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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.
Discrete Probability Distributions
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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:
Discrete Probability Distributions
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
Discrete Probability Distributions
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.10
.20
.30
.40
.50
0 1 2 3 4Values of Random Variable x (TV sales)
Probability
Discrete Probability Distributions
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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Expected Value and Variance
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
Expected Value and Variance
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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
Expected Value and Variance
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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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Binomial Distribution
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 Distribution
Binomial Probability Function
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( )!( ) (1 )!( )!
x n xnf x p px n x
Binomial Distribution
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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Binomial Distribution
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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Binomial Distribution
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
Binomial Distribution
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
Binomial Distribution
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Binomial Distribution
(1 )np p
E(x) = = np
Var(x) = 2 = np(1 p)
Expected Value
Variance
Standard Deviation
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Binomial Distribution
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
Poisson Distribution
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Poisson Distribution
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
Poisson Distribution
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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Poisson Distribution
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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Poisson Distribution
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 Distribution
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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Poisson Distribution
A property of the Poisson distribution is thatthe mean and variance are equal.
= 2
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Poisson Distribution
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
Hypergeometric Distribution
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
Hypergeometric Distribution
( )
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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Hypergeometric Distribution
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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Hypergeometric Distribution
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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Hypergeometric Distribution
( )r
E x nN
2( ) 1
1
r r N nVar x n
N N N
Mean
Variance
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Hypergeometric Distribution
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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Hypergeometric Distribution
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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Hypergeometric Distribution
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).