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176 CHAPTER 5 | Discrete Probability Distributions MEAN OF THE BINOMIAL DISTRIBUTION The mean, u, of the binomial distribution is equal to the sample size, n, multiplied by the probability of an event of interest, T. M = E(X) = NTT (5.6) On the average, over the long run, you theoretically expect A = E(X) = nu = (4) (0.1) = 0.4 tagged order form in a sample of four orders. The standard deviation of the binomial distribution can be calculated using Equation (5.7). STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION 0 = Vo? = VVar(X) = VnT(1 - ) (5.7) The standard deviation of the number of tagged order forms is 0 = V4(0.1) (0.9) = 0.60 You get the same result if you use Equation (5.3) on page 169. Example 5.4 applies the binomial distribution to service at a fast-food restaurant. EXAMPLE 5.4 Accuracy in taking orders at a drive-through window is important for fast-food chains. Peri- Computing Binomial odically, QSR Magazine publishes "The Drive-Thru Performance Study: Order Accuracy" that Probabilities for measures the percentage of orders that are filled correctly. In a recent month, the percentage of Service at a orders filled correctly at Wendy's was approximately 86.8%. Suppose that you go to the drive- Fast-Food through window at Wendy's and place an order. Two friends of yours independently place Restaurant orders at the drive-through window at the same Wendy's. What are the probabilities that all three, that none of the three, and that at least two of the three orders will be filled correctly? What are the mean and standard deviation of the binomial distribution for the number of orders filled correctly? SOLUTION Because there are three orders and the probability of a correct order is 0.868, n = 3, and 7 = 0.868, using Equation (5.5) on page 173, P(X = 3|n = 3, 1 = 0.868) = 3! 3!(3 - 3) (0.868)' (1 - 0.868) 3-3 3! 3!(3 - 3)! (0.868)(0.132) = 1 (0.868 ) (0.868) (0.868) (1) = 0.6540 P(X = 0|n = 3, 7 = 0.868) = 3! 0!(3 - 0)! (0.868) ( 1 - 0.868 ) 3-0 0!(3 - 0) (0.868)(0.132) 3 = 1(1) (0.132) (0.132) (0.132) = 0.0023 (continued)