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Homework 5 Solution

# Homework 5 Solution - Economics 506-Sections M&M1 Solution...

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Economics 506-Sections M &M1 Solution to Homework 5 (The problem numbers refer to numbers in 7 th edition) 7.1 : a. c. Answers vary. d. The histogram exhibits a mound shape. The sample mean should be close to 3.5 = μ e. The standard deviation should be close to σ/ 3 = 1.708/ 3 = .9860. f. Very similar pictures. 7.2 : a. P ( Y = 2) = P ( W = 6) = p (4, 1, 1) + p (1, 4, 1) + p (1, 1, 4) + p (3, 2, 1) + p (3, 1, 2) = p (2, 3, 1) + p (2, 1, 3) + p (1, 3, 2)+ p (1, 2, 3) + p (2, 2, 2) = 216 10 . b. Answers vary, but the relative frequency should be fairly close. c. The relative frequency should be even closer than what was observed in part b. 7.9 : a. P (| Y – μ| ≤ .3) = P (–1.2 ≤ Z ≤ 1.2) = .7698. b. P (| Y – μ| ≤ .3) = P (–.3 n Z ≤ .3 n ) = 1 – 2 P ( Z > .3 n ). For n = 25, 36, 69, and 64, the probabilities are (respectively) .8664, .9284, .9642, and .9836. c. The probabilities increase with n , which is intuitive since the variance of Y decreases with n . d. Yes, these results are consistent since the probability was less than .95 for values of n less than 43. 7.10 : a. P (| Y – μ| ≤ .3) = P (–.15 n Z ≤ .15 n ) = 1 – 2 P ( Z > .15 n ). For n = 9, the probability is .3472 (a smaller value).

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