# Why would someone plot variance normalized by the mean?

I'm reading a scientific paper where they plot the variance of particle intensity normalized by the mean of particle intensity. I'm a bit confused and don't have an intuition for how this should be helping me. I'm used to seeing standard deviation and variance, both of which reflect dispersion.

What does plot of variance normalized by the mean give me that plot of variance does not?

Here is an example of the graph I am talking about. The x-axis represents the length of an RNA molecule resulting in a distribution of particle sizes. You can just take it as a categorical variable.

edit:

This relation is apparently called the "index of dispersion".

Here they may be looking for "overdispersion" indicate an unexpectedly large grouping or clustering of these particles (which actually what they see under the microscope).

I'm still trying to wrap my head around the intuition though. I know that for a Poisson distribution the mean = variance (dispersion factor = 1), but that's just me knowing facts, and I don't want to go off thinking of this relation as measuring "how Poissonian" a distribution is.

• this is a statistical forum .. you have to provide the chemistry knowledge (why should the mean and variance of particle intensity be related) or go to another forum – seanv507 Jul 3 '17 at 7:55
• @seanv507 I don't think the OP is asking for an interpretation of the example plot specifically, but is asking for plots of variance over mean more generally. – Greenparker Jul 3 '17 at 8:08

Variance over mean is known as the Index of dispersion.

This can be useful when comparing two random variables with different means, in order to account for larger variance for larges means.

Example: suppose two fields have some sheep, and I hire 2 "counters" to count the sheep in each field. The counters count the sheep 10 times for their field and report the number of sheep counted. If the first counter gets a field with 50 sheep, then for that counter the variability across their counts will be naturally low. On the other hand, for the second counter if their field has 1000 sheep, then the variability of their counting will most definitely be higher than the first counter.

So in this case, we can't just study the variance for the two counters and declare the first counter to be better since he/she had a lower variance. We have to account for the number of sheep in their respective fields. This can be done by weighting the variance by the mean, i.e., index of dispersion.

• Any idea why var/mean and not stddev/mean? – Has QUIT--Anony-Mousse Jul 6 '17 at 6:25
• @Anony-Mousse Here is a good discussion between the two. – Greenparker Jul 10 '17 at 10:44