Ch 05 Discrete Probability Distribution
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Transcript of 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
.10
.20
.30
.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
Example: JSL Appliances
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Example: JSL Appliances
Graphical Representation of the Probability
Distribution
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.20
.30
.40
.50
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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Example: JSL Appliances
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.
Example: JSL Appliances
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Binomial Probability Distribution
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
Binomial Probability Distribution
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 Distribution
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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Example: Evans Electronics
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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Example: Evans Electronics
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
Example: Evans Electronics
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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Binomial Probability Distribution
Expected Value
E(x) = = np
Variance
Var(x) = 2
= np(1 -p) Standard Deviation
SD( ) ( )x np p 1
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Example: Evans Electronics
Binomial Probability Distribution
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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Hypergeometric Probability Distribution
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
Hypergeometric Probability Distribution
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
Hypergeometric Probability Distribution
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