Computing Binomial ProbabilitiesProperties of a Binomial DistributionComputing Binomial Probabilities by TI CalculatorsBinomial Random VariablesNormal ApproximationExampleWhen Approximation is GoodExamplesBinomial ExampleNormal ExampleChapter 4Part 2. Binomial DistributionJ.C. WangiClicker Question 4.4 Pre-lectureiClicker Question 4.4 Pre-lectureOutlineComputing Binomial ProbabilitiesProperties of a Binomial DistributionComputing Binomial Probabilities by TI CalculatorsBinomial Random VariablesNormal ApproximationExampleWhen Approximation is GoodExamplesBinomial ExampleNormal ExamplePropertiesof a binomial distribution1. Only two possible outcomes for each observation2. Probability of success is pprobability of failure is q = 1 − p3. Observations are independentNote:ISample size is fixedINumber of successes, X, is of interestExampleof binomial distributionStat 2160 multiple choice quiz has 5 questions with 4 choicesfor each questionsuccessdef .= answering a question correctlyIprobability of success = ?Iprobability of failure = ?Isample size = ?IWhat is probability that a student will answer exactly 3questions correctly, in other words,P(X = 3|n = 5, p = 0.25, q = 0.75) =?Computing Binomial ProbabilitiesP(X = 3|n = 5, p = 0.25, q = 0.75) =?IFormula5!3!(5 − 3)!p3q5−3= 0.0879Note: 0! = 1 and p0= 1and n! = 1 × 2 × ···×nIUsing TI Calculator<2nd><DISTR>↓binomialPDF(5, .25, 3)<ENTER>and get 0.0879Computing Binomial Probabilitiesusing TI calculatorkeyword: exactly20% of TV buyers at TV+More purchase the store’s extendedwarranty. Say that 10 TV sets were sold in one day. What is theprobabilities that exactly 3 extended warranties were sold, inother words,P(X = 3|n = 10, p = .2) =?<2nd><DISTR>↓binomPDF(10,.2,3)<ENTER>and get 0.2013Cumulative Probabilitiescumulative distribution functionkeyword: at most (i.e, left-tail probability)P(X ≤ j|n, p) = P(X = 0 or 1 or ···or j|n, p)= P(X = 0) + P(X = 1) + ··· + P(X = j)20% of TV buyers at TV+More purchase the store’s extendedwarranty. Say that 10 TV sets were sold in one day. What is theprobability that at most 3 extended warranties were sold, inother words,P(X ≤ 3|n = 10, p = .2) =?<2nd><DISTR>↓binomCDF(10,.2,3)<ENTER>and get 0.8791iClicker Question 4.5iClicker Question 4.5Cumulative Probabilities(right-) tail probabilitykeyword: at least (i.e, right-tail probability)P(X ≥ j|n, p) = P(X = j or j + 1 or ···or n|n, p)= P(X = j) + P(X = j + 1) + ··· + P(X = n)= 1−P(X ≤j−1|n, p)20% of TV buyers at TV+More purchase the stores extendedwarranty. Say that 10 TV sets were sold in one day. What is theprobability that at least 3 extended warranties were sold, i.e.,P(X ≥ 3|n = 10, p = .2) = 1−P(X ≤(3−1)|10, .2)= 1−P(X ≤2|10, .2) =?1− <2nd><DISTR>↓binomCDF(10,.2,2)= 0.3222Example of binomial distributionAn important part of the customer service responsibilities of atelephone company relates to the speed with which troubles inresidential service can be repaired. Suppose past data indicatethat the likelihood is 0.7 that troubles in residential service canbe repaired on the same day. For the first five troubles reportedon a given day, what is the probability that at most three will berepaired on the same day?IGiven: n = 5, p = 0.7, j = 3, P(X ≤ 3) =?IP(X ≤ 3) = binomCDF(5, 0.7, 3) = 0.4718Example of binomial distribution, continuedFor the first five troubles reported on a given day, what is theprobability that at least three will be repaired on the same day?IGiven: n = 5, p = 0.7, j = 3, P(X ≥ 3) =?IP(X ≥ 3) = 1 −P(X ≤ 2) = 1 − binomCDF(5, 0.7, 2) =0.83692iClicker Question 4.6iClicker Question 4.6Binomial Random VariablesIExpected value = E[X ] = np, i.e.expected number of successesIStandard deviation = sd(X ) =pnp(1 − p) =√npqBinomial Random VariablesTV+More ExampleSuppose that 20% of TV buyers at TV+More purchase thestore’s extended warranty. If 26 TVs were sold last week, the(expected) number of extended warranties should be aroundnp = 26 × 0.2 = 5.2 give or takepnp(1 − p) =p26 × 0.2(1 − 0.2) = 2.0396.Say that the extended warranty cost is $100, how muchrevenue will be generated?($100 × 5.2) ± ($100 × 2.0396) = $520 ± $203.96Note: Multiplication rule of location and spread.Graphical Representationof binomial probabilities— bar graph and probability histogram0.000.050.100.150.20Binomial(n=15,p=0.5)probabilityP(X=x) = bar heightP(X=x) = bar height= bar area0 5 10 150.000.050.100.150.20xP(X=x)7.50 5 100.000.050.100.150.20Binomial(n=14,p=0.5)xP(X=x)7Note: if probability of a success is 0.5, the shape is symmetric about n/2.Normal Approximation of binomial probabilitiesX~Binomial(n=26,p=0.2)np == 5.2,, npq == 2.0396Binomial Probability ≈≈ Y~Normal(µµ == 5.2, σσ == 2.0396) Probabilityactual probabilityapproximate probability0 5 10 15 20 250.000.050.100.150.20P(X=x) ≈≈ P(x−0.5<Y<x+0.5)x0 5 10 15 20 250.000.050.100.150.20P(X ≤≤ x) ≈≈ P(Y<x+0.5)x0 5 10 15 20 250.000.050.100.150.20P(X ≥≥ x) ≈≈ P(Y>x−0.5)xNormal ApproximationTV+more exampleSay n = 26 and p = .2. Using the Normal Curve; make surenp > 5 and nq > 5. Here 26 × .2 = 5.2 > 5 and26 × (1 − .2) = 20.8 > 5. Therefore, conditions hold and thestandard deviation is√npq = 2.0396Let’s say X ≤ 5, P(X ≤ 5) ≈ P(Y < 5.5) =normalCDF(−9999,5.5,5.2,2.0396) = 0.5585Normal ApproximationTV+more example, continued0 5 10 15 20 250.000.050.100.150.20P(X ≤≤ 5) ≈≈ P(Y << 5 ++ 0.5) == P(Y<5.5)Normal Approximation—TV+more examplecomparison with exact binomial probabilitySay n = 26 and p = .2 and X ≤ 5Using the Binomial ProbabilityP(X ≤ 5|n = 26, p = .2) = binomCDF(26, .2, 5) = 0.5775Note: Normal approximation gives a value close to the precisebinomial method.Conditions for Good ApproximationNormal curve gives a reasonable approximation for thebinomial probabilities whenever both np > 5 and nq > 5.Note:Inp = expected number of successesInq = expected number of failuresiClicker Question 4.4 Post-lectureiClicker Question 4.4 Post-lectureBinomial Exampletravel agents exampleThe rate of commission that commercial airlines pay travelagents has been declining for several years. In an attempt bytravel agents to raise revenue, many agents are now chargingtheir customers a ticket fee, typically between 10 and 15dollars. According to the ASTA, about 90 percent of travelagents charge customer fees when purchasing an airline ticket.Travel Agents ExamplecontinuedSuppose that a