Bài giảng Statistical Techniques in Business and Economics - Chapter 6 Discrete Probability Distributions

Chapter 6Discrete Probability Distributions1.Define the terms probability distribution and random variable. 2.Distinguish between discrete and continuous random variables.3.Calculate the mean, variance, and standard deviation of a discrete probability distribution.4.Describe the characteristics and compute probabilities using the Poisson probability distribution.Chapter GoalsWhen you have completed this chapter, you will be able to:TerminologyRandom Variableis a numerical value determined by the outcome of an experiment.Probability Distributionis the listing of all possible outcomes of an experiment and the corresponding probability.DiscreteTypes of Probability Distributions Under this distribution the random variable has a countable number of possible outcomesUnder this distribution the random variable has an infinite number of possible outcomesContinuousExamplesDiscreteContinuousExamplesStudents in a classNumber of children in a familyTypes of Probability Distributions Mortgage LoanNumber of Mortgages approved in a monthDistance driven by an executive to get to workThe length of time of a particular phone callThe length of time of an afternoon nap!Distinguishing features of a Discrete Distribution:The sum of the probabilities of the various outcomes is 1.00The probability of a particular outcome is between 0 and 1.00The outcomes are mutually exclusive Consider a random experiment in which a coin is tossed three timesQuestionHeadsLet x be the number ofTailsLet T represent the outcome ofLet H represent the outcome of aHead Determine the probability distributionListing the possibilitiesHeadsHeadsHeadsTailsHeadsHeadsTailsHeadsHeadsTailsHeadsHeadsTailsHeadsTailsTailsHeadsTailsTailsTailsTails the possible values of x (number of heads) are 0,1,2,3.TailsHeadsTails Consider a random experiment in which a coin is tossed three times. Determine the probability distribution.QuestionProbability Distributionx# of Outcomes P(x)0123133181/83/83/81/88/8 = 1What is the probability of tossing 2 heads in 3 flips?reports the central location of the dataMean of a Discrete Probability Distributionis the long-run average value of the random variablealso referred to as its expected value, E(X), in a probability distributionis a weighted average is denoted by the Greek symbol , mu xP(x)Mean of a Discrete Probability DistributionQuestionLet x be the number of heads Flip a coin three times )]([xxPS=mS)(xxPFormula 03/83/86/812/8=1.5x0123 P(x)1/83/83/81/88/8 = 1Variance of a Discrete Probability Distributionmeasures the amount of spread (variation) of a distribution.denoted by the Greek letter 2 (sigma squared).the standard deviation is the square root of 2 (X- )2m Variance of a Discrete Probability Distribution Flip a coin three times. Let x be the number of heads Q)]()[(2xPxm-S 2s= X- m- 1.5- 0.51.50.50.252.250.25(X- )2m P(x).28125.093750.75S.09375.28125( x)Pxm-Formula xP(x)03/83/86/8x0123 P(x)1/83/83/81/88/8 = 1=1.5m2.25Dan Desch, owner of College Painters, studied his records for the past 20 weeks and reported the following number of houses painted per week: P(x)5/20 = 0.256/20 = 0.307/20 = 0.352/20 = 0.1020/20 = 1.0Computing the mQuestionDetermine the Probability distribution and its mean and variance.# of Painted Houses10111213 Weeks567220x)]( [xxPS=mS)(xxP xP(x)2.53.31.34.211.3Computing the mComputing the 2sFormula P(x)6/20 = 0.307/20 = 0.352/20 = 0.1020/20 = 1.0 # of Painted Houses1112135/20 = 0.2510)]()[(2xPxm-S 2s=(x - )2m 0.091.692.890.49(x - )2m P(x).4225.02700.91S.1715.2890( x)PComputing the 2s P(x) 0.25 0.30 0.35 0.101.0x10111213xP(x)2.53.31.34.211.3)2(xm-Formula (10-11.3)2Binomial Probability DistributionTerminologyBernoulli Trialis a random experiment in which the number of possible outcomes is precisely two!HeadsTailsRightWrongCourse GradeAB or C or D or!For ExampleCharacteristicsBinomial Probability DistributionThe probability of success stays the same for each trial The trials are independent Interested in the number of successesThe experiment consists of n Bernoulli trialsThe outcomes are classified into one of two mutually exclusive categories, such as success or failuren be the number of trialsx be the number of observed successes p be the probability of success on each trial x)(n xxnppCxP -- =) 1()(Binomial Probability DistributionFormula LetQuestionThe Department of Labour reports that 20% of the workforce aged between 15 and 19 years is unemployed.From a sample of 10 workers in this age group, calculate the following probabilities: Exactly three are unemployed At least three are unemployed None are unemployed At least one is unemployedCalculations Exactly three are unemployedAt least three are unemployed x)(nxxnppCxP-- =) 1()(DATA: 20% Unemployed & Sample of 103)20(. 10-3)20.1(-3C10)3(=P(.2097)(.0080)(120)=.2013 or 20.13%=0)80(.10)20(. 10-3)80(.3310)20(.CP10C10+...+=)3(x³000.0881.2013..322 or 32.2%=+...++=More None are unemployed At least one is unemployedi.e. P (³ 1) =same as “all the time”(100%) except when “none are unemployed”0)20(. 10-0)20.1(-0C10)0(=P(.1074)(1)(1)=.1074 or 10.74%=1 – P(0) = 1 - .1074= .8926 or89.26% x)(nxxnppCxP-- =) 1()(DATA: 20% Unemployed & Sample of 1010 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 0.599 0.349 0.107 0.028 0.006 0.001 0.000 0.000 0.000 0.000 0.0000.315 0.387 0.268 0.121 0.040 0.010 0.002 0.000 0.000 0.000 0.000 0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.000 0.000 0.000 0.000 0.010 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001 0.000 0.000 0.001 0.011 0.088 0.200 0.251 0.205 0.111 0.037 0.006 0.000 0.000 0.000 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001 0.000 0.000 0.000 0.006 0.037 0.111 0.205 0.251 0.200 0.088 0.011 0.001 0.000 0.000 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.010 0.000 0.001 0.000 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.075 0.000 0.000 0.000 0.000 0.002 0.010 0.040 0.121 0.268 0.387 0.315 0.000 0.001 0.000 0.000 0.000 0.001 0.006 0.028 0.107 0.349 0.5990123456789XBinomial Probability Distributionn = 10ProbabilityTable 6.2 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95 Represents the ‘Probability’012345678910XWhere X represents the ‘Number Unemployed’0.599 0.349 0.107 0.028 0.006 0.001 0.000 0.000 0.000 0.000 0.0000.315 0.387 0.268 0.121 0.040 0.010 0.002 0.000 0.000 0.000 0.000 0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.000 0.000 0.000 0.000 0.010 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001 0.000 0.000 0.001 0.011 0.088 0.200 0.251 0.205 0.111 0.037 0.006 0.000 0.000 0.000 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001 0.000 0.000 0.000 0.006 0.037 0.111 0.205 0.251 0.200 0.088 0.011 0.001 0.000 0.000 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.010 0.000 0.001 0.000 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.075 0.000 0.000 0.000 0.000 0.002 0.010 0.040 0.121 0.268 0.387 0.315 0.000 0.001 0.000 0.000 0.000 0.001 0.006 0.028 0.107 0.349 0.599Represents the ‘ PROBABILITY’ when n = 10ExplanationsBinomial Probability Distributionn = 10ProbabilityFrom text Appendix AUsing Appendix AAlternate SolutionUsing Appendix A Binomial Probability DistributionDATA: 20% Unemployed & Sample of 10 Exactly three are unemployed.201 or 20.1% = 0.05 0.10 0.20 0.30 0.80 0.90 0.950.599 0.349 0.107 0.028 0.000 0.000 0.0000.315 0.387 0.268 0.121 0.000 0.000 0.0000.075 0.194 0.302 0.233 0.000 0.000 0.0000.010 0.057 0.201 0.267 0.001 0.000 0.0000.001 0.011 0.088 0.200 0.006 0.000 0.00001234X30.201Alternate SolutionAt least three are unemployedUsing Appendix A Binomial Probability DistributionDATA: 20% Unemployed & Sample of 10Alternate Reasoning: If ‘at least three are unemployed’ it follows that ‘at most seven are employed!’You can turn this into a problem of 80% employment if you wish0.322 or 32.2%= At least three are unemployed012345678910X0.599 0.349 0.107 0.315 0.387 0.268 0.075 0.194 0.302 0.010 0.057 0.201 0.001 0.011 0.088 0.000 0.001 0.026 0.000 0.000 0.006 0.000 0.000 0.0010.000 0.000 0.0000.000 0.000 0.0000.000 0.000 0.0000.05 0.10 0.20To account for the ‘at least 3 unemployed’, we must TOTAL the percentages from 3 to 10, inclusively0.2010.0880.0260.0060.001 0.000 0.000 0.000Using Appendix A Binomial Probability DistributionDATA: 20% Unemployed & Sample of 10Alternate Solution 0.05 0.10 0.20 0.30 0.80 0.90 0.950.599 0.349 0.107 0.028 0.000 0.000 0.0000.315 0.387 0.268 0.121 0.000 0.000 0.0000.075 0.194 0.302 0.233 0.000 0.000 0.0000.010 0.057 0.201 0.267 0.001 0.000 0.0000.001 0.011 0.088 0.200 0.006 0.000 0.00001234XAlternate SolutionDATA: 20% Unemployed & Sample of 10 None are unemployed.107 or 10.7%=00.107Using Appendix A Binomial Probability DistributionAlternate SolutionDATA: 20% Unemployed & Sample of 10 At least one is unemployed= .892 or89.2% 0.05 0.10 0.200.599 0.349 0.107 0.315 0.387 0.268 0.075 0.194 0.302 0.010 0.057 0.201 0.001 0.011 0.088 0.000 0.001 0.026 0.000 0.000 0.006 0.000 0.000 0.001 0.000 0.001 0.000 0.000 0.000 0.000 0.000 0.001 0.000 012345678910X0.2680.3020.2010.0880.0260.0060.0010.0000.0000.000Using Appendix A Binomial Probability Distributionat most nine are employed!np=)1(pnp-== 10(.20)= 2.0= 10(.20)(.80)= 1.60Mean and Varianceof a Binomial Probability Distribution 2smThe Ontario Department of Labour reports that 20% of the workforce aged between 15 and 19 years is unemployed. QmFormula 2sFormula Therefore, the standard deviation is 1.6 = 1.3From a sample of 10 workers in this age group, calculate: andms2Poisson Probability DistributionSkewed RightThe Binomial Distribution becomes more skewed to the rightPositiveas the Probability of success become smaller.Poisson Probability DistributionRecallThe limiting form of the Binomial Distribution where the probability of success p is small and n is large is called the Poisson Probability Distribution Poisson Probability Distributioncan be described mathematically using the formula:Pxex xu()!=-mPoisson Probability DistributionWhereµis the mean number of successes in a particular interval of timeeis the constant 2.71828xis the number of successesPoisson Probability DistributionThe mean number of successescan be determined in binomial situations bywhere n is the number of trialsand p the probability of a successnpmThe variance of the Poisson distribution is also equal to np2sQuestion The Daily (Statistics Canada) reports that 5% of college and bachelor degree students default on their student loan within 2 years of graduation. From a sample of 10 students, find the probability that exactly 1 will default on their loan.Poisson Probability Distribution The Daily (Statistics Canada) reports that 5% of college and bachelor degree students default on their student loan within 2 years of graduation. QPoisson Probability Distribution .0476 or 4.76%!1 .05.05 1=-e)(1PPxex xu()!=-mFormula =Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!This completes Chapter 6