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##### Probability Distributions and Expected Value

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**In previous chapters, our emphasis was on the probability of**individual outcomes. • This chapter develops models for distributions that show the probabilities for all possible outcomes of an experiment.**Random Variable (X)**• Has a single value, x, for each outcome of an experiment. To show all the possible outcomes, a chart is normally used.**Discrete variables take values that are separate (or that**can be “counted”)Continuous variables have an infinite number of possible outcomes (usually measurements that can have an unlimited decimal place)**For example:**• The number of phone calls made by a salesperson • Discrete (1,2,3,4,5…..)**For example:**• The length of time a person spends on the phone • Continuous (1 min, 1.23min …..)**Let a DRV (X) be the possible outcomes when rolling a die**The probability distribution could be written in a table This is a uniform distribution, because all the probabilities are the same.**A graph would look like this…**1 1/6 2 3 4 5 6**Expected Value: Informal**When rolling 2 dice, the sum that is generated most frequently is called the expected value. (7) This can also be calculated mathematically. Multiply each roll by it’s probability of occuring…**E(sum) = 2P(sum = 2) + 3P(sum = 3) + …**+…12P(sum = 12) = 2 X 1 / 36 + 3 X 2 / 36 … = 252 / 36 = 7**The expected value, E(X), is the predicted average of all**possible outcomes.It is equal to the sum of the products of each outcome with its probability**n**= xiP(X = xi) i = 1 Expected Value of a Discrete Random Variable The sum of the terms of the form (X)(P[X]) E(X) = x1P(X = x1) + x2P(X = x2) + … + xnP(X = xn)**Ex 1: Suppose you toss 3 coins.**• What is the likelihood that you would observe at least two heads? • What is the expected number of heads?**1**3 3 1 8 8 8 8 Represent the theoretical probability distribution as a table.The DRV, X, represents the number of heads observed. X 0 hs 1 hs 2 hs 3 hs P(X) = x**a) P(X = 2) + P(X = 3) = 3 / 8 + 1 / 8**= 1 / 2 b) The expected number of heads = 0(1 / 8) + 1(3 / 8) + 2(3 / 8) + 3(1 / 8) = 3 / 2**For a game to be fair, E(X) must be zero Consider a dice**game If you roll a 1 2 3 you win $1.00 If you roll a 4 5 6 you pay $1.00 Is this game fair? E(X) = (1)(1/6) + (1)(1/6) + (1)(1/6) + (-1)(1/6) + (-1)(1/6) + (-1)(1/6)**Page 374**1, 2(ex 2), 3a,c, 4, 9, 11,12, 19