Find the mean and variance of the mean of a random sample of 9 from a distribution having pdf 6x(1-x) 0 < x < 1, zero elsewhere.
I don't even know how to start this one.
Start with the basic definitions and formulae:
$\displaystyle \bar{X} = \frac{X_1 + X_2 + \, .... \, + X_9}{9}$.
$\displaystyle E(\bar{X}) = \frac{1}{9} \left[ E(X_1) + E(X_2) + \, .... \, + E(X_9) \right] = E(X_i)$
where $\displaystyle E(X_i) = \int_0^1 6x^2 (1 - x) \, dx$.
$\displaystyle Var(\bar{X}) = \frac{1}{81} \left[ Var(X_1) + Var(X_2) + \, .... \, + Var(X_9) \right] = \frac{1}{9} Var(X_i)$
where $\displaystyle Var(X_i) = E(X_i^2) - [E(X_i)]^2$ and $\displaystyle E(X_i^2) = \int_0^1 6x^3 (1 - x) \, dx$.