# hw3sol - 36-225 - Homework 3 SOLUTIONS1.19 points: 4...

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Unformatted text preview: 36-225 - Homework 3 SOLUTIONS1.19 points: 4 forE(Y), 5 forE(1Y), 5 forE(Y2-1), 5 forV(Y)E(Y) =XyyPY(y) = 1(.4) + 2(.3) + 3(.2) + 4(.1) = 2.E1Y=Xy1yPY(y) = 1(.4) +12(.3) +13(.2) +14(.1) =.6417E(Y2-1) =E(Y2)-1 = [1(.4) + 4(.3) + 9(.2) + 16(.1)]-1 = 5-1 = 4Using Theorem 4 from lecture #6V(Y) =E(Y2)-[E(Y)]2= 5-(2)2= 12.18 points: 6 forE(Y), 6 forE(Y2), 6 forV(Y). In each part, 2 for setup, 3 forwork, 1 for correct answer.E(Y) =kXy=1y(1k) =1k·kXy=1y=1k·k(k+ 1)2=k+ 12E(Y2) =kXy=1y2(1k) =1k·kXy=1y2=1k·k(k+ 1)(2k+ 1)6=(k+ 1)(2k+ 1)6V(Y) =E(Y2)-E2(Y) =(k+ 1)(2k+ 1)6-k+ 122=2(k+ 1)(2k+ 1)-3(k+ 1)212=(k+ 1)(k-1)123.10 pointsBy definition of the variance,V ar(aY+b) =E[aY+b-(aμ+b)]2=E[aY-aμ+b-b]2=E[a(Y-μ)]2] =E[a2(Y-μ)2].Using Theorem 2 in lecture #6, this equalsa2E(Y-μ)2which equalsa2V(Y) =a2σ2.4.22 points: (a) 13 points [4 for definingYcorrectly, 4 forE(Y), 4 forV(Y)and 1forσx] (b) 9 points [8 for work, 1 for numerical accuracy](a) From the given information is follows thatY= 12 + 8X....
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## This note was uploaded on 07/15/2011 for the course STAT 36225 taught by Professor Tom during the Summer '09 term at Carnegie Mellon.

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hw3sol - 36-225 - Homework 3 SOLUTIONS1.19 points: 4...

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