I need to compute the value of this:
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)$$
the $N$ is a gaussian noise with mean=2 and standard deviation=2.
The question:
is this equivalent to:
$$ \frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z_{3}}{f} + N(2,2)\right) $$
or
$$ \frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right) $$
On
The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. The scope is given by the expression that follows immediately the $\Sigma$ and is valid respecting the arithmetic precedence rules up to an operator with precedence level equal to '$+$' or up to the end if no such operator follows.
This implies that \begin{align*} \color{blue}{\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)}&=\sum_{n=1}^{3}\left( \frac{z_{n}}{f} \right)+ N(2,2)\\ &=\left(\sum_{n=1}^{3} \frac{z_{n}}{f}\right) + N(2,2)\\ &\,\,\color{blue}{=\frac{z_{1}}{f} + \frac{z_{2}}{f} + \frac{z_{3}}{f} + N(2,2)}\\ \end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
It could be either one. It is more likely to be the latter; at least I myself would interpret it as $$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)=\frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right),$$ because the former would more likely be written with extra braces, i.e. $$\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z_{3}}{f} + N(2,2)\right)=\frac{1}{3}\left(\sum_{n=1}^{3}\left(\frac{z_{n}}{f} + N(2,2)\right)\right),$$ but not everyone is as generous with the braces, and some may omit them to avoid clutter.
Your best chances at interpreting the expression properly is to inform with whomever gave it to you.