Standard Deviation of Weighted Average

That is not a weighted average. To perform a weighted average, you use another variable, typically a size type variable, to do the weighting. For example, if you had the average miles driven by state, you could conduct a weighted average by multiplying the average for each state by its population and then dividing by the total population of the country. What you computed is the sum of x^2 / sum x.

You can modify the standard deviation function to account for weighting, but you have to have that second size variable.

Thank you.

I'm trying to analyze fiber length distributions. I've been using the calculation I mentioned because I want to put more "weight" on the longer fiber lengths. Is there a better way to do so?

First, be sure there is a valid reason to put more weight on longer fibers. If you have four fibers of lengths 1, 2, 3 and 10, you have 16 lengths of fiber, four pieces, for an average of 4. You might really like that 10, but there is no obvious reason to give it extra weight. If you are showing these numbers to someone else, it is going to be hard to justify a made up average formula.

If these numbers are coming out of a process that is trying to make long lengths but the fibers keep breaking, you are making a Poisson distribution. Given enough data, you could model the breaks probability. From that link, you can see how to compute the standard deviation expected.

There are some simple ways to give bigger numbers more weight. You can use a geometric mean (find the square root of the average of the squares), but usually there is a solid reason to do something like that. In this case, the sum of the squares is 1 + 4 + 9 + 100 = 114. The average of that is 28.5 and the square root is 5.3. Again, you should have a solid reason related to your process to do that.