Given $n$ independent Normally distributed random variables $X_i \sim N(\mu_i, \sigma^2_i)$ and $n$ real constants $a_i \in \mathbb{R}$, I need to find an acceptable Normal approximation of the distribution of $Y$ random variable (assuming $\Pr[X_i\leq0]\approx0$, to avoid divisions by zero)
$$ Y = \frac{\sum_{i,N} a_i X_i}{\sum_{i,N} X_i} $$
I thought to split $Y$ into single components
$$ Y = a_1\frac{X_1}{\sum_{i,N} X_i} + a_2\frac{X_2}{\sum_{i,N} X_i} + \cdots+ a_n\frac{X_n}{\sum_{i,N} X_i} $$ $$ Y = a_1Y_1 + a_2Y_2 + \cdots+ a_nY_n $$ where the distribution of each $Y_i$ can be found noting that $$ Y_i = \frac{X_i}{X_i + \sum_{j \neq i} X_j} $$ and that $$ \frac{1}{Y_i} = \frac{X_i + \sum_{j \neq i} X_j}{X_i} = 1 + \frac{\sum_{j \neq i} X_j}{X_i} $$ so, calling $\sum_{j \neq i }X_j = U_i$, we can say that $X_i$ and $U_i$ are independent and, according to Díaz-Francés et Al. (https://doi.org/10.1007%2Fs00362-012-0429-2), a Normal approximation of $1/Y_i$ can be $$ \frac{1}{Y_i} \sim N\left( \mu_{1/Y_i}= 1+\frac{\mu_{U_i}}{\mu_{X_i}} \;,\; \sigma^2_{1/Y_i} = \frac{\mu_{U_i}^2}{\mu_{X_i}^2}\left[ \frac{\sigma_{U_i}^2}{\mu_{U_i}^2} + \frac{\sigma_{X_i}^2}{\mu_{X_i}^2} \right] \right) $$ and then, considering $1 \sim N(1,0)$, the r.v. $Y_i$ can be approximated to $$ Y_i \sim N\left( \frac{1}{\mu_{1/Y_i}} \;,\; \frac{1}{\mu_{1/Y_i}^2}\cdot\frac{\sigma^2_{1/Y_i}}{\mu_{1/Y_i}^2} \right) $$ and $$ a_iY_i \sim N\left( a_i\frac{1}{\mu_{1/Y_i}} \;,\; a_i^2 \frac{\sigma^2_{1/Y_i}}{\mu_{1/Y_i}^4} \right) $$
But now... I'm stuck at the sum of $a_iY_i$ because, not being independent, I don't know how to approximate the variance of their sum.
Any advice? Any more straightforward or more efficient method?