Problem image:
answering to questions (i) and (ii) i found that:
$$pdf: f(x) = x \frac{2}{\theta^2}$$
$$E(x) = \frac{2}{3} \theta$$
$$Var(x) = \frac{1}{18}\theta^2$$
And now I tried to answer questions (iii) , (iv) and (v). My solutions:
(iii):
$E(Y) = E(X^3) = \int_{0}^\theta x^3* f(x) dx = \int_{0}^\theta x^3* (x* \frac{2}{\theta^2}) dx$ ,
$\frac{2}{\theta^2}$ is constant so:
$E(X^3) = \frac{2}{\theta^2} * \int_{0}^\theta x^4 dx = \frac{2}{\theta^2}*\frac{\theta^5}{5} = \frac{2}{5}\theta^3$
(iv):
probability to draw one random sample that is less than $\frac{2}{\theta}$ :
$P(X<\frac{2}{\theta}) = \int_{0}^\frac{2}{\theta} x* \frac{2}{\theta^2} dx = \frac{2}{\theta^2} \int_{0}^\frac{2}{\theta} x dx = \frac{2}{\theta^2} * \frac{\theta^2}{8} = \frac{1}{4} $
probability to draw 3 independant sample less than $\frac{\theta}{2}$ :
$\frac{1}{4}^3 = \frac{1}{64}$
(v):
estimate of mean: $\frac{4.1+6.6+7.2+8.9+10}{5} = 7.36$
we know that $E(x) = \frac{2}{3}*\theta$
hence, estimate of $\theta$:
$\frac{2}{3}\theta = 7.36$
$\theta = 11.04$
QUESTION:
Are my calculations correct? If no, what am I doing wrong? What correct solutions should look like?