6. (20 points) This question is about using the normal distribution as an approximation to the binomial distribution. When the number of trials is large, adding up many binomial probabilities can be cumbersome, and in some cases a normal distribution can provide a close fit to the binomial distribution. In class, we saw an example of an airline that overbooks ights, knowing that not all passengers will show up. Suppose that an airline sells 165 tickets for a ight that has 150 seats. The airline believes that each passenger has a 90 percent chance of showing up for the ight, and it treats the appearance of each passenger as an independent event. Let X be a binomial random variable that counts the number of passengers who show up for the ight. 6.1. What is E(X)? 6.2. What is sd(X)? 6.3. If you wanted to compute the probability that there are more passengers than seats, P(X > 150), how many binomial pmf values would you have to sum? Note that you are not being asked to perform the probability calculation. 6.4. Let Y be a normal random variable that has the same mean and standard deviation as X, the binomial random variable. Using the binomial distribution, the probability that there are more passengers than seats is 1901 > 150) = 0.311. What is the normal distribution approximation to this probability, P(Y > 150)? 6.5. The approximation in the previous problem is not especially good. Perhaps it makes more sense to treat exactly 150 passengers showing up, X = 150, as corresponding to 149.5 5 Y 5 150.5. With this adjustment, what is the normal distribution approximation to probability that there are more passengers than seats, P(Y > 150.5)