hw6-print

# hw6-print - F(5 = P(X<= 5 = 36/36 = 1 The median of X is...

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Homework #6 (Solution) Total possible points is 15 Ex. 4.2. Since X is the absolute value of the difference of the two rolls, X = 0, 1, 2, 3, 4, 5. P(X=0)=(1/6)*(1/6)*6=1/6; P(X=1)=(1/6)*(1/6)*2+ (1/6)*(2/6)*4 = 10/36; P(X=2)=((1/6)*(1/6)*4 + (1/6)*(2/6)*2 = 2/9; P(X=3)=(1/6)*(1/6)*6 = 1/6; P(X=4)=(1/6)*(1/6)*4 = 1/9; P(X=5)=(1/6)*(1/6)*2 = 1/18; CDF of X: F(0) = P(X = 0) = 6/36 = 1/6; F(1) = P(X <= 1) = 16/36 = 4/9; F(2) = P(X <= 2) = 24/36 = 2/3; F(3) = P(X <= 3) = 30/36 = 5/6; F(4) = P(X <= 4) = 34/36 = 17/18;
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Unformatted text preview: F(5) = P(X <= 5) = 36/36 = 1; The median of X is 2 and it is unique. 4.3. X values could be 0, 1, 2, 3, 4. P(X=0) = 1/2; P(X=1) = (1/2)*(1/2) = 1/4; P(X=2) = (1/2)*(1/2)*(1/2) = 1/8; P(X=3) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16; P(X=4) = (1/2)*(1/2)*(1/2)*(1/2) = 1/16; I.23. Let P(X=0)=p1, P(X=0.5)=p2, P(X=1)=p3. E(X) = 0*p1 + 0.5*p2 + 1*p3 = 0.5 P(X <= 1) = p1 + p2 + p3 = 1 Which implies that p1 + 0.5*p2 = 0.5 and p1 = p3; i.e. P(X=0) = P(X=1). X 0 1 2 3 4 5 P(X) 1/6 10/36 2/9 1/6 1/9 1/18...
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## This note was uploaded on 04/23/2011 for the course STAT 416 taught by Professor Staff during the Spring '08 term at Purdue.

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