8-random variable and discrete probability distribution (1)

7/31/2019 8-Random Variable and Discrete Probability Distribution (1)

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Chapter 5Discrete Probability Distributions

.10

.20

.30

.40

0 1 2 3 4

Random Variables

Discrete Probability Distributions

Expected Value and Variance

Binomial Probability Distribution

Poisson Probability Distribution

Hypergeometric ProbabilityDistribution


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A random variable is a numerical description of theoutcome of an experiment.

Random Variables

A discrete random variable may assume either a

finite number of values or an infinite sequence ofvalues.

A continuous random variable may assume any

numerical value in an interval or collection ofintervals.


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Let x = number of TVs sold at the store in one day,

where x can take on 5 values (0, 1, 2, 3, 4)

Example: JSL Appliances

Discrete random variable with a finite number

of values


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Let x = number of customers arriving in one day,

where x can take on the values 0, 1, 2, . . .

Example: JSL Appliances

Discrete random variable with an infinite sequence

of values

We can count the customers arriving, but there is nofinite upper limit on the number that might arrive.


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Random Variables

Question Random Variable x Type

Family

size

x = Number of dependents

reported on tax returnDiscrete

Distance fromhome to store

x = Distance in miles fromhome to the store site

Continuous

Own dogor cat

x = 1 if own no pet;

= 2 if own dog(s) only;

= 3 if own cat(s) only;

= 4 if own dog(s) and cat(s)

Discrete


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The probability distribution for a random variabledescribes how probabilities are distributed overthe values of the random variable.

We can describe a discrete probability distributionwith a table, graph, or equation.



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The probability distribution is defined by aprobability function, denoted byf(x), which providesthe probability for each value of the random variable.

The required conditions for a discrete probabilityfunction are:


f(x) > 0

f(x) = 1


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a tabular representation of the probabilitydistribution for TV sales was developed.

Using past data on TV sales,

Number

Units Sold of Days0 80

1 50

2 40

3 104 20

200

x f(x)0 .40

1 .25

2 .20

3 .054 .10

1.00

80/200



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.10

.20

.30

.40

.50

0 1 2 3 4Values of Random Variable x (TV sales)

Probability


Graphical Representation of Probability Distribution


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Discrete Uniform Probability Distribution

The discrete uniform probability distribution is thesimplest example of a discrete probabilitydistribution given by a formula.

The discrete uniform probability function is

f(x) = 1/n

where:

n = the number of values the randomvariable may assume

the values of therandom variableare equally likely


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The expected value, or mean, of a random variableis a measure of its central location.

The variance summarizes the variability in thevalues of a random variable.

The standard deviation, , is defined as the positivesquare root of the variance.

Var(x) = 2 = (x - )2f(x)

E(x) = = xf(x)


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Expected Value

expected number ofTVs sold in a day

x f(x) xf(x)

0 .40 .00

1 .25 .25

2 .20 .403 .05 .15

4 .10 .40

E(x) = 1.20



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Variance and Standard Deviation

0

12

3

4

-1.2

-0.20.8

1.8

2.8

1.44

0.040.64

3.24

7.84

.40

.25

.20

.05

.10

.576

.010

.128

.162

.784

x - (x - )2 f(x) (x - )2f(x)

Variance of daily sales = 2 = 1.660

x

TVssquared

Standard deviation of daily sales = 1.2884 TVs



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Binomial Distribution

Four Properties of a Binomial Experiment

3. The probability of a success, denoted byp, doesnot change from trial to trial.

4. The trials are independent.

2. Two outcomes, success and failure, are possibleon each trial.

1. The experiment consists of a sequence of nidentical trials.

stationarityassumption


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Our interest is in the number of successesoccurring in the n trials.

We let x denote the number of successesoccurring in the n trials.


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where:f(x) = the probability of x successes in n trials

n = the number of trials

p = the probability of success on any one trial

( )!( ) (1 )!( )!

x n xnf x p px n x


Binomial Probability Function


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( )!( ) (1 )!( )!

x n xnf x p px n x


Binomial Probability Function

Probability of a particularsequence of trial outcomeswith x successes in n trials

Number of experimentaloutcomes providing exactly

x successes in n trials


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Example: Evans Electronics

Evans is concerned about a low retention rate foremployees. In recent years, management has seen aturnover of 10% of the hourly employees annually.Thus, for any hourly employee chosen at random,

management estimates a probability of 0.1 that theperson will not be with the company next year.


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Using the Binomial Probability Function

Choosing 3 hourly employees at random, what isthe probability that 1 of them will leave the companythis year?

f xn

x n xp px n x( )

!

!( )!( )

( )

1

1 23!(1) (0.1) (0.9) 3(.1)(.81) .2431!(3 1)!

f

Let: p = .10, n = 3, x = 1


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Tree Diagram


1st Worker 2nd Worker 3rd Worker x Prob.

Leaves

(.1)

Stays(.9)

3

2

0

2

2

Leaves (.1)

Leaves (.1)

S (.9)

Stays (.9)

Stays (.9)

S (.9)

S (.9)

S (.9)

L (.1)

L (.1)

L (.1)

L (.1) .0010

.0090

.0090

.7290

.0090

1

1

.0810

.0810

.0810

1


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Using Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

3 0 .8574 .7290 .6141 .5120 .4219 .3430 .2746 .2160 .1664 .1250

1 .1354 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .37502 .0071 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .3750

3 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

p



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(1 )np p

E(x) = = np

Var(x) = 2 = np(1 p)

Expected Value

Variance

Standard Deviation


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3(.1)(.9) .52 employees

E(x) = = 3(.1) = .3 employees out of 3

Var(x) = 2 = 3(.1)(.9) = .27

Expected Value

Variance

Standard Deviation


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A Poisson distributed random variable is oftenuseful in estimating the number of occurrencesover a specified interval of time or space

It is a discrete random variable that may assumean infinite sequence of values (x = 0, 1, 2, . . . ).

Poisson Distribution


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Examples of a Poisson distributed random variable:

the number of errors committed while typinga article

the number of vehicles arriving at atoll booth in one hour



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Two Properties of a Poisson Experiment

2. The occurrence or nonoccurrence in anyinterval is independent of the occurrence ornonoccurrence in any other interval.

1. The probability of an occurrence is the samefor any two intervals of equal length.


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Poisson Probability Function


f xe

x

x

( )!

where:f(x) = probability of x occurrences in an interval

= mean number of occurrences in an interval

e = 2.71828


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Example: Mercy Hospital

Patients arrive at the

emergency room of Mercy

Hospital at the average

rate of 6 per hour onweekend evenings.

What is the

probability of 4 arrivals in

30 minutes on a weekend evening?


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Using the Poisson Probability Function

4 33 (2.71828)(4) .1680

4!

f

= 6/hour = 3/half-hour, x = 4


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Poisson Distribution of Arrivals


Poisson Probabilities

0.00

0.05

0.10

0.15

0.20

0.25

0 1 2 3 4 5 6 7 8 9 10

Number of Arrivals in 30 Minutes

Probabilit

y

actually,the sequence

continues:11, 12,


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A property of the Poisson distribution is thatthe mean and variance are equal.

= 2


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Variance for Number of Arrivals

During 30-Minute Periods

= 2 = 3


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Hypergeometric Distribution

The hypergeometric distribution is closely relatedto the binomial distribution.

However, for the hypergeometric distribution:

the trials are not independent, and

the probability of success changes from trial

to trial.


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Hypergeometric Probability Function


n

N

xn

rN

x

r

xf )( for 0 < x < r

where: f(x) = probability of x successes in n trials

n = number of trials

N= number of elements in the populationr= number of elements in the population

labeled success


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Hypergeometric Probability Function


( )

r N r

x n xf x

Nn

for 0 < x < r

number of waysnx failures can be selectedfrom a total of Nrfailures

in the population

number of waysx successes can be selected

from a total of rsuccessesin the populationnumber of ways

a sample of size n can be selectedfrom a population of size N


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Example: Neveready

Bob Neveready has removed twodead batteries from a flashlight and

inadvertently mingled them with

the two good batteries he intended

as replacements. The four batteries look identical.Bob now randomly selects two of the four

batteries. What is the probability he selects the twogood batteries?


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Using the Hypergeometric Function

2 2 2! 2!

2 0 2!0! 0!2! 1( ) .167

4 4! 6

2 2!2!

r N r

x n xf x

N

n

where:x = 2 = number of good batteries selected

n = 2 = number of batteries selectedN= 4 = number of batteries in totalr= 2 = number of good batteries in total


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( )r

E x nN

2( ) 1

1

r r N nVar x n

N N N

Mean

Variance


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22 1

4

rn

N

2 2 2 4 2 12 1 .333

4 4 4 1 3

Mean

Variance

b


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Consider a hypergeometric distribution with n trialsand let p = (r/n) denote the probability of a successon the first trial.

If the population size is large, the term (Nn)/(N 1)

approaches 1.

The expected value and variance can be written

E(x) = np and Var(x) = np(1 p).

Note that these are the expressions for the expectedvalue and variance of a binomial distribution.

continued

H i Di ib i


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When the population size is large, a hypergeometricdistribution can be approximated by a binomialdistribution with n trials and a probability of success

p = (r/N).

8-random variable and discrete probability distribution (1)

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