Suppose that $X$, with a random sample $X_1, ..., X_n$, is a random variable with the following pdf: $f(x; t) = xte^{\frac{-x^2t}{2}}$ with the following support: $x \in [0, \infty)$. Note that $0 < t < \infty$ is unknown. (i) Determine the maximum likelihood estimator, $\hat{t}$, for $t$. (ii) Suppose that $Y_1, ..., Y_n$ are order statistics for the random sample; identify an expression for $P(Y_2 < \ln(9)t^{-1} < Y5)$.
My attempt:
(i) I found the following likelihood function
$$L = x_1 \cdot...\cdot x_n \cdot t^n\cdot e^{-\frac{t}{2}\sum_{j=1}^{n}x_j^2}$$
And then I took the ln:
$$\ln(L) = \ln(x_1 \cdot...\cdot x_n) + \ln(t^n) -\frac{t}{2}\sum_{j=1}^{n}x_j^2$$
And applying some easy calculus and maximizing gives:
$$(\ln(L))' = nt^{-1} - \frac{1}{2}\sum_{j=1}^{n}x_j^2 = 0$$
And then we have:
$$\hat{t} = 2\frac{n}{\sum_{j=1}^{n}X_j^2}$$
(ii) This probability is equivalent to saying that between $2$ and $4$ (inclusive) of $X_j$'s (for $j = 1, ..., n$) are greater than $\ln(9)t^{-1}$. Taking $W$ to be the number of $X_j$'s in this range, we can model this with a binomial distribution with $n$ and $p =$ the percentile at which $\ln(9)t^{-1}$ lies i.e. $p = F(\ln(9)t^{-1})$ where $F$ is the cdf. But:
$$F(x) = \int_{0}^{x} zte^{\frac{-z^2t}{2}} \,dz$$
$$p = F(\ln(9)t^{-1}) = \int_{0}^{\ln(9)t^{-1}} zte^{\frac{-z^2t}{2}} \,dz = 1 - e^{-\frac{2(\ln(3))^2}{t}}$$
However, since $n$ is unknown, and the value of $p$ depends on $t$, I have no idea how to proceed.
I can't figure out (ii). Also is (i) correct? Any assistance is much appreciated.