Chapter 11

Discrete Random Variables and their Probability Distributions

Random Variables

• A random variable is a numerical outcome of a random experiment

• A discrete random variable can take on only specific, isolated numerical values, – Finite discrete random variables: Discrete random variables that can

take on only finitely many values (like the outcome of a roll of a die) – Infinite discrete random variables: Discrete random variables that can

take on an unlimited number of values (like the number of stars estimated to be in the universe)

• A continuous random variable can take on any values within a continuous range or an interval (like the temperature, or the height of an athlete in centimeters.)

Random VariablesExample

• Experiment: Next four customers who enter a bank• Random variable x: the number of customers who make a

deposit (D)• x = 1 represents the event “exactly one customer makes a

deposit”x Outcomes

0 NNNN

1 DNNN,NDNN,NNDN,NNND

2 DDNN,DNDN,DNND,NDDN,NDND,NNDD

3 DDDN,DDND,DNDD,NDDD

4 DDDD

D represents a customer who makes a deposit, N represents a customer who does not.

Discrete vs. Continuous Random Variables

DISCRETE CONTINUOUS

Values that can be counted and ordered

Values that cannot be counted

Gap between consecutive values On continuous spectrum

Examples:1) Insurance claims filed in one day2) Cars sold in one month3) Employees who call in sick on a day

Examples:1)Time to check out a customer2)Weight of an outgoing shipment3)Distance traveled by a truck in a single day4)Price of a gallon of gas

Measure with a specific amount of precision

Probability Distributions of a Discrete Random Variable

• The distribution of a random variable is the collection of possible outcomes along with their probabilities. This may be described by a table, a formula, or a probability histogram.

• The probability assigned to each value of x lies in the range 0-1.

• The sum of all the probabilities of x must equal 1.

1)(0 xP

1)(xP

Probability Distributions of a Discrete Random Variable: Example

• The probability distribution of x describes a list of all the possible values that a x can assume and their corresponding probabilities.

CALCULATE the probability of :• P(Exactly one depositor in four customers) or P(x=1)• P(two or more depositors) or P(x2)• P(Fewer than four depositors) or (P(x<4)

x Outcomes P(X)

0 NNNN 1/16 = .0625

1 DNNN, NDNN, NNDN, NNND 4/16 = .2500

2 DDNN, DNDN, DNND, NDDN, NDND, NNDD 6/16 = .3750

3 DDDN, DDND, DNDD, NDDD 4/16 = .2500

4 DDDD 1/16 = .0625

Mean of a Discrete Random Variable

• The mean of a discrete random variable, , is actually the mean of its probability distribution.

• The mean is also called the expected value and is denoted by E (x).

x Outcomes P(X) xP(x)

0 NNNN 1/16 = .0625 0 (.0625)= 0

1 DNNN,NDNN,NNDN,NNND 4/16 = .2500 1 (.25)= .25

2 DDNN,DNDN,DNND,NDDN,NDND,NNDD 6/16 = .3750 2 (.375)= .75

3 DDDN,DDND,DNDD,NDDD 4/16 = .2500 3 (.25)= .75

4 DDDD 1/16 = .0625 4 (.0625)= .25

xP(x) = 2.00

)()( xxPxE

Standard Deviation of a Discrete Random Variable

• The standard deviation, , of a discrete random variable measures the spread of its probability distribution.

• A higher value indicated that x can assume values over a larger range ± the mean.

x P(X) xP(x) x xP(x)

0 1/16 = .0625 0 (.0625)= 0 0 0

1 4/16 = .2500 1 (.25)= .25 1 .25

2 6/16 = .3750 2 (.375)= .75 4 1.5

3 4/16 = .2500 3 (.25)= .75 9 2.25

4 1/16 = .0625 4 (.0625)= .25 16 1

xP(x) = 2.00 xP(x) = 5

)()( 2 xPx 22 )( xPx

1

25

)(

2

22

xPx

EXAMPLE: Discrete Random Variable Calculations

• Accidents do happen at Brown’s Manufacturing Corp. Let x be the number of accidents that occur during a month, with their probability distribution.

• Find:

X P(X)

0 .25

1 .30

2 .20

3 .15

4 .10

Binomial Distribution

A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties:•The experiment consists of n repeated trials.•Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.•The probability of success, denoted by P, is the same on every trial.•The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

•Binomial probability distributions one of the most widely used discrete probability distributions

Combinations

• Combinations: Give the number of ways x elements can be selected from n elements.

– Combinations are expressed as nCx (meaning n elements from which x elements are selected)

– Example: How many ways to combine managers A,B,C and D in groups of two? 6C2

• AB,AC,AD,BC,BD,CD (so 6. Note that order doesn’t matter).

– Calculated as:

– Calculate the above example (selecting 2 out of 6 managers) using the formula.

)!(!

!

xnx

nCxn

Binomial Probability Distribution

• The probability distribution of x in binomial experiments is called binomial (probability) distribution.

• The probability of exactly x successes in n trials is:

P(x)= nCx pxqn-x

Where n = total number of trialsp= probability of successq= probability of failure = 1-px=number of successes in n trialsn-x= number of failures in n trials

Example: 15% of engineering students work full time. Using the binomial probability formula find the probability that in a random sample of 5 students, the number who work full time is:

(a) exactly 0 (b) exactly 2 (c) exactly 1

Binomial Probability Distribution

Binomial Distribution using MINITAB

• Quality department selects n items from a shipment and observes the number of defective items. If the number of defects is not more than x the shipment is accepted.

– Oaks received motors in shipments of 500– Quality control randomly selects 20 motors to inspect– If the sample has more than 2 defective motors the shipment is rejected.– The supplier promised that only 5% of its motors are defective.

• Find the probability that a given shipment of 500 motors will be accepted. AND find the probability that a given shipment of 500 motors will be rejected.

- In MINITAB select Calc>Probability Distributions>Binomial- Choose cumulative probability

- Enter number of trials

- Enter Probability of success

- Choose to input constant (to be the value for x)

- Click ok

Binomial DistributionMean and Std Deviation

= np

Where n = total number of trials

p= probability of success

q=probability of failure = 1-p

npq

Poisson Probability Distribution

A Poisson experiment is a statistical experiment that has the following properties:

• The experiment results in outcomes that can be classified as successes or failures.

• The average number of successes (μ) that occurs in a specified region is known.

• The probability that a success will occur is proportional to the size of the region.

• The probability that a success will occur in an extremely small region is virtually zero.

Poisson Probability Distribution

• Applied to experiments with random and independent occurrences.– Each event is called an occurrence– Independence means that one occurrence of an event does not

influence successive occurrences

– Good Examples: • Accidents that occur at a company during a one-month period• Number of customers in a grocery store during a one-hour interval.

– Not good Examples:• Patients that arrive at a doctor’s office (since they have

appointments, not random).• Arrival of commercial airplanes at an airport.

Calculating Poisson

represents average number of occurrences in an interval. • x represents the actual number of occurrences• e is approximately 2.71828

EXAMPLE: Let be the average number of customers using an ATM per hour. = 5. What is the probability (x) that 8 customers will try to use the machine in an hour.

- Solve using formula

- Solve using MINITAB Calc->Probability Distributions->Poisson

!)(

x

exP

x

Poisson Probability Distribution Mean and Std Deviation

=

Poisson Probability Distribution Mean and Std Deviation

Negative Binomial Distribution

A negative binomial experiment is a statistical experiment that has the following properties:•The experiment consists of x repeated trials.•Each trial can result in just two possible outcomes, a success and a failure.•The probability of success, denoted by P, is the same on every trial.•The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.•The experiment continues until r successes are observed, where r is specified in advance.

Negative Binomial Distribution

• A negative binomial random variable is the number X of repeated trials to produce r successes in a negative binomial experiment.

• The negative binomial distribution is also known as the Pascal distribution.

Negative Binomial Distribution

x: The number of trials required to produce r successes in a negative binomial experiment.

r: The number of successes in the negative binomial experiment.

p: The probability of success on an individual trial.

q: The probability of failure on an individual trial. (This is equal to 1 - P.)

rxrrx qpCxP 11)(

Negative Binomial DistributionMean and Variance

p

r

: the average no. of trials required to produce r successes

22 )1(

p

pr

Negative Binomial DistributionsExample

• Bob is a high school basketball player. He is a 70% free throw shooter. That means his probability of making a free throw is 0.70. During the season, what is the probability that Bob makes his third free throw on his fifth shot?

• Solution: The probability of success (p) is 0.70, the number of trials (x) is 5, and the number of successes (r) is 3.

1852.)3(.)7(.)( 232411

CqpCxP rxrrx

Geometric Distribution

• The geometric distribution is a special case of the negative binomial distribution. It deals with the number of trials required for a single success. Thus, the geometric distribution is negative binomial distribution where the number of successes (r) is equal to 1.

• Outcomes are either success/failure. Trial continues until success (defect) occurs for the first time.– Useful for manufacturing where the line will be shut down for

recalibration upon first defect.

Geometric Distribution

• Negative Binomial Distribution:

• Geometric Distribution:

x: The number of trials required to produce 1 success in a geometric experiment.

p: The probability of success on an individual trial.

q: The probability of failure on an individual trial. (This is equal to 1 - P.)

rxrrx qpCxP 11)(

1)( xpqxP

Geometric DistributionMean and Variance

p

1

: the average no. of trials required to produce 1 success

22 1

p

p

Geometric DistributionExample

Bob is a high school basketball player. He is a 70% free throw shooter. That means his probability of making a free throw is 0.70. What is the probability that Bob makes his first free throw on his fifth shot?

Solution:

Probability of success (p) is 0.70, the number of trials (x) is 5, and the number of successes (r) is 1. We enter these values into the geometric formula.

00567.)3)(.7(.)( 41 xpqxP

Geometric DistributionExample

Military contractor is producing nuts that must be within .04 mm of specified diameter. If nut exceeds the limit the line must be shut down and adjusted. The probability that the diameter of a nut will exceeds the allowable error is .0014.

• What is the probability the machine will be shut down exactly after the 100th nut is produced?

• What is the probability the machine will be shut down exactly after the 200th nut is produced?

Hypergeometric Probability Distribution

• A sample of size n is randomly selected without replacement from a population of N items.

• In the population, r items can be classified as successes, and N - r items can be classified as failures.

• A hypergeometric random variable, x, is the number of successes that result from a hypergeometric experiment

Hypergeometric Probability Distribution

Where N = total number of elements in the population

r = number of success in the population

N-r = number of failures in the population

n = number of trials (sample size)

x = number of successes in trial

n-x = number of failures in n trials

nN

xnrNxr

C

CCxP

)(

Hypergeometric DistributionMean and Variance

np

Where p= r/N

)1

)(1(2

N

nNpnp

Hypergeometric Probability DistributionExample

2215.)0(552

539013

C

CCxP

Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts?Solution: N = 52; since there are 52 cards in a deck.r = 13; since there are 13 hearts in a deck.n = 5; since we randomly select 5 cards from the deck.x = 0 to 2; since our selection includes 0, 1, or 2 hearts.We plug these values into the hypergeometric formula as follows:

4114.)1(552

439113

C

CCxP

2743.)2(552

339213

C

CCxP

Hypergeometric Probability in MINITAB

• Acceptance testing of ice cream cones Ice cream parlor checks a batch of 400 waffle cones by checking 50 of them. They will not buy them if more than 3 cones are broken.

• What is the probability that the parlor will buy the cones if 35 of the 400 cones are broken.– Define , n, r, N-r, x

– In MINITAB select: Calc-> Probability Distributions -> Hypergeometric