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hw9-sol

# hw9-sol - AMS 311(Fall 2009 Joe Mitchell PROBABILITY THEORY...

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AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY THEORY Homework Set # 9 – Solution Notes (1). (20 points) The average number of bank failures in the USA is two per week. (a). Estimate the probability, p , that at least five bank failures occur in the USA next week. (What inequality are you using?) Let X be the number of bank failures that occur in the USA next week. We know that E ( X ) = 2. We use Markov’s inequality to get a bound on p : 0 p = P ( X 5) E ( X ) 5 = 2 5 (b). Assume now (for parts (b), (c), and (d)) that you are told that the variance of the number of bank failures in the USA in any one week is 4. Now give an improved estimate of p (using an inequality). Now we know that var ( X ) = 4. So we apply the one-sided Chebyshev inequality, with μ = E ( X ) = 2 and a = 3: 0 P ( X 5) = P ( X 2 + 3) 4 4 + 3 2 = 4 13 (You could have used a 2-sided Chebyshev, which would result in a weaker bound as follows: P ( X 5) P ( X 5 or X ≤ - 1) = P ( | X - 2 | 3) 4 3 2 = 4 9 .) (c). Give a Central Limit Theorem estimate for the probability q that during the academic year (considered to be 31 weeks total, fall and spring) there are more than 75 bank failures in the USA. Let Y i be the number of bank failures in week i , for i = 1 , . . . , 31. Let Y = 31 i =1 Y i denote the total number of bank failures during the academic year. Then, E ( Y ) = 31 · E ( Y i ) = 31 · 2 = 62 and var ( Y ) = 31 · var ( Y i ) = 124 (using the fact that the Y i ’s are independent). Then, using the Central Limit Theorem estimate, we get q = P ( Y > 75) = P ( Y - 62 124 > 75 - 62 124 ) P ( Z > 1 . 17) = 1 - Φ (1 . 17) = 1 - . 879 = . 121 (d). Use an inequality to get the best bounds you can on the probability q estimated in part (c). We can apply the one-sided Chebyshev inequality to get 0 P ( Y > 75) = P ( Y > 62 + 13) 124 124 + 13 2 = . 4232 (Note that Markov inequality gives P ( Y > 75) 62 75 , which is weaker than what we get above.) (2). (15 points) The cash register is broken at the bagel shop, so transactions are done by hand and recorded. Your lazy employee, Joe, charges customers exactly (e.g., if they owe \$ 3.61, he makes sure they pay exactly \$ 3.61); however, lazy Joe only records in the notebook the dollar amount of each transaction (e.g., \$ 3 for the \$ 3.61 transaction), omitting the cents. Assume that Joe serves 20 customers today. Find an upper bound on the probability that the total shown in Joe’s notebook is at least \$ 15 less than the actual amount of money those 20 customers paid to Joe. Let X i be the amount (in dollars) Joe fails to record when he writes the i th customer’s purchase in the notebook. Then, X i is Uniform(0,1). (Actually, X i is discrete uniform on the set { 0 , 0 . 01 , . . . , 0 . 99 } , but we often approximate it with a continuous uniform. You can readily compute E ( X i ) = 99 i =0 i 100 1 100 = 99 200 0 . 5 and var ( X i ) = E ( X 2 i ) - [ E ( X i )] 2 = 99 i =0 ( i 100 ) 2 1 100 - [ E ( X i )] 2 1 / 12 ) Let X = X 1 + · · · + X 20 be the total amount (in dollars) that Joe fails to record for the 20 transactions; then, his notebook ends up showing \$ X dollars less than it should (i.e., less than what

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