Random variablesDiscrete Random Variables IIJuana [email protected] Department of StatisticsJuly 11J. Sanchez Stat 100A Intro ProbabilityAnnouncementsHomework 3 is being discussed in the TA session. No turning in.Midterm 1 is on Tuesday, JULY 16th, 10:00-11:30 (regular exam).11:35-11:50 simulation required.ROOM: BROAD, 2160E (Noticechange of room) DURATION OF EXAM: 1 hour and 30 minutes.SIMULATION: 15 minutes. No regular lecture or TA session on July16th.No graphics calculators. You may have a scientific calculator. Phonecalculators and any other electronic device is not allowed. Honorcode applies.See syllabus.Cheat sheet (one side of 11x8) allowed. Only formulas anddefinitions. No drawings, no numerical examples, no proofs nointermediate steps. I will post a cheat sheet from old students that isinappropriate, and one that is appropriate in the web site. Pointsdeducted for each unacceptable item.J. Sanchez Stat 100A Intro ProbabilityTodayReference reading: Ross(8th edition), chapter 4 (4.1-4.7); Chapter 7,section 7.7 (discrete variables cases only).I. Problem unsolved in last lecture. Review of expectations andexpectations of functions of random variables.II. Some proofs of expectations and variance of linear functions of arandom variable. Other proofs left as exercises.III. Alternative way of computing the variance.IV. The Moment generating function of a discrete random variable.V. The Poisson Random VariableVI. The Bernoulli r.v.VII. The Binomial r.v.J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureExampleThe manager of a stock room in a factory knows from his study ofrecords that the daily demand (number of times used) for a certain toolhas the following probability distribution:Quantity demanded (X) 0 1 2Probability 0.1 0.5 0.4(In other words, 50% of thedaily records show that the tool was used one time). Let X denote thedaily demand.(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?(c) Suppose that it costs the factory $10 each time the tool is used.What is the expected daily cost of using the tool? How much is thiscost expected to vary from one day to another?J. Sanchez Stat 100A Intro ProbabilityNotesNotesNotesNotesClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?E (X ) =XxxP(X = x) = 0 × 0.1 + 1 ×0.5 + 2 ×0.4 = 1.3The Expected value of X is 1.3, the manager should expect the toolto be used tomorrow 1.3 times..J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?E (X ) =XxxP(X = x) = 0 × 0.1 + 1 ×0.5 + 2 ×0.4 = 1.3The Expected value of X is 1.3, the manager should expect the toolto be used tomorrow 1.3 times..J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?Var (X ) = E (X − µ)2=Xx(x − E (X ))2P(X = x)= (0 − 1.3)2(0.1) + (1 − 1.3)2(0.5) + (2 − 1.3)2(0.4) = 0.41SD(X ) =√0.41 = 0.6303Because the standard deviation of X is 0.6303, he may be off by0.6303. That is, the use of the tool could be expected to be 1.3times give or take 0.6303.J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?Var (X ) = E (X − µ)2=Xx(x − E (X ))2P(X = x)= (0 − 1.3)2(0.1) + (1 − 1.3)2(0.5) + (2 − 1.3)2(0.4) = 0.41SD(X ) =√0.41 = 0.6303Because the standard deviation of X is 0.6303, he may be off by0.6303. That is, the use of the tool could be expected to be 1.3times give or take 0.6303.J. Sanchez Stat 100A Intro ProbabilityNotesNotesNotesNotesClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?Var (X ) = E (X − µ)2=Xx(x − E (X ))2P(X = x)= (0 − 1.3)2(0.1) + (1 − 1.3)2(0.5) + (2 − 1.3)2(0.4) = 0.41SD(X ) =√0.41 = 0.6303Because the standard deviation of X is 0.6303, he may be off by0.6303. That is, the use of the tool could be expected to be 1.3times give or take 0.6303.J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?Var (X ) = E (X − µ)2=Xx(x − E (X ))2P(X = x)= (0 − 1.3)2(0.1) + (1 − 1.3)2(0.5) + (2 − 1.3)2(0.4) = 0.41SD(X ) =√0.41 = 0.6303Because the standard deviation of X is 0.6303, he may be off by0.6303. That is, the use of the tool could be expected to be 1.3times give or take 0.6303.J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded (X) 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?(c) Suppose that it costs the factory $10 each time the tool is used.What is the expected daily cost of using the tool? How much is thiscost expected to vary from one day to another?Let C equal cost.ThenC = 10X, a linear function of X. Use expectation of the function of a randomvariable.E (10X ) =Xx(10X )P(X ) = 10XxXP(X ) = 10E(X ) = 10(1.3) = $13Var (10X ) = 102Var (X ) = 100(0.41) = 41 SD(X ) =p(41) = 6.4031The expected cost is $13 per day give or take $6.4031.J. Sanchez Stat 100A Intro ProbabilityClass assigment at the end of last lectureQuantity demanded (X) 0 1 2Probability 0.1 0.5 0.4(a) How much can he expect the tool to be used tomorrow?(b) By how much could he be off?(c) Suppose that it costs the factory $10 each time the tool is used.What is the expected daily cost of using the tool? How much is thiscost expected to vary from one day to another? Let C equal cost.ThenC = 10X, a linear function of X. Use expectation of the function of a randomvariable.E (10X ) =Xx(10X )P(X ) = 10XxXP(X ) = 10E(X ) = 10(1.3) = $13Var (10X ) = 102Var (X ) = 100(0.41) = 41 SD(X ) =p(41) = 6.4031The expected cost is $13 per day give or take $6.4031.J. Sanchez Stat 100A Intro ProbabilityNotesNotesNotesNotesWe have used in last example: Definitions of ExpectedValue, variance and standard deviation of a discreterandom variable and a function of a
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