Problem 9.3d ONLY.

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Problem 9.3 (Polling: smoking ban in NYC). Now that you are all experts on the Normal approximation of the binomial distribution, here is a rather challenging probability problem for this assignment. Assume that fifty-two percent of the residents of New York are in favor of outlawing cigarette smoking in publicly owned areas (so, take p = 0.52). Using a Normal approximation, compute the probability that more than half of a random sample of n people from New York are in favor of this prohibition when the size of the sample is: (a) n = 11, (b) n = 101, (c) n = 1001. Hint #1: Let Na, No, and Ne be the number of people in your three samples that are in favor of the smoking ban (these are three different random variables!). You have to approximate: P(N. 2 6), P(N, 2 51), and P(Nc > 501). You will note that, as the sample size n increases, so does the probability that more than half of the people in the random sample are in favor of the smoking prohibition. (d) The challenging part! How large does n have to be, to make this probability be equal to 95%? More precisely: find n such that P(N > 5) ~ 0.95, using a Normal approximation for the distribution of N. Hint #2: For part (d), the computation actually simplifies considerably if you do not introduce the +. "correction for continuity" when approximation Binomial distribution with a Normal probability density. We suggest that you proceed this way (which, in this part, will not change the result significantly)