# Sum of a set normalize by total items in set

This might be a be simple but I just want to make sure I didn't use the wrong notation.

If I have a set of weighted terms, ${w_1, w_2, \dots, w_n}$ and the score is the sum of $w_1$ to $w_n$ normalize by count of $w$. Count of $w$ is just the total number of terms, i.e. $n$. so, can I write:

$score = \frac{\sum{w_i}}{\left | w \right |}$ ?

p/s: $W = \{w_1, w_2, \dots, w_n \}$ but I don't really want to introduce $W$ for the sake of showing the count using $|W|$.

Thanks.

-
I am cool to get a downvote like I said this is a simple question and I am clarifying it. But for the nice guy who downvoted would you be nice again and explain why do I get a downvote? –  cherhan Oct 27 '13 at 13:25
For the guys who downvoted, that's very helpful. Thanks very much. –  cherhan Oct 27 '13 at 13:28
I don't think one has to be sarcastic about that. I think the downvoter just needs you to edit your post and clarify the meaning of $w$ ;-) –  BIS HD Oct 27 '13 at 13:29
@BISHD I am more that happy to edit and learn and improve the answer so that it benefits everyone else too. But NO information provided to advise how/what should I improve? –  cherhan Oct 27 '13 at 13:31
If information is missing out you could start with the information I gave you in my answer. Tell the world, what $w$ is. ;-) Otherwise you are right about the downvoter not giving any information on why he down voted it. –  BIS HD Oct 27 '13 at 13:33

If $w$ is the set of weighted terms, you are right.
$w$ is a set of weighted terms. Count of $w$ is total number of terms, i.e. $n$ –  cherhan Oct 27 '13 at 13:24
I understand that, and $w$ is the individual item in $W$, what I want is to avoid using $W$ just because I want to show $\left | W \right |$ if it's possible –  cherhan Oct 27 '13 at 13:26