Suppose that $X$, with a random sample $X_1, \ldots, X_n$, is a random variable with the following pdf: $f(x; t) = xte^{\frac{-x^2t}{2}}$ with the following support: $x \in [0, \infty)$, where $0 < t < \infty$ is unknown. If $Y_1, \ldots, Y_n$ are order statistics and $\hat{t} = 2\frac{n}{\sum_{j=1}^{n}X_j^2}$ is the maximum likelihood estimator for $t$ (done in part i, not shown here), find $P(Y_2 < \sqrt{\ln(9)t^{-1}} < Y_5)$.
My attempt:
This probability is equivalent to saying that between $2$ and $4$ (inclusive) of $X_j$'s (for $j = 1, \ldots, n$) are greater than $\sqrt{\ln(9)t^{-1}}$. 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 $\sqrt{\ln(9)t^{-1}}$ lies i.e. $p = F(\sqrt{\ln(9)t^{-1}})$ where $F$ is the cdf. But
$$F(x) = \int_0^x zte^{\frac{-z^2t}{2}} \,dz$$
$$p = F(\sqrt{\ln(9)t^{-1}}) = \int_0^{\sqrt{\ln(9)t^{-1}}} zte^{\frac{-z^2t}{2}} \,dz = \frac{2}{3}.$$
$W$ is binomial with $n$ and $p = \frac{2}{3}$ so
$$P\left(Y_2 < \sqrt{\ln(9)t^{-1}} < Y_5\right) = \sum_{k=2}^4 {}_nC_k \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{n-k}$$
Is this correct? I do not know if I can go further, or find an exact value for the probability or if I am supposed to use the MLE. Any assistance is appreciated.
One more typo: "between 2 and 4 (inclusive) of the $X_j$ are less than $\sqrt{\ln(9)/t}$." (I am still confused why the problem statement mentions the MLE if it is irrelevant to the question.)
But otherwise I don't see any issues with your work. Since the answer involves a sum of 3 terms, you can expand it out as $$\begin{align}\frac{1}{3^n}\left(\binom{n}{2} 2^2 + \binom{n}{3} 2^3 + \binom{n}{4} 2^4\right) &= \frac{1}{3^n}\left[2n(n-1) + \frac{4}{3}n(n-1)(n-2) + \frac{2}{3} n(n-1)(n-2)(n-3) \right] \\ &=\frac{2}{3^{n+1}}n(n-1) \left[3 + 2(n-2) + (n-2)(n-3)\right] \\ &=\frac{2}{3^{n+1}}n(n-1) (n^2 - 3n + 5) \end{align}$$ which is the same as the expression in JimB's comment.