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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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 Spring '08
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