Chapter 5 Discrete Probability Distributions

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Chapter 5

Discrete Probability Distributions

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Chapter Outline

Random Variables Discrete Probability Distributions Expected Value and Variance Binomial Probability Distribution

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Random Variables

A random variable is a numerical description of the outcome of an experiment.

A discrete random variable assumes numerical values that have gaps or jumps between them.

A A continuous random variablecontinuous random variable assumes numerical assumes numerical values that have NO gaps or jumps between them.values that have NO gaps or jumps between them.

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Random Variables

Experiment Random Variable xx Type

Take a quiz with 10 Ture/False questions

xx = Number of correct

answers

Discrete

Run 5K x = x = time to finish a 5k run

Continuous

Weigh a sample of 36 cans of coffee (labeled as 3lbs

x = x = the average weight of a sample of 36 cans of coffee

Continuous

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The The probability distributionprobability distribution for a random variable for a random variable describes how probabilities are distributed over the values describes how probabilities are distributed over the values of the random variable.of the random variable.

We can describe a discrete probability distribution with a We can describe a discrete probability distribution with a table, graph, or formula.table, graph, or formula.

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The probability distribution is defined by a The probability distribution is defined by a probability probability functionfunction, denoted by , denoted by ff((xx), which provides the probability ), which provides the probability for each value of the random variable.for each value of the random variable.

The required conditions for a discrete probability function The required conditions for a discrete probability function are:are:

ff((xx) ) >> 0 0ff((xx) ) >> 0 0

ff((xx) = 1) = 1ff((xx) = 1) = 1

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Example: Probabilities of the # of correct answers to a quiz of 4 Example: Probabilities of the # of correct answers to a quiz of 4 True/False questions.True/False questions. We can use a table to represent the probability distribution.We can use a table to represent the probability distribution. The random variable The random variable xx represents the number of correct answers. represents the number of correct answers.

xx ff((xx)) 00 .10 .10 11 .25 .25 22 .35 .35 33 .20 .20 44 .10 .10

1.001.00

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Discrete Uniform Probability Distribution

f(x) = 1/nf(x) = 1/n

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

the values of the

random variable

are equally likely

The discrete uniform probability distribution is the simplest example of a discrete probability distribution given by a formula.

The discrete uniform probability function is

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Expected Value

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

The expected value is a weighted average of the values the The expected value is a weighted average of the values the random variable may assume. The weights are the random variable may assume. The weights are the probabilities.probabilities.

The arithmetic mean introduced in Chapter 3 can be viewed as a special case of weighted average, where the weights for all the values are the same, i.e. 1/n.

EE((xx) = ) = = = xfxf((xx))EE((xx) = ) = = = xfxf((xx))

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Variance and Standard Deviation

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

The variance is a weighted average of the squared The variance is a weighted average of the squared

deviations of a random variable from its mean. The deviations of a random variable from its mean. The weights are the probabilities.weights are the probabilities.

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

Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx))Var(Var(xx) = ) = 22 = = ((xx - - ))22ff((xx))

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Expected Value

Example: Take a quiz (# of correct answers)

xx ff((xx)) xfxf((xx))

00 .10 .10 .00 .00

11 .25 .25 .25 .25

22 .35 .35 .70 .70

33 .20 .20 .60 .60

44 .10 .10 .40.40

EE((xx) = 1.95) = 1.95

Expected number of correct answers

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Variance

Example: Take a quiz (# of correct answers)

00

11

22

33

44

-1.95-1.95

-0.95-0.95

0.050.05

1.051.05

2.052.05

3.80253.8025

0.90250.9025

0.00250.0025

1.10251.1025

4.20254.2025

.10.10

.25.25

.35.35

.20.20

.10.10

.3803.3803

.2256.2256

.0009.0009

.2205.2205

.4203.4203

x - x -

Variance of daily sales = Variance of daily sales = 22 = 1.2476 = 1.2476

Standard deviation of daily sales = Standard deviation of daily sales = = 1.11 correct answers = 1.11 correct answers

((x - x - ))22 ff((xx)) ((xx - - ))22ff((xx))xx

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Binomial Probability Distribution

Properties of a Binominal Experiment:

1.1. The experiment consists of a sequence of The experiment consists of a sequence of nn identical identical trials;trials;

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

3.3. The probability of a success, denoted by The probability of a success, denoted by pp, does not , does not change from trial to trial;change from trial to trial;

4.4. The trials are independent from one another.The trials are independent from one another.

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

We let We let xx denote the number of successes occurring in the denote the number of successes occurring in the nn trials. trials.

Either outcome can be named as ‘Success’. We need to Either outcome can be named as ‘Success’. We need to make sure that in the calculation, the probability make sure that in the calculation, the probability pp is is matched with the definition of the random variable matched with the definition of the random variable xx..

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Binomial Probability Function

( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x

where:where:

xx = the number of successes = the number of successes

pp = the probability of a success on one trial = the probability of a success on one trial

nn = the number of trials = the number of trials ff((xx) = the probability of ) = the probability of xx successes in successes in nn trials trials nn! = ! = nn((nn – 1)( – 1)(nn – 2) ….. (2)(1) – 2) ….. (2)(1)

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Binomial Probability Function

( )!( ) (1 )

!( )!x n xn

f x p px n x

( )!( ) (1 )

!( )!x n xn

f x p px n x

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

Number of experimentalNumber of experimental outcomes providing exactlyoutcomes providing exactly

xx successes in successes in nn trials trials

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with x successes in with x successes in nn trials trials

Probability of a particularProbability of a particular sequence of trial outcomessequence of trial outcomes with x successes in with x successes in nn trials trials

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Binomial Probability Distribution Example: Purchasing a pair of shoes

Based on recent sales data, a shoe store manager estimates that the probability a customer makes a purchase is 30%. For the next three customers, what is the probability that exactly 1 of them will make a purchase?

Analysis: Is this example a binomial experiment? If so, which outcome is to be named ‘Success’? And what is the probability of Success?

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Does the example satisfy the properties of a binomial distribution?

N trials? – Yes, 3 trials ( 3 customers)

Two outcomes for each trial? – Yes, purchase or not

Probability of success stays the same – 30% chance for making a purchase can be assumed to be the same for all the customers.

Independent trials – Assume the three customers are independent in their decision on making a purchase.

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How many favorable outcomes are there where exactly ONE of the next three customers makes a purchase? With the success representing ‘making a purchase’ and the three customers assumed to be independent, we should have the following outcomes and their probabilities:

ExperimentalExperimentalOutcomeOutcome

((SS, , FF, , FF))((FF, , SS, , FF))((FF, , FF, , SS))

Probability ofProbability ofExperimental OutcomeExperimental Outcome

pp(1 – (1 – pp)(1 – )(1 – pp) = (.3)(.7)(.7) ) = (.3)(.7)(.7) = .147= .147

(1 – (1 – pp))pp(1 – (1 – pp) = (.7)(.3)(.7) ) = (.7)(.3)(.7) = .147= .147

(1 – (1 – pp)(1 – )(1 – pp))pp = (.7)(.7)(.3) = (.7)(.7)(.3) = = .147.147

Total = .441Total = .441

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f xn

x n xp px n x( )

!!( )!

( )( )

1f xn

x n xp px n x( )

!!( )!

( )( )

1

LetLet: p: p = .30, = .30, nn = 3, = 3, xx = 1 = 1Using theUsing theprobabilityprobabilityfunctionfunction

Using theUsing theprobabilityprobabilityfunctionfunction

441.049.03.037.03.0!13!1

!3)1( 21

f

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11stst Customer Customer 11stst Customer Customer 22ndnd Customer Customer22ndnd Customer Customer 33rdrd Customer Customer33rdrd Customer Customer xxxx Prob.Prob.Prob.Prob.

PurchasePurchase (.3)(.3)PurchasePurchase (.3)(.3)

Not Not PurchasePurchase (.7)(.7)

Not Not PurchasePurchase (.7)(.7)

3333

2222

0000

2222

2222

P (.3)P (.3)P (.3)P (.3)

P (.3)P (.3)P (.3)P (.3)

NP (.7)NP (.7)NP (.7)NP (.7)

NP (.7)NP (.7)NP (.7)NP (.7)

NP (.7)NP (.7)NP (.7)NP (.7)

NP(.7)NP(.7)NP(.7)NP(.7)

NP (.7)NP (.7)NP (.7)NP (.7)

NP (.7)NP (.7)NP (.7)NP (.7)

P (.3)P (.3)P (.3)P (.3)

P (.3)P (.3)P (.3)P (.3)

P (.3)P (.3)P (.3)P (.3)

P (.3)P (.3)P (.3)P (.3) .027.027.027.027

.063.063.063.063

.063.063.063.063

.343.343.343.343

.063.063.063.063

1111

1111

.147.147.147.147

.147.147.147.147

.147.147.147.147

1111

Example: Purchasing a pair of shoes Using a tree diagramUsing a tree diagramUsing a tree diagramUsing a tree diagram

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Binomial Probability Table Statisticians have developed tables that give probabilities

and cumulative probabilities for a binomial random variable. In the appendix of our textbook, you can locate the binomial probability tables.

For our example, the table is presented as below (where x represents the number of success):

xx ff((xx)) 00 .343 .343 11 .441 .441 22 .189 .189 33 .027.027 1.001.00

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Binomial Probability Distribution We can apply the formulas of expected value and variance

for a binomial probability distribution. However, those formulas can be further simplified as follows:

(1 )np p (1 )np p

EE((xx) = ) = = = npnp

Var(Var(xx) = ) = 22 = = npnp(1 (1 pp))

Expected ValueExpected Value

VarianceVariance

Standard DeviationStandard Deviation

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Expected ValueExpected Value

VarianceVariance

Standard DeviationStandard Deviation

EE((xx) = ) = npnp = 3(.3) = .9 customers out of 3 = 3(.3) = .9 customers out of 3

Var(Var(xx) = ) = npnp(1 – (1 – pp) = 3(.3)(.7) = .63) = 3(.3)(.7) = .63

79.063.07.03.03 customers

Chapter 5 Discrete Probability Distributions

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