## Sampling distribution - weighted mean

I have a problem of outlier detection, So let $X = \{x_1, \cdots, x_N \}$ be a data set containing $N$ observations, for example, an Human Height dataset of size $N = 10$:
$X = {180, 170, 172, 190, 255, 210, 183, 178, 185, 175}$

I have two categories of outlier detection algorithm.

The first category will give me a subset $S \subset X$, $|S| = n$, containing the observations considered outliers. Let consider $n$ fixed for all the possible outlier detection solution, for example, $n = 3$, and we are looking only for the outliers in the upper bound of the data, for example, in the dataset given above, I would be interested only in the tallest people. Different solutions could be:
$S_1 = \{180, 190, 183\}$
$S_2 = \{170, 255, 178\}$
$S_3 = \{255, 190, 210\}$

The example above is only an illustrative example, I know that $n$ and $N$ are too small to CLT holds true, but in the real problem both values will be large enough to it holds true. I don't have a specific data, but the data are usually multidimensional (e.g. iris dataset).
So, assuming that the CLT will hold true, if I average the $n$ values contained in $S$ I know from the distribution of the sample means that all possibles realization of $S$ will follow a Normal $\mathcal{N}(E, Var)$, so I can statically evaluate all the solutions and compare to a random solution.

My problem is in the second category of algorithms, instead of give a subset $S \subset X$, it will give a set of scores $w = \{ w_1, \cdots, w_N \}$, $w_i$ in th interval $$0, 1$$, where $w_i$ represents the outlier score (probability) associated with the observation $x_i$. Different solutions could be:
$w_1 = \{ 0, 0, 0, 1, 0.9, 0.5, 0, 1, 0.3, 0.1 \}$
$w_2 = \{ 0, 0, 0, 1, 1, 1, 0, 0, 0, 0 \}$
$w_3 = \{ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 \}$

Here I am averaging the values contained in the solution again, but now I am using a weighted average instead ($\frac{1}{\sum_{i = 1}^N w_i}\sum_{i = 1}^N w_i \cdot x_i$), the idea the higher the value the better the solution.

My question is: Can I in some way still use the CLT in the second case? in order to statically evaluate the solutions? If my weights were constants I would say yes, but the weights are a random variable itself.