finalans

finalans - Answers to Final Exam Review Questions Question...

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Answers to Final Exam Review Questions Question 1 : This problem is a Bayes Law problem. The forward tree branches first on “will” and “won’t” date George; the second branch is “is free” and “isn’t free”. When you reverse the tree you find that the () P will date George says"free" = + = . .. . 075 075 225 25 . Question 2 : Here we have X = 13, Median = 14, s = 592 . ; the 98% C.I. is 729 1871 ., . A normal quantile plot, sometimes called a normal probability plot, is the technique I am looking for. This is the one that creates a straight line if the data is normally distributed. Question 3 : Part (a) is hypergeometric, with N r n x or == = = 10 6 3 2 3 ,,, . The answer is two thirds. Part (b) is poisson with µ = 33 5 / and x = 1. The answer is .0787. Question 4 : This is a straightforward normal probability calculation. The answer for part (a) is z = =− 0 5 31 16 . . . and Pz >− = 16 5636 . The answer to part (b) is −= z X . . . . 35 39 305 31 , or X = 293 Question 5 : In part (a) the mean is xf x 25 and the variance is ( ) σµ 2 2 105 = xf x . In part (b) you use the formula for the mean and variance of a linear transformation of a random variable; that is Y ab X X =+ ≡ .10 . Therefore the mean of the phone bill is $2.50, and the variance is 1.05. In part (c) we have a problem involving the distribution of the mean. The question asks the probability that
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() Px P x n x i i >= = 72 72 30 240 . . We know that Ex == µ 250 . and that σ x n = 105 30 187 . . . Therefore the z value is z = =− 240 250 187 53 .. . . and the answer is Pz >− = 53 7019 . In part (d) the answer is yes; because the sample size is large (n=30) the sample means have a normal distribution even though the original x’s obviously did not. We needed the normal distribution to do our probability calculation. Question 6 Part (a) is a binomial with n = 5 and p = 18/38. Being a net winner means winning 3, 4, or 5 of your 5 bets. The probability of this is 5 3 474 526 5 4 474 526 5 5 474 526 4514 32 41 50 + + = . For part (b) you are asked to do a normal approximation to a binomial with n=25 and p = 18/38 = .474. The distribution of the number of successes in 25 trials is approximately normal with () ( ) 25 .474 11.84 EX n p = and 1 25 .474 .526 2.497 X np p = × × = Being a net winner means winning 13 or more of your bets. The z value is 12.5 11.84 .26 2.497 z . The answer is ( ) .26 .5000 .1026 .3974 > =−= .
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This note was uploaded on 03/01/2012 for the course ECON 371 taught by Professor Staff during the Spring '08 term at UVA.

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finalans - Answers to Final Exam Review Questions Question...

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