solutionsproblemset1

# solutionsproblemset1 - Solutions to Problem Set #1 1. A)...

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Solutions to Problem Set #1 1. A) let X= # heads; heads = H, tails = T. X=0 X=1 X=2 X=3 X=4 TTTT HTTT HHTT HHHT HHHH THTT HTHT HHTH TTHT HTTH HTHH TTTH THTH THHH T H H T T T H H B) From part a) we see that each sample point should have the same probability = 1/16; the probability distribution for the # of heads is P(X=0) = 1/16, P(X=1) = 4/16 =1/4, P(X=2) = 6/16 = 3/8, P(x=3) = 4/16 = 1/4, P(x=4) = 1/16. C) P(X=3) = 1/4 from part b). 2. A) Marginal probability distribution of X: P(X=0) = P(X=0, Y=0) + P(X=0, Y=1) = .3 +.1 = .4 P(X=1) = P(X=1,Y=0) + P(X=1,Y=1) = .4+.2 = .6 Marginal probability distribution of Y; P(Y=0) = P(Y=0, X=0) + P(Y=0, X=1) = .3 + .4 = .7 P(Y=1) = P(Y=1, X=0) + P(Y=1, X=1) = .1 + .2 = .3

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B) E(X) = (0)P(X=0)+(1)P(X=1) = .6 E(X 2 ) = (0 2 )P(X=0)+(1 2 )P(X=1) = .6 Var(X) = E(X 2 )-[E(X)] 2 = .6 – (.6 2 ) = .24 E(Y) = (0)P(Y=0) + (1)P(Y=1) = .3 E(Y 2 ) = (0 2 )P(Y=0) + (1 2 )P(Y=1) = .3 Var(Y) = E(Y 2 )-[E(Y)] 2 = .21 C) 3 / 1 3 . 1
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## This note was uploaded on 03/30/2008 for the course ECG 561 taught by Professor Wohlgenant during the Spring '08 term at N.C. State.

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solutionsproblemset1 - Solutions to Problem Set #1 1. A)...

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