ch 05 discrete probability distribution

8/22/2019 Ch 05 Discrete Probability Distribution

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Slides Prepared by

JOHN S. LOUCKSSt. Edwards University

2002 South-Western/Thomson Learning


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Chapter 5Discrete Probability Distributions

Random Variables

Discrete Probability Distributions

Expected Value and Variance

Binomial Probability Distribution

Poisson Probability Distribution Hypergeometric Probability Distribution

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.20

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.40

0 1 2 3 4


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Random Variables

A random variable is a numerical description of the

outcome of an experiment.

A random variable can be classified as being eitherdiscrete or continuous depending on the numericalvalues it assumes.

A discrete random variable may assume either afinite 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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Example: JSL Appliances

Discrete random variable with a finite number of

values

Let x= number of TV sets sold at the store in one day

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

Discrete random variable with an infinite sequence ofvalues

Let x= number of customers arriving in one day

where xcan take on the values 0, 1, 2, . . .

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


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Discrete Probability Distributions

The probability distribution for a random variable

describes how probabilities are distributed over thevalues of the random variable.

The probability distribution is defined by aprobability function, denoted byf(x), which provides

the probability for each value of the random variable. The required conditions for a discrete probability

function are:

f(x) > 0

f(x) = 1

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


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Using past data on TV sales (below left), a tabular

representation of the probability distribution for TVsales (below right) was developed.

Number

Units Sold of Days x f(x)0 80 0 .40

1 50 1 .25

2 40 2 .20

3 10 3 .05

4 20 4 .10

200 1.00



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Graphical Representation of the Probability

Distribution

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0 1 2 3 4

Values of Random Variable x(TV sales)

Probabi

lity


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Discrete Uniform Probability Distribution

The discrete uniform probability distribution is the

simplest example of a discrete probabilitydistribution given by a formula.

The discrete uniform probability function is

f(x) = 1/nwhere:

n= the number of values the random

variable may assume

Note that the values of the random variable areequally likely.


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Expected Value and Variance

The expected value, or mean, of a random variable is

a measure of its central location.Expected value of a discrete random variable:

E(x) = = xf(x)

The variance summarizes the variability in the valuesof a random variable.

Variance of a discrete random variable:

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

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


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Expected Value of a Discrete Random Variable

x f(x) xf(x)0 .40 .00

1 .25 .25

2 .20 .403 .05 .15

4 .10 .40

E(x) = 1.20

The expected number of TV sets sold in a day is 1.2


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Variance and Standard Deviation

of a Discrete Random Variable

x x - (x - )2 f(x) (x- )2f(x)0 -1.2 1.44 .40 .576

1 -0.2 0.04 .25 .0102 0.8 0.64 .20 .128

3 1.8 3.24 .05 .162

4 2.8 7.84 .10 .784

1.660 =

The variance of daily sales is 1.66 TV sets squared.

The standard deviation of sales is 1.2884 TV sets.



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Properties of a Binomial Experiment

The experiment consists of a sequence of nidentical trials.

Two outcomes, success and failure, are possible oneach trial.

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

The trials are independent.


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Example: Evans Electronics


Evans is concerned about a low retention rate foremployees. On the basis of past experience,management has seen a turnover of 10% of thehourly employees annually. Thus, for any hourly

employees chosen at random, management estimatesa probability of 0.1 that the person will not be withthe company next year.

Choosing 3 hourly employees a random, what is

the probability that 1 of them will leave the companythis year?

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


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Binomial Probability Function

where:

f(x) = the probability of xsuccesses in ntrialsn= the number of trials

p= the probability of success on any one trial

f x n

x n xp px n x( )

!

!( )!( )( )

1


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Using the Binomial Probability Function

= (3)(0.1)(0.81)

= .243

f x n

x n xp px n x( )

!

!( )!( )( )

1

f( )

!

!( )!( . ) ( . )1

3

1 3 1 0 1 0 9

1 2


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Using the Tables of Binomial Probabilities

p

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

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

1 .2430 .3251 .3840 .4219 .4410 .4436 .4320 .4084 .3750

2 .0270 .0574 .0960 .1406 .1890 .2389 .2880 .3341 .37503 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250


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Using a Tree Diagram


FirstWorker

SecondWorker

ThirdWorker

Valueof x Probab.

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

1

.0810

.0810

.0810


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Expected Value

E(x) = = np

Variance

Var(x) = 2

= np(1 -p) Standard Deviation

SD( ) ( )x np p 1


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Expected Value

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

Variance

Var(x) = 2

= 3(.1)(.9) = .27 Standard Deviation

employees52.)9)(.1(.3)(SD x


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Poisson Probability Distribution

Properties of a Poisson Experiment

The probability of an occurrence is the same forany two intervals of equal length.

The occurrence or nonoccurrence in any interval isindependent of the occurrence or nonoccurrence

in any other interval.


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Poisson Probability Distribution

Poisson Probability Function

where:

f(x) = probability of xoccurrences in an interval = mean number of occurrences in an interval

e= 2.71828

f x e

x

x

( )!


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Example: Mercy Hospital

Using the Poisson Probability Function

Patients arrive at the emergency room of MercyHospital at the average rate of 6 per hour onweekend evenings. What is the probability of 4arrivals in 30 minutes on a weekend evening?

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

f( ) ( . )

!.4

3 2 71828

41680

4 3


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x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0

0 .12 25 .110 8 .1003 .0907 .0821 .0743 .0 672 .06 08 .05 50 .0 498

1 .25 72 .243 8 .2306 .2177 .2052 .1931 .1 815 .17 03 .15 96 .1 494

2 .27 00 .268 1 .2652 .2613 .2565 .2510 .2 450 .23 84 .23 14 .2 240

3 .18 90 .196 6 .2033 .2090 .2138 .2176 .2 205 .22 25 .22 37 .2 240

4 .0 992 .1 082 .11 69 .12 54 .133 6 .141 4 .1488 .1557 .1622 .1 680

5 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .1008

6 .0 146 .0 174 .02 06 .02 41 .027 8 .031 9 .0362 .0407 .0455 .0 504

7 .0 044 .0 055 .00 68 .00 83 .009 9 .011 8 .0139 .0163 .0188 .0 216

8 .0 011 .0 015 .00 19 .00 25 .003 1 .003 8 .0047 .0057 .0068 .0 081

Example: Mercy Hospital

Using the Tables of Poisson Probabilities


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Hypergeometric Probability Distribution

The hypergeometric distribution is closely related to

the binomial distribution. With the hypergeometric distribution, the trials are

not independent, and the probability of successchanges from trial to trial.


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n

N

xn

rN

x

r

xf )(

Hypergeometric Probability Function

for 0 < x< r

where: f(x) = probability of xsuccesses in ntrials

n= number of trials

N= number of elements in the population

r= number of elements in the population

labeled success


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Example: Neveready


Bob Neveready has removed two dead batteriesfrom a flashlight and inadvertently mingled themwith the two good batteries he intended asreplacements. The four batteries look identical.

Bob now randomly selects two of the fourbatteries. What is the probability he selects the twogood batteries?


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Example: Neveready


where:

x= 2 = number of good batteries selected

n= 2 = number of batteries selected

N= 4 = number of batteries in totalr= 2 = number of good batteries in total

167.6

1

!2!2

!4

!2!0

!2

!0!2

!2

2

4

0

2

2

2

)(

n

N

xn

rN

x

r

xf


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End of Chapter 5

ch 05 discrete probability distribution

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