Discrete Random Variables

Discrete Random Discrete Random VariablesVariables

G. Baker, Department of StatisticsG. Baker, Department of StatisticsUniversity of South Carolina; Slide University of South Carolina; Slide 22

Random Variables – What We Random Variables – What We KnowKnow

A random variable A random variable is a function that is a function that associates a unique numerical value with associates a unique numerical value with every outcome of an experiment.every outcome of an experiment.

The value of a random variable changes The value of a random variable changes from occurrence to occurrence.from occurrence to occurrence.

Although we don’t know the next value of Although we don’t know the next value of a random variable, a random variable, we do know the we do know the overall pattern of the values assumed overall pattern of the values assumed by a random variable over many by a random variable over many repetitions.repetitions.


Categorical Random Categorical Random VariablesVariables

Molding Defects:Molding Defects:TypeType FreqFreqBlk SpotsBlk Spots 100100ScratchesScratches 40 40SplaySplay 35 35SinksSinks 15 15Flow LinesFlow Lines 12 12OtherOther 20 20

Black Spot LocationsBlack Spot LocationsLocationLocation FreqFreqTopTop 7070MidMid 1010BottomBottom 1010LeftLeft 5 5RightRight 5 5

0

20

40

60

80

100

120

Blk Spots Scratches Splay Sinks Flow Line Other

Fre

qu

ency

0

10

20

30

40

50

60

70

80

Top Mid Bottom Left Right

Fre

qu

ency


Numerical Random Numerical Random VariablesVariables

Assume numerical values from a Assume numerical values from a meaningful numerical scale.meaningful numerical scale.

Example:Example:– Age is a numerical variable.Age is a numerical variable.– Zip code is a categorical random Zip code is a categorical random

variable.variable.

Life of a Light Bulb in HoursLife of a Light Bulb in Hours



NotationNotation We will use capital letters from the We will use capital letters from the

end of the alphabet to represent a end of the alphabet to represent a random variable.random variable.– Usually X or YUsually X or Y

The corresponding lower case letter The corresponding lower case letter will represent a particular value of will represent a particular value of the random variable.the random variable.– P(Y = y) is the probability that the P(Y = y) is the probability that the

random variable Y is equal to the value random variable Y is equal to the value y. y.


Discrete and Continuous Discrete and Continuous Random VariablesRandom Variables

A A continuous random variablecontinuous random variable can take any value in an interval of can take any value in an interval of the real number line.the real number line.– Usually measurementsUsually measurements

A A discrete random variablediscrete random variable can can take one of a countable list of take one of a countable list of distinct values.distinct values.– Usually countsUsually counts


Numeric Random VariablesNumeric Random Variables

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0 1 2 3 4 5 6

Rel

ativ

e F

req

uen

cy

OuncesNumber of Active Pumps

16.0 16.2 16.4

ContinuousContinuous DiscreteDiscrete


Discrete and Continuous Random Discrete and Continuous Random VariablesVariables

Discrete or continuous?Discrete or continuous?– Time until a projectile returns to earth.Time until a projectile returns to earth.– The number of times a transistor in The number of times a transistor in

computer memory changes state in one computer memory changes state in one operation.operation.

– The volume of gasoline that is lost to The volume of gasoline that is lost to evaporation during the filling of a gas tank.evaporation during the filling of a gas tank.

– The outside diameter of a machined shaft.The outside diameter of a machined shaft.– The number of chapters in your text book.The number of chapters in your text book.


Discrete Numeric Random Discrete Numeric Random VariablesVariables

A small airport in New Zealand is interested A small airport in New Zealand is interested in the number of late aircraft arrivals per in the number of late aircraft arrivals per day. Every day for a year it counts the daily day. Every day for a year it counts the daily number of late arrivals.number of late arrivals.

Let random variable Y = Number of aircraft in one day that arrive late.

YY 00 11 22 33 4+4+

P(Y=y) 0.120.12 0.190.19 0.300.30 0.260.26 0.130.13



What is the probability that on any What is the probability that on any randomly chosen day that 3 aircraft randomly chosen day that 3 aircraft are late?are late?

What is the chance that on any What is the chance that on any randomly chosen day at least 1 aircraft randomly chosen day at least 1 aircraft is late?is late?

YY 00 11 22 33 4+4+P(Y=y) 0.120.12 0.190.19 0.300.30 0.260.26 0.130.13



An assembly consists of three An assembly consists of three mechanical components. Suppose that mechanical components. Suppose that the probabilities that the first, second, the probabilities that the first, second, and third components meet and third components meet specifications are 0.90, 0.95 and 0.99 specifications are 0.90, 0.95 and 0.99 respectively. Assume the components respectively. Assume the components are independent.are independent.

Let SLet Sii = the event that a component is = the event that a component is within specification.within specification.


Possible Outcomes for One Possible Outcomes for One assemblyassembly

SS11SS22SS33 SS11SS22SS33 SS11SS22SS33 SS11SS22SS33 SS11SS22SS33 SS11SS22SS33 SS11SS22SS33 SS11SS22SS33

0.84640.846455

0.00040.000455

0.00090.000955

0.00850.008555

0.04450.044555

0.09400.094055

Let Y = Number of components within specification in a randomly chosen assembly

Y 0 1 2 3

P(Y=y) 0.00005 0.00635 0.14715 0.84645

0.00490.004955

0.00000.000055


Discrete Random VariablesDiscrete Random Variables

Probability or Distribution Probability or Distribution FunctionFunction

p(y) = P(Y = y)p(y) = P(Y = y)

Note:Note:

0 0 << p(y) p(y) << 1 for every value of y 1 for every value of y

andand 1)(yp


Probability Function for a Discrete Probability Function for a Discrete Random Variable can be expressed as Random Variable can be expressed as a table or graph and sometimes as a a table or graph and sometimes as a formula.formula.Y 0 1 2 3

P(Y=y) 0.00005 0.00635 0.14715 0.84645

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 1 2 3


Cumulative Probability or Cumulative Probability or Distribution FunctionDistribution Function

If Y is a random variable, then the If Y is a random variable, then the cumulative distribution function is cumulative distribution function is denoted by F(y).denoted by F(y).

F(y) = P(Y F(y) = P(Y << y) y)


Number of Components within Number of Components within Specification for a Randomly chosen Specification for a Randomly chosen AssemblyAssembly

Y 0 1 2 3

P(Y=y) 0.00005 0.00635 0.14715 0.84645

Distribution Function:

Y 0 1 2 3

P(Y<y) 0.00005 0.00640 0.15355 1.0000

Cumulative Distribution Function:


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 1 2 3

Distribution Function

Cumulative Distribution Function

P(Y = y) P(Y < y)


Expected ValueExpected Value Note, this distribution is the Note, this distribution is the

distribution for adistribution for a population population

The The expected valueexpected value of the of the population is the population mean, population is the population mean, µ, µ, for the distribution.for the distribution.

Y 0 1 2 3

P(Y=y) 0.00005 0.00635 0.14715 0.84645

y

yypYE )()(


Number of Components within Number of Components within Specification on a Randomly Specification on a Randomly

Chosen AssemblyChosen Assembly

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 1 2 3


VarianceVariance

The The variancevariance of the population is of the population is the expected value of (Y - the expected value of (Y - µ)µ)22..

])[()var( 22 YEY

Y 0 1 2 3

P(Y=y) 0.00005 0.00635 0.14715 0.84645

y

ypyY )()()var( 22


Computational Formula for Computational Formula for VarianceVariance

])[()var( 22 YEY

222 )()var( YEY

y y

ypyypyY

2

22 )]([)]([)var(


Population Standard Population Standard DeviationDeviation

2

Standard Deviation is the Standard Deviation is the positive square root of the positive square root of the variance.variance.


We receive 75% of a certain part We receive 75% of a certain part from Supplier A and 25% from from Supplier A and 25% from supplier B. Two of these parts are supplier B. Two of these parts are used in an assembly. Let Y = number used in an assembly. Let Y = number of parts from Supplier A in the of parts from Supplier A in the assembly. What is the expected assembly. What is the expected value of Y?value of Y?


A project manager for an engineering A project manager for an engineering firm submitted bids on three projects. firm submitted bids on three projects. The following table summarizes the The following table summarizes the firm’s chances of having the three firm’s chances of having the three projects accepted.projects accepted.

Project A B C

Prob of accept 0.30 0.80 0.10Assuming the projects are independent of

one another, what is the probability that the firm will have all three projects accepted?


What is the probability of What is the probability of having at least one project having at least one project accepted?accepted?

Project A B C

Prob of accept 0.30 0.80 0.10


Let Y = number of projects Let Y = number of projects accepted.accepted. y 0 1 2 3

p(y) 0.126 0.572 0.278 0.024

What is the expected number of projects accepted?


Let Y = number of projects Let Y = number of projects accepted.accepted. y 0 1 2 3

p(y) 0.126 0.572 0.278 0.024

What is the variance for number of projects accepted?


Rules of ExpectationRules of Expectation

1)1) E(c) = cE(c) = c

2)2) E(cY) = cE(Y)E(cY) = cE(Y)

3)3) E(X + Y) = E(X) + E(Y)E(X + Y) = E(X) + E(Y)


Consider the following Consider the following process:process:

5511 11


Rules of VarianceRules of Variance

1)1) Var(c) = 0Var(c) = 0

2)2) Var(cY) = cVar(cY) = c22Var(Y)Var(Y)

3)3) Var(X + Y) = Var(X) + Var(Y)Var(X + Y) = Var(X) + Var(Y)


Binomial Random VariableBinomial Random Variable

Suppose we toss a fair coin 20 times, Suppose we toss a fair coin 20 times, counting the number of heads.counting the number of heads.

This is an example of a binomial This is an example of a binomial experiment.experiment.

The possible outcomes can be The possible outcomes can be represented with a binomial random represented with a binomial random variable which follows a binomial variable which follows a binomial distribution.distribution.


Binomial ExperimentBinomial Experiment

1)1) Y represents the outcome of Y represents the outcome of nn identical trials.identical trials.

2)2) There are two possible outcomes There are two possible outcomes for each trial: success and failure.for each trial: success and failure.

3)3) Outcomes are independent from Outcomes are independent from trial to trial.trial to trial.

4)4) Probability of success, Probability of success, pp, remains , remains constant from trial to trial.constant from trial to trial.


Binomial ExperimentBinomial Experiment

Each toss of the coin is a trail. We Each toss of the coin is a trail. We performed 20 trials. So performed 20 trials. So n = 20n = 20..

The two possible outcomes of a trials The two possible outcomes of a trials are heads or tails. Since we are are heads or tails. Since we are counting heads, heads is a success, counting heads, heads is a success, while tails is a failure.while tails is a failure.

Are the outcomes of our trials Are the outcomes of our trials independent of one another?independent of one another?

Probability of success (heads) is 0.50Probability of success (heads) is 0.50


Binomial ExperimentBinomial Experiment A manufacturer of water filters for A manufacturer of water filters for

refrigerators monitors the process for refrigerators monitors the process for defective filters. Historically, this process defective filters. Historically, this process averages 5% defective filters. averages 5% defective filters.

Suppose five filters are randomly selected for Suppose five filters are randomly selected for testing and we count number of defective testing and we count number of defective filters.filters.

Define a trial.Define a trial. Calculate Calculate nn.. Define a “success”.Define a “success”. Are the outcomes independent from trial to Are the outcomes independent from trial to

trial?trial? Calculate Calculate pp..


Water FiltersWater Filters A manufacturer of water filters for A manufacturer of water filters for

refrigerators monitors the process for refrigerators monitors the process for defective filters. Historically, this defective filters. Historically, this process averages 5% defective filters. process averages 5% defective filters. Five filters are randomly selected.Five filters are randomly selected.

Find the probability that all five filters Find the probability that all five filters are defective.are defective.

Find the probability that no filters are Find the probability that no filters are defective.defective.

Find the probability that exactly 1 Find the probability that exactly 1 filter is defective.filter is defective.


A manufacturer of water filters for A manufacturer of water filters for refrigerators monitors the process for refrigerators monitors the process for defective filters. Historically, this defective filters. Historically, this process averages 5% defective process averages 5% defective filters. Five filters are randomly filters. Five filters are randomly selected.selected.

Find the probability that exactly 2 Find the probability that exactly 2 filters are defective.filters are defective.

A.A. 5(.05)5(.05)22(.95)(.95)33

B.B. 10(.05)10(.05)22(.95)(.95)33

C.C. 15(.05)15(.05)22(.95)(.95)33


Remember CombinationsRemember Combinations

Number of ways to choose r Number of ways to choose r distinct objects from n distinct distinct objects from n distinct objects:objects:

)!(!

!

rnr

n

r

n

“n choose r”


Probability Function for a Probability Function for a Binomial Random VariableBinomial Random Variable

yny ppy

nyp

)1()(

for y = 0, 1, 2, …,n


If Y follows a Binomial If Y follows a Binomial DistributionDistribution

npYE )(

)1(2 pnp

)1( pnp

BinomiaBinomial meanl mean

BinomiaBinomial l variancvariancee BinomiaBinomial l standarstandard d deviatiodeviationn


Suppose we are producing Suppose we are producing 15% defective filters. Let Y = 15% defective filters. Let Y = Number of Defective Filters in Number of Defective Filters in a Sample of 10a Sample of 10

Binomial n=10 p=0.15

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1 2 3 4 5 6 7 8 9 10

Rela

tive F

req

uen

cy


What is the approximate What is the approximate probability that there will be at probability that there will be at least 1 defective filter in a sample least 1 defective filter in a sample of 10?of 10?

Binomial n=10 p=0.15

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 1 2 3 4 5 6 7 8 9 10

Rela

tive F

req

uen

cy


Historically, 10% of homes in Florida Historically, 10% of homes in Florida have radon levels higher than that have radon levels higher than that recommended by the EPA. In a recommended by the EPA. In a random sample of 20 homes, find the random sample of 20 homes, find the probability that exactly 3 have radon probability that exactly 3 have radon levels higher than the EPA levels higher than the EPA recommendation. Assume home are recommendation. Assume home are independent of one another.independent of one another.


If a manufacturing process has a If a manufacturing process has a 0.03 defective rate, what is the 0.03 defective rate, what is the probability that at least one of the probability that at least one of the next 25 units inspected will be next 25 units inspected will be defective?defective?A.A. (0.03)(0.03)11 * (0.97) * (0.97)2424

B.B. 1 – (0.03)1 – (0.03)11 * (0.97) * (0.97)2424

C.C. 1 – (0.03)1 – (0.03)00 * (0.97) * (0.97)2525

D.D. 1 – (0.03)1 – (0.03)2525

1

25

0

25

1

25

0

25


A manufacturing process has a A manufacturing process has a 0.03 defective rate. If we 0.03 defective rate. If we randomly sample 25 unitsrandomly sample 25 units

a)a) What is the probability that less than What is the probability that less than 6 will be defective?6 will be defective?

b)b) What is the probability that 4 or What is the probability that 4 or more are defective?more are defective?

c)c) What is the probability that between What is the probability that between 2 and 5, inclusive, are defective?2 and 5, inclusive, are defective?


Golf ResortGolf Resort

At a particular golf resort, if a person At a particular golf resort, if a person hits a hole-in-one, the player gets to hits a hole-in-one, the player gets to select 4 slips of paper from a bin of select 4 slips of paper from a bin of 100 slips. Of the 100 slips of paper, 2 100 slips. Of the 100 slips of paper, 2 are good for a brand new driver are good for a brand new driver valued at $300. So if the person valued at $300. So if the person selects 1 or 2 slips good for the selects 1 or 2 slips good for the driver, he gets the driver.driver, he gets the driver.

Is this a binomial experiment?Is this a binomial experiment? We can get the answer using straight We can get the answer using straight

probability.probability.


Golf ResortGolf Resort

Let Y = number of slips chosen that Let Y = number of slips chosen that are good for a free driver.are good for a free driver.

To win, the player must draw 1 or 2 To win, the player must draw 1 or 2 of the 2 slips which entitle him to a of the 2 slips which entitle him to a free driver.free driver.

P(Y = 1 or 2) = P(Y = 1) + P(Y = 2)P(Y = 1 or 2) = P(Y = 1) + P(Y = 2) Recall:Recall:

outcomes possible all of #

occurcan A waysof #)( AP


Probability of Winning Probability of Winning DriverDriver

4

100

2

2

2

98

4

100

1

2

3

98

)21(YP


Hypergeometric DistributionHypergeometric Distribution Models the number of successes out of Models the number of successes out of nn

trials when sampling trials when sampling without without replacement from a finite populationreplacement from a finite population of of NN objects that contains exactly objects that contains exactly rr successes.successes.

n

N

yn

rN

y

r

yYP )( for y = 0, 1, 2,…,n


Hypergeometric DistributionHypergeometric Distribution

Expected Value:Expected Value:

N

rnYE )(


If a shipment of 100 generators If a shipment of 100 generators contains 5 faulty generators, what is contains 5 faulty generators, what is the probability that we can select 10 the probability that we can select 10 generators from the shipment and not generators from the shipment and not get a faulty one?get a faulty one?

10

100

10

95

0

5

9

100

9

95

0

5

10

100

9

95

1

5

10

100

0

95

10

5

A B C D


Insulated WireInsulated Wire Consider a process that produces Consider a process that produces

insulated copper wire. Historically insulated copper wire. Historically the process has averaged 2.6 breaks the process has averaged 2.6 breaks in the insulation per 1000 meters of in the insulation per 1000 meters of wire. We want to find the probability wire. We want to find the probability that 1000 meters of wire will have 1 that 1000 meters of wire will have 1 or fewer breaks in insulation?or fewer breaks in insulation?

Is this a binomial problem?Is this a binomial problem? A hypergeometric problem?A hypergeometric problem?


Poisson DistributionPoisson Distribution

Poisson distribution can be used to Poisson distribution can be used to model the number of events model the number of events occurring in a continuous time or occurring in a continuous time or space.space.

Let Y = number of breaks in 1000 Let Y = number of breaks in 1000 meters of wire.meters of wire.

P(Y P(Y << 1) = P(Y = 0) + P(Y = 1) 1) = P(Y = 0) + P(Y = 1)


Poisson DistributionPoisson Distribution

!

)()(

y

etyp

ty

for y = 0, 1, 2,…

where λ is the average number of occurrences per base unit.

and t is the number of base units inspected.

Further: 2)( tYE


Insulated WireInsulated Wire

!1

6.2

!0

6.2)10(

6.216.20

ee

YP

Let Y = Number of breaks in 1000 meters of wire. λ = 2.6 and t = 1

The expected number of breaks in 1000 meters of wire is 2.6.


Insulated WireInsulated Wire

If we were inspecting 2000 meters of If we were inspecting 2000 meters of wire, wire, λλt = 2.6*2 = 5.2t = 2.6*2 = 5.2

If we were inspecting 500 meters of If we were inspecting 500 meters of wire, wire, λλt = 2.6*0.5 = 1.3t = 2.6*0.5 = 1.3


Conditions for a Poisson Conditions for a Poisson DistributionDistribution

1)1) Areas of inspection are Areas of inspection are independent of one another.independent of one another.

2)2) The probability of the event The probability of the event occurring at any particular point occurring at any particular point in space or time is negligible.in space or time is negligible.

3)3) The mean remains constant The mean remains constant over all areas of inspection.over all areas of inspection.


Suppose we average 5 Suppose we average 5 radioactive particles passing a radioactive particles passing a counter in 1 millisecond. What counter in 1 millisecond. What is the probability that exactly 3 is the probability that exactly 3 particles will pass in one particles will pass in one millisecond?millisecond?


Suppose we average 5 radioactive Suppose we average 5 radioactive particles passing a counter in 1 particles passing a counter in 1 millisecond. What is the millisecond. What is the probability that exactly 10 probability that exactly 10 particles will pass in particles will pass in three three milliseconds?milliseconds?

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Discrete Random Variables