For each of the following distributions,find = E( X ), E[ X(X - 1)], a 2 = E[X(X
ID: 2918884 • Letter: F
Question
For each of the following distributions,find = E( X ), E[ X(X - 1)], a 2 = E[X(X -1)] + E(X) - 2: a) f(x)= 3! (1/4)x (3/4)3-x x = 0,1,2,3. x!(3 - x)!b) f(x)= 4! (1/2)4 x = 0,1,2,3,4. x!(4 - 3)! For each of the following distributions,find = E( X ), E[ X(X - 1)], a 2 = E[X(X -1)] + E(X) - 2: a) f(x)= 3! (1/4)x (3/4)3-x x = 0,1,2,3. x!(3 - x)!
b) f(x)= 4! (1/2)4 x = 0,1,2,3,4. x!(4 - 3)! x!(4 - 3)!
Explanation / Answer
(a) X~Binomial (n=3, p=1/4) E(X)=n*p=3*(1/4)=3/4 ^2=n*p*(1-p)=3*(1/4)*(3/4)=9/16 E(X(X-1)) = E(X^2) - E(X) = ^2+ {E(X)}^2 - E(X)=(9/16)+(3/4)^2 - (3/4) = 0.375 (Since ^2=E(X^2)- {E(X)}^2 ) (b) X~Binomial(n=4, p=1/2) E(X)=n*p=4*(1/2)=2 ^2=n*p*(1-p)=4*(1/2)*(1/2)=1 E(X(X-1)) = E(X^2) - E(X) = ^2+ {E(X)}^2 - E(X) =1+(2)^2 - 2= 3 (Since ^2=E(X^2)- {E(X)}^2 )
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