Random variablesDiscrete Random Variables IIIJuana [email protected] Department of StatisticsJ. Sanchez Stat 100A Intro ProbabilityThis lectureRecommended reading: Ross, chapter 4 (4.1-4.7); Chapter 7, section 7.7(discrete variables cases only).I. The Bernoulli r.v.II. The Binomial r.v.III. Poisson approximation to the Binomial distributionJ. Sanchez Stat 100A Intro ProbabilityI The Bernoulli Random VariableA Bernoulli random variable X has the following probability distribution:Bernoulli Random Variable with parameter pP(X ) = px(1 − p)1−xX = 0, 1.X P(X)0 (1-p)1 pE (X ) =PxXP(X ) = 0(1 − p) + 1p = pVar(X ) =Px(X − E (X ))2P(X )= (0 − p)2(1 − p) + (1 − p)2p = p2− p3+ p + p3− 2p2= p(1 − p)J. Sanchez Stat 100A Intro ProbabilityI I. The Binomial Random Variable Bin(n, p)Suppose that a certain machine producing millions (large number) ofitems produces a defective item with a probability of p and producesa nondefective item with a probability of q = 1 − p.Suppose further that n independent items produced by the machineare examined (drawn without replacement), but assuming the samep for each (because drawing from large lot), and let X denote thenumber of these items that are defective.The r.v. X will have a discrete distribution, and the possible valuesof X will be 0, 1, 2, ....., n.J. Sanchez Stat 100A Intro ProbabilityFor X = 0, 1, ...., n, the probability of obtaining any particular orderedsequence of n items containing exactly x defectives and n − xnonedefectives is pxqn−x. Since there arenxdifferent orderedsequences of this type, it follows thatBinomial r.v. Probability mass functionP(X = x) =nxpxqn−xfor x = 0, 1, 2, ...., n0 elsewhereA random variable X has a binomial probability mass function withparameters n and p if X has the p.m.f. given above. In this distribution,n > 0 and 0 ≤ p ≤ 1.J. Sanchez Stat 100A Intro ProbabilityThe binomial distribution is of fundamental importance in probability andstatistics because of the following result: Suppose thatThe outcome of an experiment can be either success or failure;The experiment is performed n times independently;The probability of the success of any given performance is p .If X denotes the total number of successes in the n performances, then Xhas a binomial distribution with parameters n and p.J. Sanchez Stat 100A Intro ProbabilityMany experimental situations involve r.v.’s that can be adequatelymodeled by the binomial distribution.The number of defectives in a sample of n itemsthe number of employees in a sample of n employees from a bigcompany l who favor a certain retirement policythe number of electronic systems sold this week out of the n thatwere manufacturedthe number of jury members out of n that think a defendant is guilty.etc. etc..J. Sanchez Stat 100A Intro ProbabilityExp ectation and Variance of the Binomial r.v.Expected value of a binomial r.v.E (X ) =XxXP(X = x) = npVariance of a binomial r.v.Var (X ) =Xx(X − E (X ))2P(X = x) = np(1 − p)J. Sanchez Stat 100A Intro ProbabilityExampleSuppose that a large lot of fuses contains 10 percent defectives. If fourfuses are randomly sampled from the lot,(a) find the probability that exactly one fuse is defective.(b) Find the probability that at least one fuse in the sample of four isdefective.(c) Suppose that the four fuses sampled from the lot were shipped to acustomer, before being tested, on a guarantee basis. Assume thatthe cost of making the shipment good is given by C = 3X2, where Xdenotes the number of defectives in the shipment of four. Find theexpected repair cost.J. Sanchez Stat 100A Intro ProbabilityExampleSuppose that a large lot of fuses contains 10 percent defectives. If fourfuses are randomly sampled from the lot,(a) Find the probability that exactly one fuse is defective:P(X = 1 | n = 4, p = 0.1) =41(0.1)1(0.9)3= 0.2916(b) Find the probability that at least one fuse in the sample of four isdefective.P(X ≥ 1) = 1 − P(X = 0) = 1 −40(0.1)0(0.9)4= 0.3439J. Sanchez Stat 100A Intro ProbabilityExampleSuppose that a large lot of fuses contains 10 percent defectives. If fourfuses are randomly sampled from the lot,(a) Find the probability that exactly one fuse is defective:(b) Find the probability that at least one fuse in the sample of four isdefective.(c) Suppose that the four fuses sampled from the lot were shipped to acustomer, before being tested, on a guarantee basis. Assume thatthe cost of making the shipment good is given by C = 3X2, where Xdenotes the number of defectives in the shipment of four. Find theexpected repair cost.E (C ) = 3E (X2) = 3[Var(X ) + [E(X )]2]= 3[4(0.1)(0.9) + (4(0.1))2] = 1.56dollarsJ. Sanchez Stat 100A Intro ProbabilityThe binomial can arise in situations where the underlying probleminvolves a continuous (that is, non-discrete) r.v.ExampleIn a study of life lengths for a certain type of battery, researchers foundthat the probability that a battery life, X, will exceed four hours is 0.135.If three such batteries are in use in independently operating systems, findthe probability that only one of the batteries will last 4 hours or more.J. Sanchez Stat 100A Intro ProbabilityExampleA bakery has 5 ovens. At least 4 ovens must be working in order to meetcustomer demand on a given day. The probability of a particular ovenworking is 0.9. We want to find out the probability of meeting customerdemand.(a) The random variable in this problem, X , is:The number of ovens working (out of 5)(b) The probability distribution model for X can be assumed to be:Binomial (n=5, p=0.9)(c) Use that model to show how to find the probability of meetingcustomer demand.P(X ≥ 4 | n = 5, p = 0.9) =540.94(0.1)+540.95(0.1)0= 0.91854J. Sanchez Stat 100A Intro ProbabilityMoment Generating Function of the Binomial DistributionM(t) = E (etx)=nPk=0etknkpk(1 − p)n−k=nPk=0nk(pet)k(1 − p)n−k= (pet+ 1 − p)nJ. Sanchez Stat 100A Intro ProbabilityWe used the binomial theorem to obtain the result in the last slide. Forany x and any y and any n,(x + y )n=nPk=0nkxkyn−kJ. Sanchez Stat 100A Intro ProbabilityBinomial distribution: proofsE(X)=npE (X ) =Pnx=0xnxpx(1 − p)n−xProof:E (Xk) =nXx=0xknxpx(1 − p)n−x=nXx=1xknxpx(1 − p)n−xNotice:xnx= nn − 1x − 1=nXx=1xnxxk−1px(1 − p)n−x=nXx=1nn − 1x − 1xk−1px(1 − p)n−xcontinues next pageJ. Sanchez Stat 100A Intro Probability= npn−1Xy=0(y +
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