaparticularpointisapparently
h
l h
d
l
l
zero.
12
UniformProbabilityDistribution
UniformProbabilityDistribution
Arandomvariableissaidtohaveauniform
distribution iftheprobabilityisproportionalto
thelengthoftheinterval.
h l
h f h i
l
Thepdf ofauniformlydistributedrandomvariable
is defined as:
isdefinedas:
y Expectedvalue:
E(x) = (a + b)/2
f (x) = 1/(b – a) for a < x < b
=0
otherwise
th
i
y Variance:
whereaisthelowestvaluethatx cantake(the
(
lowerboundoftheinterval),andb isthelargest
value(theupperboundoftheinterval).
13
Var(x) = (b - a)2/12
14
UniformProbabilityDistribution:Example
UniformProbabilityDistribution:Example
Slatercustomersarechargedfortheamountofsalad
theytake.Samplingsuggeststhattheamountof
saladtakenisuniformlydistributedbetween5ounces
and15ounces.
Uniform Probability Density Function is:
UniformProbabilityDensityFunctionis:
Expectedvalue:
f(x) = 1/10 for 5 < x < 15
=0
otherwise
Variance:
wherewedenotetheamountofsaladasx.
Yourtaskistofindtheexpectedamountofsaladthat
youmaygetonaverage,anditsvariance.
15
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33
16
UniformProbabilityDistribution:Graphical
representation
p
Anotherexamplesoftdrink
Basedonthepreviousexample,whatistheprobabilitythat
theamountofsaladyougetwillbebetween12and15
h
f l d
ill b b
12 d 15
ounces?
y You’reproductionmanagerofasoftdrinkbottling
p
g
g
company.Youbelievethatwhenamachineisset
todispense12oz.,itreallydispenses11.5to12.5
oz.inclusive.Supposetheamountdispensedhas
auniformdistribution.Whatistheprobability
that less than 11 8 oz is dispensed?
thatlessthan11.8oz.isdispensed?
f(x)
P(12 < x < 15) = 1/10(3) = .3
1/10
x
5
10 12
15
Salad Weight (oz.)
(oz )
17
E( ) = (a
E(x)
( + b)/2
= (5 + 15)/2
= 10
18
Normaldistribution
ExampleAnswerkey
y Describe many phenomena
f(x)
1.0
1
d c
y Goodapproximationtobinomialandhypergeometric
1
12.5 11.5
1
1.0
1
P(11.5d x d 11.8)
y Limiting distribution of sample averages
y Basis for Classical Statistical Inference
x
11.5
11.8
y Two Parameters : mean (), standard deviation ()
12.5
y X ~ normal distribution with mean & variance 2
=(Base)(Height)
=(11.8 11.5)(1)=0.30
1
e
2S V
f ( x; P , V )
19
( xP )2
2V 2
f x f
20
NormalDistribution
Normaldistribution(Youden)
Anormalprobabilitydistribution,whenplotted,givesa
bellshapedcurvesuchthat:
1 Thetotalareaunderthecurveis1.
1.
The total area under the curve is 1
2. Thecurveissymmetricaboutthemeanofthe
distribution.
3 Thetwotailsofthecurveextendindefinitely.
3.
Th t t il f th
t d i d fi it l
Standarddeviation=
THE
NORMAL
NORMAL
LAWOFERROR
STANDSOUTINTHE
EXPERIENCE OF MANKIND
EXPERIENCEOFMANKIND
ASONEOFTHEBROADEST
GENERALIZATIONSOFNATURAL
PHILOSOPHY+ITSERVESASTHE
GUINDINGINSTRUMENTINRESEARCHERS
INTHEPSYCHICALANDSOCIALSCIENCESAND
INMEDICINEAGRICULTUREANDINGENEERING+
ITISANINDISPENSABLETOOLFORTHEANALYSISANDTHE
INTERPRETATIONOFTHEBASICDATAOBTAINEDBYOBSERVATIONANDEXPERIMENT
21
Normaldistributionwithmean andstandarddeviation.
22
NormalDistribution
NormalDistribution
Eachofthetwoshadedareasis
0.5(or50%)
.5
Normaldistributionwithdifferentmeans(canbenegativeaswell)butthesame
(
g
)
standarddeviation
.5
x
Ɋ=-5
Normaldistributionwiththesame
Eachofthetwoshaded
areasisclosetozero
mean,butdifferentstandard
deviations: largervaluesresult
inwiderandflattercurves
23
Ɋ Ȃ 3ɐ
Ɋ + 3ɐ
24
Ɋ=5
=5
=10
=16
16
Ɋ=10
TheStandardNormalDistribution
Standardizinganormaldistribution
Definition
Thenormaldistributionwith=0and=1iscalledthe
standardnormaldistribution.
Foranormalrandomvariablex,aparticularvalue
l
d
bl
l
l
ofx canbeconvertedtoitscorrespondingz value
by using the formula
byusingtheformula
x P
z
V
=1
where
where
and
and arethemeanandstandarddeviation
are the mean and standard deviation
ofthenormaldistributionofx,respectively.
The density function forastandardnormalrandom
Thedensityfunction
for a standard normal random
variableisdefinedas
μ=0
-3
3
-2
2
-11
0
1
2
3
Theunitsmarkedonthehorizontalaxisofthestandardnormal
curvearedenotedbyz andarecalledthez valuesor z scores.A
specificvalueofz
ifi
l
f givesthedistancebetweenthemeanandthe
i
th di t
b t
th
d th
pointrepresentedbyz intermsofthestandarddeviation.
25
where z = (x – P)/V
where
26
Example:P(3.8X5),X~N(5,102)
“Standardization”meansonetable!!!
Z
X P
V
X P
V
Z
Standardized Normal
Distribution
V
3.8 5
10
.12
Standardized Normal
Distribution
V= 1
V = 10
V=1
.0478
P
P= 0
X
Z
27
3 8 P= 5
3.8
-.12
12 P = 0
X
28
Let’sconsiderr.v.XdistributedN(5,102)
Probabilitieswithin1;2;3sigma
P(2.9 X 7.1) = P(-0.21 Z 0.21) = 0.1664
X P
Z
V
X P
Z
V
2.9 5
10
7.1 5
10
.21
99.72%
V=1
95.44%
68.26%
.21
-.21 0 .21
Z
P(X > 10) = P( Z >0.50)
>0 50) = 1 – P(Z<0.5)
P(Z<0 5) = 1-0.6915
1 0 6915
Z
29
X P
V
10 5
10
P – 3V
.50
30
P – 2V
P – 1V
P
P + 1V
P + 3V
P + 2V
x
Z
NormalApproximation
ofBinomialProbabilities
of
Binomial Probabilities
Example– Binomialapproximation
y Whenthenumberoftrials,n,becomeslarge,
31
evaluatingthebinomialprobabilityfunctionby
l i
h bi
i l
b bili f
i b
handorwithacalculatorisdifficult
y Thenormalprobabilitydistributionprovidesan
The normal probability distribution provides an
easytouseapproximationofbinomial
probabilitieswheren>20,np >5,andn(1 p)>5.
Set
P np and V
npq
y Addandsubtract0.5(acontinuitycorrection
f t )b
factor)becauseacontinuousdistributionisbeing
ti
di t ib ti i b i
usedtoapproximateadiscretedistribution.For
example,
p ,
P(x=10)isapproximatedbyP(9.5<x<10.5)
y Supposex isabinomialrandomvariablewith
n =30andp =.4.Usingthenormalapproximation
tofindP(x 10).
n = 30 p = .4 q = .6
np = 12 nq = 18
The
h normall
approximation
is ok!
Calculate
P np 30(.4) 12
V
npq
Approximation– answerkey
10.5 12
)
2.683
P ( z d .56) .2877
33
P( x 9.5)
P(x t 5 )
P ( x t 4.5 )
P(x ! 5 )
P(x ! 5.5)
P(( 5 x 100 )
P(( 5.5 x 9.5 )
P(( 5 d x 10 )
P(( 4.5 x 9.5 )
34
Yetanotherexampleonapproximation
y AproductionlineproducesAAbatterieswitha
reliabilityrateof95%.Asampleofn =200
batteriesisselected.Findtheprobabilitythatat
l t 195 f th b tt i
least195ofthebatterieswork.
k
Success = working battery n = 200
p = .95
np = 190 nq = 10
P ( x t 195) | P( z t
The normal
approximation
is ok!
194.5 190
)
200(.95)(.05)
Standardnormalprobabilitydistribution:Example
PepZonesellsautopartsandsuppliesincludingapopular
multigrademotoroil.Whenthestockofthisoildropsto
20gallons,areplenishmentorderisplaced.
Thestoremanagerisconcernedthatsalesarebeinglostdueto
stockoutswhilewaitingforanorder.Ithasbeendetermined
thatdemandduringreplenishmentleadtimeisnormally
distributedwithameanof15gallonsandastandarddeviation
of 6 gallons The manager would like to know the probability of
of6gallons.Themanagerwouldliketoknowtheprobabilityof
astockout,P(x >20).
P ( z t 1.46) 1 .9278 .0722
35
2.683
Anotherexamples– binomialapproximation
P( 10 )
P(x
P ( x d 10) | P( z d
30(.4)(.6)
32
36
SolvingfortheStockoutprobability
(continued)
Standardnormalprobabilitydistribution:Solvingforthe
p
y
Stockoutprobability
Step1: Convertx tothestandardnormaldistribution.
z = ((x - P)/V = ((20 - 15)/6
)/ = .83
Step3: Computetheareaunderthestandardnormalcurve
to the right of z =.83.
totherightofz
= 83
P(z > .83) = 1 – P(z < .83)
= 1- .7967
Step2: Findtheareaunderthestandardnormal
curvetotheleftofz=.83.
z
,00
,01
,02
,03
,04
,05
,06
,07
,08
,09
.
.
.
.
.
.
.
.
.
.
.
,5
,6915
,6950
,6985
,7019
,7054
,7088
,7123
,7157
,7190
,7224
,6
,7257
,7291
,7324
,7357
,7389
,7422
,7454
,7486
,7517
,7549
,,7
,,7580
,,7611
,,7642
,,7673
,,7704
,,7734
,,7764
,,7794
,,7823
,,7852
,8
,7881
,7910
,7939
,7967
,7995
,8023
,8051
,8078
,8106
,8133
,9
,8159
,8186
,8212
,8238
,8264
,8289
,8315
,8340
,8365
,8389
.
.
.
.
.
.
.
.
.
.
.
P(z < .83)
37
Probability
of a stock-out:
P(x > 20)
Area = 1 - .7967
Area = .7967
2033
= .2033
38
0
z
.83
Estimatinglifeofalightbulb
Answerkey
y YouworkinQualityControlforGE.Lightbulblife
y P(X<1470)=P(Z<2.65)=P(Z>2.65)=1P(Z<=2.65)
hasanormaldistributionwith
=2000hours&=200hours.What’sthe
probabilitythatabulbwilllast
b bilit th t b lb ill l t
y A.between2000&2400
hours?
y B.lessthan1470hours?
=0.040
Z
39
X P
V
1470 2000
200
2.65
y P(2000<X<2400)=P(0<Z<2)=0.472
Z
X P
V
2400 2000
200
2 .0
40
Exercises– Normaldistribution
ExponentialProbabilityDistribution
y Z=astandardnormalR.V.
• Theexponentialprobabilitydistributionisuseful
indescribingthetimeittakestocompleteatask.
P(Z 1.43)
P(Z ! 0.89)
• Theexponentialrandomvariablescanbeusedto
P(2.16 Z 0.65)
Exe: X ~ normal,
(a)
(b) P ( X ! 17 )
(c) P ( X 22 )
P
describesucheventsas:
30 V
6
¾ Timebetweenvehiclearrivalsatatollbooth
¾ Timerequiredtocompleteaquestionnaire
Time required to complete a questionnaire
¾ Distancebetweenmajordefectsinahighway
P (32 X 41)
41
= .2033
42
ExponentialProbabilityDistribution
ExponentialProbabilityDistribution:Example
y Densityfunction:
f ( x)
1
P
ex / P
ThetimebetweenarrivalsofcarsatAl’sfullservicegas
pumpfollowsanexponentialprobabilitydistributionwith
f ll
i l
b bili di ib i
ih
ameantimebetweenarrivalsof3minutes.Alwouldlike
toknowtheprobabilitythatthetimebetweentwo
successivearrivalswillbe2minutesorless.
for x > 0, P > 0
whereP isthemeanofthedistribution.
y Cumulativeprobabilities:
P ( x d x0 ) 1 e xo / P
wherex0 issomespecificvalueofx.
43
44
ExponentialProbabilityDistribution:some
properties
Graphicalrepresentation
• Apropertyoftheexponentialdistributionisthatthe
mean and standard deviation ɐ,areequal.
mean,,andstandarddeviation,
are equal
f(x)
P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866
.4
Thus, the standard deviation, s, and variance, s 2, for
the time between arrivals at Al’s full-service p
pump
p are:
.33
= = 3 minutes
.2
2 = (3)2 = 9
.11
x
• Theexponentialdistributionisskewedtotheright.
1 2 3 4 5 6 7 8 9 10
Time Between Successive Arrivals (mins.)
• Theskewness measurefortheexponentialdistributionis2.
45
46
Relationshipamongdistributions
Normal (X)
P ,V 2
X
ln Y
eX
Y
X P
Z
¦
F2
n
i 1
0, 1
Zi
2
Chi-square ( F)
n
Lognormal (Y)
P ,V
Normal (Z)
V
2
D
Uniform(X)
n / 2, E
2
D, E
X D
E D
D
E 1
Beta
D, E
47
X
( E D )U D
Uniform(U)
0,1
J,E
n=2
D 1
Gamma
U
Weibull
D, E
X
E ln
l U
J
1
Exponential(X)
E