They ask me to estimate any parameter and I do not if the solution is correct:
The life time X of a battery is considered to be a random variable with density function
$f (x; Θ) =\frac{ 2 }{ Θ²} (Θ - x)$ if 0 < x <Θ
Be X1, X2, ..., Xn a simple random sample of X. Knowing that $E(X) = \frac{Θ}{ 3}$ and $E(X²) = \frac{Θ²}{6}$, determine:
(a) An unbiased estimator for Θ, Θ ^
(b) The variance of the estimator Θ ^
For (A) I have
$\mu= \frac{Θ}{3}$
$ Θ=3 \mu$
$ Θ= 3 X$
$E(Θ)= E(3X) $
$ 3 E(X)= 3 *\mu$
$3 * \mu =3 * \frac{Θ}{3} = Θ$
(that if it is unbiased)
For (B) I have:
$Var(Θ)= Var(3X) = 3^2 Var(X) = 9 σ^2 /n$
Strictly speaking, you have a sample from $X$, so your chosen estimator should be based on the sample $X1,X_2,\ldots,X_n$ rather than on the underlying $X$
Although you did not say so, you seem to have chosen $3\bar{X}=\frac3n\sum\limits_{i=1}^n X_i$ as your estimator of $\Theta$. This does indeed have an expectation of $3 E[\bar{X}] =3 E[X] = \Theta$ so is unbiased.
You may have miscalculated the variance of $X$ which is $\text{Var}(X) = E\left[X^2\right]-\left(E[X]\right)^2 = \frac{\Theta^2}{6}-\left(\frac{\Theta}{3}\right)^2 = \frac{\Theta^2}{18}$ so the variance of your estimator is $\text{Var}(3\bar{X}) = 9\text{Var}(\bar{X}) = \frac{9}{n}\text{Var}(X) = \frac{\Theta^2}{2n}$
Alternatively you might have chosen $3X_1$ or something similar as your estimator. That would have mean $3 E[X_1] =3 E[X] = \Theta$ so is also unbiased. Its variance is $\text{Var}(3X_1) = 9\text{Var}(X_1) = 9\text{Var}(X) = \frac{\Theta^2}{2}$