discrete random variables

Slide 1

Econ10/Mgt 10 Stuffler1Discrete Random Variables Discrete Probability Distributions Binomial Distribution Poisson Distribution

Hypergeometric Distribution

Econ10/Mgt 10 Stuffler2Discrete Probability Distributions Discrete Random Variables Sample Space includes all mutually exclusive outcomes

Probabilities from subjective, frequency or subjective methods Two conditions apply:

Econ10/Mgt 10 Stuffler3 Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household (millions) from US survey data:

Discrete Probability Distributions

Econ10/Mgt 10 Stuffler4Discrete Probability Distributions Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data:

1,218 101,501 = 0.012

Econ10/Mgt 10 Stuffler5Discrete Probability Distributions Probability distributions can be estimated from relative frequencies. Consider the discrete (countable) number of televisions per household from US survey data:

EX: P(X=4) = P(4) = 0.076 = 7.6%

Econ10/Mgt 10 Stuffler6Discrete Probability Distributions What is the probability there is at least one televisionbut no more than three in any given household?

Econ10/Mgt 10 Stuffler7Discrete Probability Distributions What is the probability there is at least one television but no more than three in any given household?

Econ10/Mgt 10 Stuffler8Discrete Probability Distributions What is the probability there is at least one television but no more than three in any given household?

at least one television but no more than threeP(1 X 3) = P(1) + P(2) + P(3) = .319 + .374 + .191 = .884

Econ10/Mgt 10 Stuffler9Discrete Probability Distributions Assume a mutual fund salesman knowsthat there is 20% chance of closing a sale on each call he makes.

Econ10/Mgt 10 Stuffler10Discrete Probability Distributions What is the probability distribution of the number of sales if he plans to call three customers?

Let S denote success, making a sale: P(S) = .20,

then S, not making a sale: P(S) = .80

Econ10/Mgt 10 Stuffler11Discrete Probability Distributions Developing a Probability Distribution TreeP(S)=.2P(S)=.8Sales Call 1

Econ10/Mgt 10 Stuffler12Discrete Probability Distributions Developing a Probability Distribution TreeP(S)=.2P(S)=.8P(S)=.2P(S)=.8P(S)=.8P(S)=.2Sales Call 1Sales Call 2

Econ10/Mgt 10 Stuffler13Discrete Probability Distributions Developing a Probability Distribution TreeP(S)=.2P(S)=.8P(S)=.2P(S)=.2P(S)=.2P(S)=.2P(S)=.8P(S)=.8P(S)=.8P(S)=.8S S S

S S SS S S

S S SS S S

S S SS S S

S S SP(S)=.2P(S)=.8P(S)=.8P(S)=.2Sales Call 1Sales Call 2Sales Call 3

Econ10/Mgt 10 Stuffler14Discrete Probability Distributions Developing a Probability Distribution TreeP(S)=.2P(S)=.8P(S)=.2P(S)=.2P(S)=.2P(S)=.2P(S)=.8P(S)=.8P(S)=.8P(S)=.8S S S

S S SS S S

S S SS S S

S S SS S S

S S SP(S)=.2P(S)=.8P(S)=.8P(S)=.2Sales Call 1Sales Call 2Sales Call 3

Econ10/Mgt 10 Stuffler15Discrete Probability Distributions Developing a Probability DistributionP(S)=.2P(S)=.8P(S)=.2P(S)=.2P(S)=.2P(S)=.2P(S)=.8P(S)=.8P(S)=.8P(S)=.8S S S

S S SS S S

S S SS S S

S S SS S S

S S SP(S)=.2)=.8P(S)=.8P(S)=.2XP(x).23 = .0083(.032)=.0963(.128)=.3840.83 = .512Sales Call 1Sales Call 2Sales Call 3

Econ10/Mgt 10 Stuffler16Discrete Probability Distributions Explain how to derive .032 and .128P(S)=.2P(SC)=.8P(S)=.2P(S)=.2P(S)=.2P(S)=.2P(SC)=.8P(SC)=.8P(SC)=.8P(SC)=.8S S S

S S SCS SC S

S SC SCSC S S

SC S SCSC SC S

SC SC SCP(S)=.2P(SC)=.8P(SC)=.8P(S)=.2XP(x).23 = .0083(.032)=.0963(.128)=.3840.83 = .512Sales Call 1Sales Call 2Sales Call 3

Econ10/Mgt 10 Stuffler17Discrete Probability Distributions The P(X=2) is:P(S)=.2P(SC)=.8P(S)=.2P(S)=.2P(S)=.2P(S)=.2P(SC)=.8P(SC)=.8P(SC)=.8P(SC)=.8S S S

S S SCS SC S

S SC SCSC S S

SC S SCSC SC S

SC SC SCP(S)=.2P(SC)=.8P(SC)=.8P(S)=.2XP(x).23 = .0083(.032)=.0963(.128)=.3840.83 = .512(.2)(.2)(.8)= .032Sales Call 1Sales Call 2Sales Call 3

Econ10/Mgt 10 Stuffler18Discrete Probability DistributionsA discrete probability distribution represents a population

Example: Population of number of TVs per household Example: Population of sales call outcomes

Since we have populations, we can describe them by computing various parameters:

Population Mean and Population Variance

Econ10/Mgt 10 Stuffler19Discrete Probability DistributionsPopulation Mean (Expected Value) - Weighted average of all values with the weights being the probabilities

- Expected value of X, E(X)

Econ10/Mgt 10 Stuffler20Discrete Probability Distributions Population variance - weighted average of the squared deviations from the mean.

Short cut formula for the variance

Standard deviation formula

Econ10/Mgt 10 Stuffler21Discrete Probability Distributions Find the mean, variance, and standard deviation for the population of the number of color televisions per household.

= 0(.012) + 1(.319) + 2(.374) + 3(.191) + 4(.076) + 5(.028) = 2.084

Econ10/Mgt 10 Stuffler22Discrete Probability DistributionsFind the mean, variance, and standard deviation for the population of the number of color televisions per household.

= (0 2.084)2(.012) + (1 2.084)2(.319)++(5 2.084)2(.028)= 1.107

Econ10/Mgt 10 Stuffler23Discrete Probability Distributions Find the mean, variance, and standard deviation for the population of the number of color televisions per household.

= 1.052

Econ10/Mgt 10 Stuffler24Discrete Probability DistributionsSpecial application of Expected Value

Suppose the probability that an insurance agent makes a sale is .20 and after costs earns a commission of $525.

If he/she does not make a sale, they must pay $75 in costs.

What is their expected value from a sales call?

Does the benefit exceed the cost or is the opposite true?

Econ10/Mgt 10 Stuffler25Discrete Probability DistributionsSpecial application of Expected Value

Let X be the discrete random variable of making a sale call

x P(x) xP(x) Sale $ 525 .20 $ 105 No Sale - $ 75 .80 - $ 60

E(x) = = xP(x) = $ 45

Econ10/Mgt 10 Stuffler26Binomial Distribution Binomial distribution is the probability distribution that results from doing a binomial experiment which have the properties:

Fixed number of identical trials, represented as n.Each trial has two possible outcomes: a success or failureFor all trials, the probability of success, P(success)=p, and the probability of failure, P(failure)=1p=q, are constant. The trials are independent

Econ10/Mgt 10 Stuffler27Several Binomial Distributions

Econ10/Mgt 10 Stuffler28Binomial DistributionSuccess and failure: labels for binomial experiment outcomes, no value judgment is implied.

EX: Coin flip results in either heads or tails.

If we define heads as success, then tails is considered a failure.

Other binomial examples: An election candidate wins or loses An employee is male or female A worker is employed or unemployed

Econ10/Mgt 10 Stuffler29Binomial Distribution

where x = number of successes in n trials,n x = number of failures in n trials,px = the probability of success raised to the number of successes, and qn-x = probability of failure raised to the number of failures

Binomial Formula:

Econ10/Mgt 10 Stuffler30Binomial DistributionThe random variable of a binomial experiment is defined as the number of successes in the n trials, and is called the binomial random variable.

EX: Flip a fair coin 10 times1) Fixed number of trials n=102) Each trial has two possible outcomes {heads (success), tails (failure)} 3) P(success)= 0.50; P(failure)=10.50 = 0.50 4) The trials are independent (i.e. the outcome of heads on the first flip will have no impact on subsequent coin flips).

Flipping a coin ten times is a binomial experiment since all conditions are met.

Econ10/Mgt 10 Stuffler31Binomial DistributionAnother sales call example: Assume a mutual fund salesman knows that there is20% chance of closing a sale on each call he makes. We want to determine the probability of making two sales in three calls: P(sale) = .2 P(no sale) = .8

P(X=2) = 3! = .096 2!(3-2)!

.22.8.8

Econ10/Mgt 10 Stuffler32Binomial DistributionAnother sales call example: Assume a mutual fund salesman knows that there is20% chance of closing a sale on each call he makes We want probability of making two sales in three calls P(sale) = .2 P(no sale) = .8

GOOD NEWS!

MegastatProbabilityDiscrete DistributionsBinomial

Econ10/Mgt 10 Stuffler33Binomial Distribution

Econ10/Mgt 10 Stuffler34Pat Statsly Pat Statsly is a student (not a good student) taking astatistics course. Pats exam strategy is to rely on luck for the first test. The test consists of 10 multiple-choice questions. Each question has five possible answers, only one of which is correct. Pat plans to guess the answer to each question.

What is the probability that Pat gets no answers correct?

What is the probability that Pat gets two answers correct?

Econ10/Mgt 10 Stuffler35Pat Statslyn=10

P(correct) = p = 1/5 = .20P(wrong) = q = .80

Is this a binomial experiment?

Check the conditions

Econ10/Mgt 10 Stuffler36Pat Statslyn=10P(correct) = p = 1/5 = .20P(wrong) = q = .80

Is this a binomial experiment? Check the conditions: There is a fixed finite number of trials (n=10). An answer can be either correct or incorrect. The probability of a correct answer (P(success)=.20) does not change from question to question. Each answer is independent of the others.

Econ10/Mgt 10 Stuffler37Pat Statslyn=10, and P(success) = .20

What is the probability that Pat gets no answers correct? EX: P(x=0)

Whats the interpretation of this result?


What is the probability that Pat gets no answers correct? EX: P(x=0)

Pat has about an 11% chance of getting no answers correctusing the guessing strategy.

Econ10/Mgt 10 Stuffler39Pat Statsly


What is the probability that Pat gets two answers correct? EX: P(x=2)

Econ10/Mgt 10 Stuffler41Pat Statsly We have been using the binomial probability distribution to find probabilities for individual values of x.

To answer the question: Find the probability that Pat fails the quiz

requires a cumulative probability, that is, P(X x)

If a grade on the quiz is less than 50% (i.e. 5 questions out of 10), thats considered a failed quiz.

Econ10/Mgt 10 Stuffler42Pat Statsly We have been using the binomial probability distribution to find probabilities for individual values of x.

To answer the question: Find the probability that Pat fails the test

requires a cumulative probability, that is, P(X x)

If a grade on the test is less than 50% (i.e., 5 questions out of 10), thats considered a failed test. We want to know what is: P(X 4) to answer

Econ10/Mgt 10 Stuffler43Pat StatslyP(X 4) = P(0) + P(1) + P(2) + P(3) + P(4) = .9672

What is the interpretation of this result?

Econ10/Mgt 10 Stuffler44Pat Statsly Its about 97% probable that Pat will fail the test using the luck strategy and guessing at answers.

Econ10/Mgt 10 Stuffler45Binomial Table Calculating binomial probabilities by hand is tedious and error prone. There is an easier way. Refer to Table E-6 in the Appendices. For the Pat Statsly example, n=10, so go to the n=10 table:

Look in the Column, p=.20 and substitute into:

P(X 4) = P(0) + P(1) + P(2) + P(3) + P(4)

P(X 4) = .1074 + .2684 + .3020 + .2013 + .0881 = .9672

The probability of Pat failing the test is 96.72%

Econ10/Mgt 10 Stuffler46Poisson Distribution Named for Simeon Poisson, Poisson distribution -

Discrete probability distribution There there are no trials Number of independent events (successes) occurring in a fixed time period or region of space that occur with a known average rate such as, arrivals, departures, or accidents, number of baskets in a quarter, etc.

Econ10/Mgt 10 Stuffler47Poisson DistributionFor example: The number of cars arriving at a service station in 1 hour. (The interval of time is 1 hour.)

The number of flaws in a bolt of cloth. (The specific region is a bolt of cloth.)

The number of accidents in 1 day on a particular stretch of highway. (The interval is defined by both time, 1 day, and space, the particular stretch of highway.)

Econ10/Mgt 10 Stuffler48Poisson Distribution Poisson random variable - number of successes that occur in a period of time or an interval of space

EX: On average, 96 trucks arrive at a border crossing every hour.

EX: Number of typographical errors in a new textbook edition averages 1.5 per 100 pages.

Econ10/Mgt 10 Stuffler49Poisson Distribution The Poisson random variable is the number of successes that occur in a period of time or an interval of space in a Poisson experiment.

EX: On average, 96 trucks arrive at a border crossing every hour.

E.g. The number of typographic errors in a new textbook edition averages 1.5 per 100 pages.successestime periodsuccesses (?!)interval

Econ10/Mgt 10 Stuffler50Poisson DistributionPoisson experiment has four defining characteristics or properties:

The number of successes that occur in any interval is independent of the number of successes that occur in any other intervalThe probability of a success in an interval is the same for all equal-size intervals The probability of a success is proportional to the size of the interval Only one value is required to determine the probability of a designated number of events occurring during an interval

Econ10/Mgt 10 Stuffler51Poisson Distribution The probability that a Poisson random variable assumes a value of x is given by:

= n(p) = number of successes times the probability of success

The unit of time or the interval should be short enough so that the mean rate is not large ( < 20)

YOUR TEXTBOOK USES (lambda) REPRESENT THE MEAN NUMBER OF SUCCESSES IN THE INTERVAL

Econ10/Mgt 10 Stuffler52Poisson Distribution EXAMPLE

The number of typographical errors in new editions of textbooks varies considerably from book to book. After some analysis she concludes that the number of errors is Poisson distributed with a mean of 1.5 per 100 pages.

The instructor randomly selects 100 pages of a new book. What is the probability that there are no typos?

MEGASTATPROBABILITYDISCRETE DISTRIBUTIONSPOISSON AND ENTER THE MEAN VALUE: 1.5, CLICK OK

Econ10/Mgt 10 Stuffler53Poisson Distribution

Econ10/Mgt 10 Stuffler54Mean and Variance of a Poisson Random VariableIf x is a Poisson random variable with parameter , then Mean number of events per unit of time or space = Variance = V(X) = 2 =

Econ10/Mgt 10 Stuffler55Poisson Distribution As mentioned on the first Poisson Distribution slide:

The probability of a success is proportional to the size of the interval

Thus, knowing an error rate of 1.5 typos per 100 pages, we can determine a mean value for a 400 page book as:

= n(p) = 1.5(4) = 6 typos / 400 pages.

Econ10/Mgt 10 Stuffler56Poisson Distribution For a 400 page book, what is the probability that there are no typos?

P(X=0) =

Interpretation?

Econ10/Mgt 10 Stuffler57GOOD NEWS!

Go to Excel, Megastat, Probability, Discrete, Poisson.

Type in the mean, , 6, then OK.

For a mean of 6,the Excel result is 0.00248

Econ10/Mgt 10 Stuffler58Poisson Distribution For a 400 page book, what is the probability that there are five or less typos?

P(X5) = P(0) + P(1) + + P(5)

Another alternative is to refer to Table E-7 in the Appendices

For x = 5, =6, and P(X 5) = P(0) + + P(5) = .4456

There is about a 45% chance that there will be 5 or less typos

Econ10/Mgt 10 Stuffler59Hypergeometric DistributionIf a sample is taken from a finite population without replacement, the probability of X number of successes, follows the hypergeometric distribution.

NOTE: When you see without replacement, this indicates that hypergeometric trials are dependent.

Econ10/Mgt 10 Stuffler60Hypergeometric DistributionFor both the Binomial Distribution and the Hypergeometric Distribution, there are two possible outcomes for the random discrete variable: Success or Failure

Econ10/Mgt 10 Stuffler61Hypergeometric DistributionThe difference is that for Hypergeometric each trial is without replacement, so the probability changes from one draw to the next

For Binomial, the probability of success and failure remain constant since each trial or draw is independent

Econ10/Mgt 10 Stuffler62Hypergeometric Distribution1. Trials are NOT independent2. Finite population3. Sample without replacementProbability of success and failure changes from trial to trial

Econ10/Mgt 10 Stuffler63Hypergeometric Distribution r N r = n rP(x) = x n x N N n 2 = r (N-r) n (N-n) N2 (N-1) where N = Total number in the population n = number in the sample, x = number selected or drawn, r = number of successes in N

Econ10/Mgt 10 Stuffler64Hypergeometric DistributionEXAMPLE: Assume that 3 stocks are chosen randomly from a list of 10 stocks and that of the 10 stocks 4 stocks pay dividends. The number (x) of the three selected stocks that pay a dividend is a hypergeometric random variable.

Econ10/Mgt 10 Stuffler65Hypergeometric DistributionN=10, n=3, r = 4 and x = number of the 3 stocks selected that pay a dividend

Calculate the mean, the variance and the probability that none of the 3 stocks pay dividends?

Econ10/Mgt 10 Stuffler66Hypergeometric Distribution 4 10 4 4! 6!P(0) = 0 3-0 = 0!(4-0)! 3!(6-3)! = 1 10 10! 6 3 3!(10-3)!

= (3)(4) = 1.210

2 = 4(10-4)3(10-3) = .56, = .75 (10)2 (10-1)

Econ10/Mgt 10 Stuffler67Hypergeometric DistributionGOOD NEWS!

Go to ExcelMegastatProbability Discrete Probability Distributions Hypergeometric, enter the required data

Econ10/Mgt 10 Stuffler68Hypergeometric Distribution

Econ10/Mgt 10 Stuffler69Discrete ProbabilityWHICH DISTRIBUTION?

Sony issued a recall for their plasma screen TVs. They choose 25 TVs and 7 do not operate. If the inspector chooses 3 at random from the 25, what is the probability that 2 of the 3 do not operate?


When Wet Water drills a well, their success rate is .55 and their profit is $2,550. If they do not find water, they lose $1,500.


The US unemployment rate is calculated from survey data of 60,000 households. What is the probability that no more than 5,000 respondents are unemployed?


US Transportation and Safety Board reports the annual average number of plane crashes is 14.6. For a given month, what is the probability that at 2 crashes will occur? That 3 crashes will occur?


The US Statistical Abstract of the US tracks many data items such as, the average number of automobiles per household. The average for 2009 is 2.5 autos. What is the probability that the 2010 average is at least 3 autos?

Econ10/Mgt 10 Stuffler74Discrete Random Variables Two Types of Random Variables Discrete Probability Distributions The Binomial Distribution The Poisson Distribution The Hypergeometric Distribution

Summary:

OutputBinomial distribution3n0.2pcumulativeXP(X)probability00.512000.5120010.384000.8960020.096000.9920030.008001.000001.000000.600expected value0.480variance0.693standard deviation

Output

XP(X)Binomial distribution (n = 3, p = 0.2)

Sheet1

Sheet2

Sheet3

OutputBinomial distribution3n0.2pcumulativeXP(X)probability00.512000.5120010.384000.8960020.096000.9920030.008001.000001.000000.600expected value0.480variance0.693standard deviationBinomial distribution10n0.2pcumulativeXP(X)probability00.107370.1073710.268440.3758120.301990.6778030.201330.8791340.088080.9672150.026420.9936360.005510.9991470.000790.9999280.000071.0000090.000001.00000100.000001.000001.000002.000expected value1.600variance1.265standard deviation

Output0.5120.3840.0960.008


Sheet1


Sheet2

Sheet3

OutputBinomial distribution3n0.2pcumulativeXP(X)probability00.512000.5120010.384000.8960020.096000.9920030.008001.000001.000000.600expected value0.480variance0.693standard deviationBinomial distribution10n0.2pcumulativeXP(X)probability00.107370.1073710.268440.3758120.301990.6778030.201330.8791340.088080.9672150.026420.9936360.005510.9991470.000790.9999280.000071.0000090.000001.00000100.000001.000001.000002.000expected value1.600variance1.265standard deviationPoisson distribution1.5mean rate of occurrencecumulativeXP(X)probability00.223130.2231310.334700.5578320.251020.8088530.125510.9343640.047070.9814250.014120.9955460.003530.9990770.000760.9998380.000140.9999790.000021.00000100.000001.00000110.000001.00000120.000001.000001.000001.500expected value1.500variance1.225standard deviation

Output0.5120.3840.0960.008


Sheet10.10737418240.2684354560.3019898880.2013265920.0880803840.02642411520.0055050240.0007864320.0000737280.0000040960.0000001024


Sheet2

XP(X)Poisson distribution ( = 1.5)

Sheet3

OutputBinomial distribution3n0.2pcumulativeXP(X)probability00.512000.5120010.384000.8960020.096000.9920030.008001.000001.000000.600expected value0.480variance0.693standard deviationBinomial distribution10n0.2pcumulativeXP(X)probability00.107370.1073710.268440.3758120.301990.6778030.201330.8791340.088080.9672150.026420.9936360.005510.9991470.000790.9999280.000071.0000090.000001.00000100.000001.000001.000002.000expected value1.600variance1.265standard deviationPoisson distribution1.5mean rate of occurrencecumulativeXP(X)probability00.223130.2231310.334700.5578320.251020.8088530.125510.9343640.047070.9814250.014120.9955460.003530.9990770.000760.9998380.000140.9999790.000021.00000100.000001.00000110.000001.00000120.000001.000001.000001.500expected value1.500variance1.225standard deviationHypergeometric distribution10N, population size3S, number of possible occurrences4n, sample sizecumulativeXP(X)probability00.166670.1666710.500000.6666720.300000.9666730.033331.000001.000001.200expected value0.560variance0.748standard deviation

Output0.5120.3840.0960.008


Sheet10.10737418240.2684354560.3019898880.2013265920.0880803840.02642411520.0055050240.0007864320.0000737280.0000040960.0000001024


Sheet20.22313016010.33469524020.25102143020.12551071510.04706651820.01411995540.00352998890.00075642620.00014182990.00002363830.00000354570.0000004835

XP(X)Poisson distribution ( = 1.5)

Sheet3

XP(X)Hypergeometric distribution (N = 10, S = 3, n = 4)

discrete random variables

Documents

probability distributionps

number of sales

given household

household millions

us survey data

relative frequencies

subjective methods

mutual fund salesman