Discrete Probability Distributions (Random Variables and Discrete Probability Distributions)

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Discrete Probability Distributions(Random Variables and

Discrete Probability Distributions)

Chapter 5BA 201

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RANDOM VARIABLES

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

Random Variables

A discrete random variable may assume either a finite number of values or an infinite sequence of values. Example: 1, 2, 3, …

A continuous random variable may assume any numerical value in an interval or collection of intervals. Example: 1.0, 1.1, 1.2, …

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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)

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Discrete Random Variablewith a Finite Number of Values

We can count the TVs sold, and there is a finiteupper limit on the number that might be sold (whichis the number of TVs in stock).

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Let x = number of customers arriving in one day, where x can take on the values 0, 1, 2, . . .

Discrete Random Variablewith an Infinite Sequence of Values

We can count the customers arriving, but there isno finite 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

x = Distance in miles from home 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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DISCRETE PROBABILITY DISTRIBUTIONS

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

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

Discrete Probability Distributions

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The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable.

The required conditions for a discrete probability function are:

Discrete Probability Distributions

f(x) > 0

f(x) = 1

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Discrete Probability DistributionsJSL Appliances

Units Sold(x)

Number of Days f(x)

0 80 0.401 50 0.252 40 0.203 10 0.054 20 0.10

200 1.00

80/200

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.10.20.30.40.50

0 1 2 3 4Values of Random Variable x (TV sales)

Prob

abilit

y

Discrete Probability DistributionsJSL Appliances

Graphicalrepresentationof probabilitydistribution

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Expected Value

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

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

The expected value does not have to be a value the random variable can assume.

E(x) = = x f(x)

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Variance and Standard Deviation

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

The variance is a weighted average of the squared deviations of a random variable from its mean. The weights are the probabilities.

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

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

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Expected ValueJSL Appliances

Units Sold(x)

Number of Days f(x) x f(x)

0 80 0.40 0.001 50 0.25 0.252 40 0.20 0.403 10 0.05 0.154 20 0.10 0.40

E(x) 1.20expected number

of TVs sold in a day

0 * 0.40

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Standard deviation of daily sales = = 1.2884 TVs

VarianceJSL Appliances

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

f(x)0 0.40 0.00 -1.20 1.44 0.5761 0.25 0.25 -0.20 0.04 0.0102 0.20 0.40 0.80 0.64 0.1283 0.05 0.15 1.80 3.24 0.1624 0.10 0.40 2.80 7.84 0.784

Variance of daily sales = 2 1.660

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Discrete Uniform Probability Distribution

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

The discrete uniform probability function is

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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PRACTICEDISCRETE PROBABILITY DISTRIBUTIONS

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Scenario

Amount(x)

Number of

Donors5 35

10 2515 2020 1525 5

A non-profit sends donation solicitations with pre-printed amounts. Based on past years, the non-profit knows approximately how many people will donate each amount. Compute the expected value of x, variance, and standard deviation.

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BINOMIAL PROBABILITY DISTRIBUTION

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Binomial Probability 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. Two outcomes, success and failure, are possible on each trial.

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

stationarity

assumption

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Binomial Probability Distribution

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

We let x denote the number of successes occurring in the n trials.

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

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

x n x

Binomial Probability Distribution Binomial Probability Function

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

x n x

Binomial Probability Distribution Binomial Probability

Function

Probability of a particular sequence of trial outcomes with x successes in n trials

Number of experimental outcomes providing exactly

x successes in n trials

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Binomial Probability Distribution

Evans Electronics Evans Electronics is concerned about a

lowretention rate for its employees. In recent

years,management has seen a turnover of 10% of

thehourly employees annually.

Choosing 3 hourly employees at random, what is

the probability that 1 of them will leave the company

this year?

Thus, for any hourly employee chosen at random,

management estimates a probability of 0.1 that the

person will not be with the company next year.

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Binomial Probability Distribution

Evans ElectronicsThe probability of the first employee leaving

and thesecond and third employees staying, denoted

(S, F, F),is given by

p(1 – p)(1 – p)With a 0.10 probability of an employee leaving on any

one trial, the probability of an employee leaving on

the first trial and not on the second and third trials is

given by (0.10)(0.90)(0.90) = (0.10)(0.90)2 =

0.081

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Binomial Probability Distribution

Evans ElectronicsTwo other experimental outcomes also result in

onesuccess and two failures. The probabilities for

allthree experimental outcomes involving one

successfollow.Experimental

Outcome(S, F, F)(F, S, F)(F, F, S)

Probability ofExperimental Outcome

p(1 – p)(1 – p) = (0.1)(0.9)(0.9) = 0.081

(1 – p)p(1 – p) = (0.9)(0.1)(0.9) = 0.081

(1 – p)(1 – p)p = (0.9)(0.9)(0.1) = 0.081

Total = 0.243

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L (0.1)

S (0.9)

Binomial Probability Distribution

1st Worker 2nd Worker 3rd Worker x Prob.

L (0.1)

S (0.9)

3

2

0

22

0.0010

0.0090

0.0090

0.7290

0.0090

11

0.0810

0.08100.0810

S (0.9)

L (0.1)

L (0.1)

S (0.9)

S (0.9)

L (0.1)

S (0.9)

L (0.1)

1

Evans Electronics Using a tree diagram

S (0.9)

L (0.1)

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Binomial Probability Distribution

f x nx n x

p px n x( ) !!( )!

( )( )

1

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

Let: p = 0.10, n = 3, x = 1

Evans ElectronicsUsing theprobabilityfunction

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Binomial Probability Distribution

(1 )np p

E(x) = = np

Var(x) = 2 = np(1 p)

Expected Value

Variance

Standard Deviation

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Binomial Probability Distribution

3(.1)(.9) .52 employees

E(x) = np = 3(.1) = 0.3 employees out of 3

Var(x) = np(1 – p) = 3(0.1)(0.9) = 0.27

• Expected Value

• Variance

• Standard Deviation

Evans Electronics

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PRACTICEBINOMIAL PROBABILITY DISTRIBUTIONS

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ScenarioKearn Manufacturing (KM) submits bids for sales a number of times each month. Based on an analysis of past bids, the sales director knows that on average 34% of bids result in sales. KM Last week the sales director submitted three bids. The sales director would like to know how likely it is that two of the bids will result in sales.

1. What are n, x (that two bids result in sales), and p.

2. What is f(x)?3. Draw a tree diagram illustrating this scenario.4. Compute the expected value of x.5. Compute the variance and standard deviation.

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Binomial as a Discrete Probability Distribution

x f(x) x*f(x) (x-) (x-)2 (x-)2*f(x)0 0.2875 0.0000 -1.0200 1.0404 0.29911 0.4443 0.4443 -0.0200 0.0004 0.00022 0.2289 0.4578 0.9800 0.9604 0.21983 0.0393 0.1179 1.9800 3.9204 0.1541

1.0200 0.6732

)()1(**)!(!

!)( xnx ppxnx

nxf

E(x)= 1.0200 =n*pVar(x) 0.6732 =n*p*(1-p)

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POISSON PROBABILITY DISTRIBUTION

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A Poisson distributed random variable is often useful in estimating the number of occurrences over a specified interval of time or space

It is a discrete random variable that may assume an infinite sequence of values (x = 0, 1, 2, . . . ).

Poisson Probability Distribution

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Examples of a Poisson distributed random variable:

the number of knotholes in 14 linear feet of pine board

the number of vehicles arriving at a toll booth in one hour

Poisson Probability Distribution

Bell Labs used the Poisson distribution to model the arrival of phone calls.

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Poisson Probability Distribution Two Properties of a Poisson Experiment

2. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

1. The probability of an occurrence is the same for any two intervals of equal length.

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Poisson Probability Function

Poisson Probability Distribution

f x ex

x( )

!

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

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Poisson Probability Distribution Poisson Probability Function

In practical applications, x will eventually becomelarge enough so that f(x) is approximately zero andthe probability of any larger values of x becomesnegligible.

Since there is no stated upper limit for the numberof occurrences, the probability function f(x) isapplicable for values x = 0, 1, 2, … without limit.

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Poisson Probability Distribution

Mercy Hospital Patients arrive at the emergency room of

MercyHospital at the average rate of 6 per hour onweekend evenings. What is the probability of 4 arrivals in 30

minuteson a weekend evening?

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Poisson Probability Distribution

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

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

Mercy Hospital Using theprobabilityfunction

Note: yy

nn 1

91

313 2

2 Example:

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Poisson Probability 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 10Number of Arrivals in 30 Minutes

Prob

abili

ty

actually, the

sequencecontinues:11, 12, …

Mercy Hospital

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Poisson Probability Distribution

A property of the Poisson distribution is thatthe mean and variance are equal.

= 2

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Poisson Probability Distribution

Variance for Number of ArrivalsDuring 30-Minute Periods

= 2 = 3

Mercy Hospital

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PRACTICE POISSON PROBABILITY DISTRIBUTION

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ScenarioOn weekdays between 11:00 a.m. and 2:00 p.m., Joe’s Pizza receives approximately one order every two minutes.

1. What is the expected number of orders per hour?

2. What is the probability three orders will arrive in a five-minute period?

3. What is the probability no orders will arrive in a five-minute period?

4. What is the variance of the number of orders arriving?

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OVERVIEW

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Discrete Probability Distributions• x takes discrete values.• Values for f(x) may be subjectively assigned,

assigned based on frequency of occurrence, or uniform.o If f(x) is uniform, f(x) = 1/n where n is the

number of values x may assume.

E(x) = μ = [x * f(x)]

Var(x) = 2 = [(x – μ)2 * f(x)]

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Binomial Probability Distributions• Four properties: (1) n identical trials, (2)

success or failure, (3) p does not change, and (4) trials are independent.

• x is the number of successes.

E(x) = μ = n * p

Var(x) = 2 = n * p * (1-p)

)()1(**)!(!

!)( xnx ppxnx

nxf

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Poisson Probability Distributions• Number of occurrences over a specified

interval of time or in space.• (1) Probability of occurrence e is same for any

two intervals of equal length and (2) occurrences are independent.

• x is the number of occurrences in the interval.

μ = 2

!*)(xexf

x

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