Chapter 5: Probability Distributions: Discrete Probability Distributions

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2009 University of Minnesota-Duluth, Econ-2030 (Dr. Tadesse)

Chapter 5: Probability Distributions:

Discrete Probability Distributions

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Learning Objectives

Identifying Types of Discrete Probability Distribution and their Respective Functional Representations

Calculating the Mean and Variance of a Discrete Random Variable with Each of the Different Discrete Probability Distribution.

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Random Variable

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Random Variables Any variable that is used to represent the outcomes any

experiment of interest to us is called Random Variable.

A random variable can assume (take) any value (Positive, negative, zero; finite, infinite; continuous and discrete values).

Depending upon the values they take, we can identify two types of random variables:

1. Discrete Variables2. Continuous Variables.

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Both types of variables can assume either a finite number of values or an infinite sequence of values.

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Let x = number of TVs sold at a given store in one day. The number of TV units that can be sold in a given day is finite. It is also discrete: (0, 1, 2, 3, 4).

X can be considered as Discrete Random Variable

Example: JSL Appliances

Discrete random variable with a finite number of values

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Let Y = number of customers arriving in a store in one day. Y 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. However, there is nofinite upper limit on the number that might arrive.

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Random Variables

Question Random Variable x Type

Familysize

X = Number of dependents reported on tax return

Discrete

Distance fromhome to store

Y = Distance in miles from home to the store site

Continuous

Own dogor cat

Z = 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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Probability Distributions

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The probability distribution for a random variable is a distributionthat describes the values that the random variable of interest takes.

The probability distribution is defined by a probability function, denoted by f(x)---the probability of the values of the random variable.

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For any probability function the following conditions must besatisfied:

1. f(x) > 0

2. f(x) = 1

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A Discrete Probability Distribution is a tabular, graphic orFunctional representation of a Random Variable with discrete outcomes that follows the principle of probability distribution

f(x) > 0

f(x) = 1

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• Using past data on sales, a tabular representation of the probability distribution for TV sales was developed.

Number Units Sold of Days

0 80 1 50 2 40 3 10 4 20

x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10 1.00

80/200

Discrete Probability Distributions--Example

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1. Uniform Distribution

2. Binomial Distribution

3. Poisson Distribution

A Discrete Probability Distribution can assume one or more of the Following Distributions:

4. Hyper-Geometric Distribution

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5.1) Discrete Uniform Probability Distribution

The probability distribution of a discrete probability distribution is given by the following formula.

f(x) = 1/n

where:n = the number of values the random variable may assume

the values of the

random variable

are equally likely

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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 p

5.2) Binomial Distribution

The Probability distribution of a Binomial Distribution is given by the following function

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The Probability of a Poisson Distribution is given by the following function

5.3) Poisson Distribution

f x ex

where:f(x) = probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828

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The Probability of a Hyper-geometric Distribution is given by the following Function

5.4) Hypergeometric Distribution

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 population r = number of elements in the population

labeled success

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5.1) Discrete Uniform Probability Distribution

The probability distribution of a discrete probability distribution is given by the following formula.

f(x) = 1/n

where:n = the number of values the random variable may assume

the values of the

random variable

are equally likely

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Expected Value (Mean) and Variance of Discrete Probability Distributions

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Expected Value (Mean) for ……..Discrete Uniform Distribution

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

E(x) = = xf(x)

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Variance and Standard Deviation of Discrete Uniform Probability Distribution

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

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

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

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• The likelihood of selling TV sets in any given day is considered equally likely (Uniform). The following table summarizes sales data on the past 200 days.

Number Units Sold of Days

0 80 1 50 2 40 3 10 4 20

x f(x) 0 .40 1 .25 2 .20 3 .05 4 .10

Example-TV Sales in a Given Store

Given this data what is the Average Number of TVs sold in a day?

x F(x) 0 .40 1 .65 2 .85 3 .90 4 1.00

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Given the data, we can find the Expected Value (Mean Number) of TVs sold in a day as follows

expected number of TVs sold in a day

x f(x) xf(x) 0 .40 .00 1 .25 .25 2 .20 .40 3 .05 .15 4 .10 .40

E(x) = 1.20

Expected Value (MEAN) and Variance

E(x) = = xf(x)

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-1.2-0.2 0.8 1.8 2.8

1.440.040.643.247.84

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

TVs squaredStandard deviation of daily sales = 1.2884 TVs

Find the Variance and Standard Deviation of the Number of TV Sold in a given day.

Var(x) = 2 = (x - )2f(x)= 1.66

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

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5.2) The Binomial Distribution

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5.2) Binomial Distribution

Four Properties of a Binomial Experiment

3. The probability of a success, denoted by p, does not change from trial to trial.

4. The trials are independent.

2. Only two outcomes, success and failure, are possible on each trial.

1. The experiment consists of a sequence of n identical trials.

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Typical Examples of a Binomial Experiment:

• Lottery: Win or Lose

• Election: A Candidate Wins or Loses• Gender of an Employee: is Male or Female• Flipping a coin: Heads or Tails

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Our interest is in the number of successes occurring in the n trials.

let X denote the number of successes occurring in the n trials.

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where:n = the number of trials p = the probability of success on any one trial f(x) = the probability of x successes in n trials

( )!( ) (1 )!( )!x n xnf x p p

Binomial Probability Function

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( )!( ) (1 )!( )!x n xnf x p p

!!( )!

nx n x

( )(1 )x n xp p

Binomial Probability Function

Probability of a particular sequence of outcomes

with x successesNumber of experimental

outcomes providing exactlyx successes in n trials

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Binomial Distribution Example: Evans Electronics

A local Electronics company is concerned about it low retention of employees. In recent years, management has seen an annual turnover of 10% in its hourly employees. Thus, for any hourly employee chosen at random, the company estimates that there is 0.1 probability that the person will leave the company in a year time.

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Binomial Distribution

Given the above information, if we randomly select 3 hourly employees, what is the probability that 1 of them will leave the company in one year ?

f x nx n x

p px n x( ) !!( )!

( )( )

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

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

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Using Tables of Binomial Probabilities

n x .05 .10 .15 .20 .25 .30 .35 .40 .45 .503 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 .37503 .0001 .0010 .0034 .0080 .0156 .0270 .0429 .0640 .0911 .1250

Binomial Distribution

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Mean and Variance of A Binomial Distribution

(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

Given that p=0.1, for the 3 randomly selected hourly employees, what is the expected number and variance of workers who might leave the company this year?

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5.3) Poisson Distribution

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Poisson distribution refers to the probability distribution of a trial that involves cases of rare events that occur over a fixed time interval or within a specified region

6.4. Poisson Distribution

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Examples….

• The number of errors a typist makes per page• The number of cars entering a service station per hour• The number of telephone calls received by a switchboard per

• The number of Bank Failures During a given Economic Recession.

• The number of housing foreclosures in a given city during a given year.

• The number of car accidents in one day on I-35 stretch from the Twin Cities to Duluth

6.4. Poisson Distribution

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We use a Poisson distribution to estimate the number of occurrences of such discrete incidents over a specified interval of time or space

Thus a Poisson distributed random variable is discrete; Often times it assumes an infinite sequence of values (x = 0, 1, 2, . . . ).

Poisson Distribution

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The number of successes (events) that occur in a certain time interval is independent of the number of successes that occur in another time interval.

The probability of a success in a certain time interval is• the same for all time intervals of the same size,• proportional to the length of the interval.

The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller.

Properties of a Poisson Experiment

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Poisson Probability Function

f x ex

where:f(x) = probability of x occurrences in an interval = mean number of occurrences in an interval e = 2.71828

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On Average 6 Patients per hour arrive at the emergency room of Mercy Hospital on

weekend evenings.

What is the probability of 4 arrivals in30 minutes on a weekend evening?

Poisson Distribution--Example

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Poisson Distribution-Example

Using the Poisson Probability Function

4 33 (2.71828)(4) .16804!f

m= 6/hour = 3/half-hour, P(x = 4)?

f x ex

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Using Poisson Probability Tables

x 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.00 .1225 .1108 .1003 .0907 .0821 .0743 .0672 .0608 .0550 .04981 .2572 .2438 .2306 .2177 .2052 .1931 .1815 .1703 .1596 .14942 .2700 .2681 .2652 .2613 .2565 .2510 .2450 .2384 .2314 .22403 .1890 .1966 .2033 .2090 .2138 .2176 .2205 .2225 .2237 .22404 .0992 .1082 .1169 .1254 .1336 .1414 .1488 .1557 .1622 .16805 .0417 .0476 .0538 .0602 ..0668 .0735 .0804 .0872 .0940 .10086 .0146 .0174 .0206 .0241 .0278 .0319 .0362 .0407 .0455 .05047 .0044 .0055 .0068 .0083 .0099 .0118 .0139 .0163 .0188 .02168 .0011 .0015 .0019 .0025 .0031 .0038 .0047 .0057 .0068 .0081

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Mean and Variance of Poisson Distribution

Another special property of the Poisson distribution is that the mean and variance are equal.

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Poisson Distribution MERCY

Variance for Number of ArrivalsDuring 30-Minute Periods

= 2 = 3

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5.4) Hyper-Geometric Distribution

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Hyper-geometric Distribution

The hyper-geometric distribution is closely related to the binomial distribution.

However, for the hyper-geometric distribution:

the trials are not independent, and

the probability of success changes from trial to trial.

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Hyper-geometric Probability Function

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 population r = number of elements labeled as success

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Hyper-geometric Probability Function

r N rx n x

for 0 < x < r

number of waysn – x failures can be selectedfrom a total of N – r failures

in the populationnumber of ways

x successes can be selectedfrom a total of r successes

in the population number of waysa sample of size n can be selected

from a population of size N

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( ) rE x nN

2( ) 11

r r N nVar x nN N N

Variance

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Hyper-geometric Distribution: Inspection

Electric fuses produced by a given company are packed in boxes. Each box carries 12 units of electric fuses. The role of the inspector in the company that manufactures the fuses is to make sure that all fuses in each box are in good condition.

Consider the following scenario: A worker inadvertently places 5 defective items in a box. As part of her job, the inspector randomly selects 3 fuses from the box that contains the defective fuses.

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1. What is the probability that none of the three randomly selected fuses are defective?

2. What is the probability that the inspector finds only one of the three randomly selected fuses to be defective?

3. What is the probability that the inspector finds at least one of the three fuses defective?

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1. What is the probability that none of the three randomly selected fuses are defective?

Solution:N=12; r=5; n=3; x=0;

P(x=0)?

1591.022035

!9!3!12!4!3!7

!5!0!5

)())(()0( 12

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2. What is the probability that the inspector finds only one of the three randomly selected fuses to be defective?

Solution:N=12; r=5; n=3; x=1;

P(x=1)?

4773.0220215

!9!3!12!5!2!7

!4!1!5

)())(()1( 12

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2. What is the probability that the inspector finds at least one of the three fuses defective?

Solution:N=12; r=5; n=3; x 1

8409.01591.01

)0(1?)1(

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Example: NevereadyBob has removed two dead batteries from a flashlight and inadvertently mingled them with the two good batteries that he intended to use as replacements. The four batteries look identical.

Then Bob randomly selects two of the four batteries. What is the probability that he selects the two good batteries?

ZAPZAPZAP

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Using the Hyper-geometric Function

2 2 2! 2!2 0 2!0! 0!2! 1( ) .1674 4! 6

2 2!2!

r N rx n x

f x Nn

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

x = 2 =number of good batteries to be selected n = 2 = number of batteries selected

167.061

!2!2!4!2!0!2

!0!2!2

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When the population size is large, a hyper-geometric distribution can be approximated by a binomial distribution with n trials and a probability of success p = (r/N).

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Consider a hyper-geometric distribution with n trials and let p = (r/n) denote the probability of a success on the first trial. If the population size is large, the term (N – n)/(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 expected value and variance of a binomial distribution.

Chapter 5: Probability Distributions: Discrete Probability Distributions

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