Question 440589
{{{mu = int(x*f(x), dx, a, b) = int(x*1, dx, 0,1) = (x^2/2)[0]^1 = 1/2 - 0 = 1/2}}}
==> {{{E(sum(x[k], k = 1, 12)/12) = (1/12)sum(E(x[k]), k = 1, 12) 


= (1/12)sum(1/2, k = 1, 12)  = (1/12)*6 = 1/2 }}} is the mean of sample means of size 12.

{{{E(X^2)= int(x^2*f(x), dx, a, b) = int(x^2*1, dx, 0,1) = (x^3/3)[0]^1 = 1/3 - 0 = 1/3}}}

==> {{{Var(X) = E(X^2) - mu^2 = 1/3 - 1/4 = 1/12}}}

==> {{{Var(sum(x[k], k = 1, 12)/12) = (1/144)sum(Var(X[k]), k = 1, 12)


= (1/144)sum(1/12, k = 1, 12)  = (1/144)*1 = 1/144 }}},assuming i.i.d.