Unformatted text preview: 6 ] = 6 E[X i ] = 6 (5/6) n Also we can find the number of distinct faces rolled, 6 – X Var(X) = Var(X 1 + … + X 6 ) = Var(X 1 ) + … + Var(X 6 ) + 2 Σ i<j Σ Cov(X i , X j ) = 6 Var(X i ) + 2*15 Cov(X i , X j ) since there are (6C2) = 15 pairs i < j = 6(5/6) n (1 – (5/6) n ) + 30 ((4/6) n – (5/6) 2 n ) You could simplify, but you don’t have to. The neat thing to look at is what these expressions look like as functions of n . Can you think of a logical explanation why the mean and variance behave that way as n increases?...
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 Winter '08
 Chisholm
 Probability

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