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Unformatted text preview: Solutions to Homework 6, ISyE 2027 Fall 2006 Problem 1 (a) geometric (b) Bernoulli (c) binomial (d) Poisson Problem 2 (a) We know that 1 = ∑ 4 k =1 P ( X = k ), so 1 = 4 X k =1 P ( X = k ) = c 4 X k =1 k = 10 c so c = 1 / 10. (b) The expected value (or mean) of X is just E [ X ] = 4 X k =1 kP ( X = k ) = (1 / 10) 4 X k =1 k 2 = 30 / 10 = 3 . (c) To compute the probability mass function of Y , we first need to determine the range of Y , or the set of possible values Y can take. It’s clear that the range of Y is { 1 , 4 , 9 , 16 } . Therefore, the probability mass function of Y is P ( Y = 1) = P ( X = 1) = 1 / 10 P ( Y = 4) = P ( X = 2) = 2 / 10 P ( Y = 9) = P ( X = 3) = 3 / 10 P ( Y = 16) = P ( X = 4) = 4 / 10 (d) The expected value of Y = X 2 is just E [ Y ] = (1) P ( Y = 1) + (4) P ( Y = 4) + (9) P ( Y = 9) + (16) P ( Y = 16) = 1 / 10 + 8 / 10 + 27 / 10 + 64 / 10 = 100 / 10 = 10 . (e) Using the “law of the unconscious statistician”, we see that E [ X 2 ] = 4 X k =1 k 2 P ( X = k ) = 4 X k =1 (1 / 10) k 3 = 10 . (f) Here E [ X ( X 1)] = E [ X 2 X ] = E [ X 2 ] E [ X ] = 10 3 = 7 . (g) The variance of X is just V ar ( X ) = E [ X 2 ] ( E [ X ]) 2 = 10 9 = 1 . (h) The standard deviation is just the square root of the variance, so in this case it’s 1. (i) If we condition on the event that { X ≤ 2 } , then X is either 1 or 2. 1 2 P ( X = 1  X ≤ 2) = P ( X = 1 ,X ≤ 2) P ( X ≤ 2) = P ( X = 1) P ( X ≤ 2) = (1 / 10) (3 / 10) = 1 / 3 ....
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This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn

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