3Probability Models Probability Models A random (or stochastic) process is a repeatable random experiment.For example, each call arriving at the L.L. Bean order center is a random experiment in which the variable of interest is the amount of the order.Probability can be used to analyze random (or stochastic) processes and to understand business processes.
4Discrete Distributions Random VariablesA random variable is a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment.Nomenclature: - Capital letters are used to represent random variables (e.g., X, Y). - Lower case letters are used to represent values of the random variable (e.g., x, y).A discrete random variable has a countable number of distinct values.
5Discrete Distributions Probability DistributionsA discrete probability distribution assigns a probability to each value of a discrete random variable X.To be a valid probability, each probability must be between0 P(xi) 1and the sum of all the probabilities for the values of X must be equal to unity.
6Discrete Distributions Example: Coin FlipsWhen you flip a coin three times, the sample space has eight equally likely simple events. They are:1st Toss 2nd Toss 3rd TossH H HH H TH T HH T TT H HT H TT T HT T T
7Discrete Distributions Example: Coin FlipsIf X is the number of heads, then X is a random variable whose probability distribution is as follows:Possible EventsxP(x)TTT1/8HTT, THT, TTH13/8HHT, HTH, THH2HHH3Total
8Discrete Distributions Example: Coin FlipsNote that the values of X need not be equally likely. However, they must sum to unity.Note also that a discrete probability distribution is defined only at specific points on the X-axis.
9Discrete Distributions Expected ValueThe expected value E(X) of a discrete random variable is the sum of all X-values weighted by their respective probabilities.If there are n distinct values of X,The E(X) is a measure of central tendency.
10Discrete Distributions Example: Service CallsThe probability distribution of emergency service calls on Sunday by Ace Appliance Repair is:xP(x)0.0510.1020.3030.2540.205Total1.00What is the average or expected number of service calls?
11Discrete Distributions Example: Service CallsFirst calculate xiP(xi):xP(x)xP(x)0.050.0010.1020.300.6030.250.7540.200.8050.50Total1.002.75The sum of the xP(x) column is the expected value or mean of the discrete distribution.
12Discrete Distributions Example: Service CallsThis particular probability distribution is not symmetric around the mean m = 2.75.However, the mean is still the balancing point, or fulcrum.m = 2.75Because E(X) is an average, it does not have to be an observable point.
13Discrete Distributions Application: Life InsuranceExpected value is the basis of life insurance.For example, what is the probability that a 30-year-old white female will die within the next year?Based on mortality statistics, the probability is and the probability of living another year is =What premium should a life insurance company charge to break even on a $500,000 1-year term policy?
14Discrete Distributions Application: Life InsuranceLet X be the amount paid by the company to settle the policy.The total expected payout isEventxP(x)xP(x)Live.999410.00Die500,000.00059295.00TotalSource: Centers for Disease Control and Prevention, National Vital Statistics Reports, 47, no. 28 (1999).So, the premium should be $295 plus whatever return the company needs to cover administrative overhead and profit.
15Discrete Distributions Application: Raffle TicketsExpected value can be applied to raffles and lotteries.If it costs $2 to buy a ticket in a raffle to win a new car worth $55,000 and 29,346 raffle tickets are sold, what is the expected value of a raffle ticket?If you buy 1 ticket, what is the chance you will129,34629,34529,346win =lose =
16Discrete Distributions Application: Raffle TicketsNow, calculate the E(X):E(X) = (value if you win)P(win) + (value if you lose)P(lose)= (55,000)129,346+ (0)29,345= (55,000)( ) + (0)( ) = $1.87The raffle ticket is actually worth $ Is it worth spending $2.00 for it?
17Discrete Distributions Variance and Standard DeviationIf there are n distinct values of X, then the variance of a discrete random variable is:The variance is a weighted average of the dispersion about the mean and is denoted either as s2 or V(X).The standard deviation is the square root of the variance and is denoted s.
18Discrete Distributions Example: Bed and BreakfastThe Bay Street Inn is a 7-room bed-and-breakfast in Santa Theresa, Ca.xP(x)0.05120.0630.1040.1350.2060.1570.26Total1.00The probability distribution of room rentals during February is:
19Discrete Distributions Example: Bed and BreakfastFirst find the expected valuexP(x)0.05120.0630.1040.1350.2060.1570.26Total1.00x P(x)0.000.050.120.300.521.000.901.82m = 4.71= 4.71 rooms
20Discrete Distributions Example: Bed and BreakfastThe E(X) is then used to find the variance:xP(x)x P(x)0.050.00120.060.1230.100.3040.130.5250.201.0060.150.9070.261.82Totalm = 4.71[xm]27.34412.92410.50410.08411.66415.2441[xm]2 P(x)s2 == rooms2The standard deviation is:s == rooms
21Discrete Distributions Example: Bed and BreakfastThe histogram shows that the distribution is skewed to the left and bimodal.The mode is 7 rooms rented but the average is only 4.71 room rentals.s = 2.06 indicates considerable variation around m.
22Uniform Distribution Characteristics of the Uniform Distribution The uniform distribution describes a random variable with a finite number of integer values from a to b (the only two parameters).Each value of the random variable is equally likely to occur.Consider the following summary of the uniform distribution:
23Random data generation in Excel Uniform DistributionParametersa = lower limitb = upper limitPDFRangea x bMeanStd. Dev.Random data generation in Excel=a+INT((b-a+1)*RAND())CommentsUsed as a benchmark, to generate random integers, or to create other distributions.
24Uniform Distribution Example: Rolling a Die The number of dots on the roll of a die form a uniform random variable with six equally likely integer values: 1, 2, 3, 4, 5, 6What is the probability of rolling any of these?PDF for one dieCDF for one die
25Uniform Distribution Example: Rolling a Die The PDF for all x is: Calculate the mean as:Calculate the standard deviation as:
26Uniform Distribution Application: Pumping Gas PDF CDF On a gas pump, the last two digits (pennies) displayed will be a uniform random integer (assuming the pump stops automatically).PDFCDFThe parameters are: a = 00 and b = 99
27Uniform Distribution Application: Pumping Gas The PDF for all x is: Calculate the mean as:Calculate the standard deviation as:
28Bernoulli Distribution Bernoulli ExperimentsA random experiment with only 2 outcomes is a Bernoulli experiment.One outcome is arbitrarily labeled a “success” (denoted X = 1) and the other a “failure” (denoted X = 0).p is the P(success), 1 – p is the P(failure).“Success” is usually defined as the less likely outcome so that p < .5 for convenience.Note that P(0) + P(1) = (1 – p) + p = 1 and 0 < p < 1.
29Bernoulli Distribution Bernoulli ExperimentsConsider the following Bernoulli experiments:Bernoulli ExperimentPossible OutcomesProbability of “Success”Flip a coin1 = heads0 = tailsp = .50Inspect a jet turbine blade1 = crack found0 = no crack foundp = .001Purchase a tank of gas1 = pay by credit card0 = do not pay by credit cardp = .78Do a mammogram test1 = positive test0 = negative testp = .0004
30Bernoulli Distribution Bernoulli ExperimentsThe expected value (mean) of a Bernoulli experiment is calculated as:The variance of a Bernoulli experiment is calculated as:The mean and variance are useful in developing the next model.
31Binomial Distribution Characteristics of the Binomial DistributionThe binomial distribution arises when a Bernoulli experiment is repeated n times.Each Bernoulli trial is independent so the probability of success p remains constant on each trial.In a binomial experiment, we are interested in X = number of successes in n trials. So,X = x1 + x xnThe probability of a particular number of successes P(X) is determined by parameters n and p.
32Binomial Distribution Characteristics of the Binomial DistributionThe mean of a binomial distribution is found by adding the means for each of the n Bernoulli independent events:p + p + … + p = npThe variance of a binomial distribution is found by adding the variances for each of the n Bernoulli independent events:p(1-p)+ p(1-p) + … + p(1-p) = np(1-p)The standard deviation isnp(1-p)
33Binomial Distribution Parametersn = number of trialsp = probability of successPDFExcel function=BINOMDIST(k,n,p,0)RangeX = 0, 1, 2, . . ., nMeannpStd. Dev.Random data generation in ExcelSum n values of =1+INT(2*RAND()) or use Excel’s Tools | Data AnalysisCommentsSkewed right if p < .50, skewed left if p > .50, and symmetric if p = .50.
34Binomial Distribution Application: Uninsured PatientsOn average, 20% of the emergency room patients at Greenwood General Hospital lack health insurance.In a random sample of 4 patients, what is the probability that at least 2 will be uninsured?X = number of uninsured patients (“success”)P(uninsured) = p =20% or .20P(insured) = 1 – p =1 – .20 = .80n =4 patientsThe range is X = 0, 1, 2, 3, 4 patients.
35Binomial Distribution Application: Uninsured PatientsWhat is the mean and standard deviation of this binomial distribution?Mean = m = np =(4)(.20) =0.8 patientsStandard deviation = s == 4(.20(1-.20)= 0.8 patients