Imagine I put people into different groups based on a uniformly distributed random variable $y = f(x)$ (e.g. microseconds of their arrival to a website). After a while, I observe how many people are in group 1, group 2, etc. Say I have 100 groups, one for each of 100 microsecond timestamp values.
If I sampled 10000 people in total, I would expect 100 in each group, following a uniform distribution. Now, what differences in group size should I expect by random chance alone?
My mind is drawing a blank here - been out of probability classes for too long! Thinking of the $y$ as $U(1,100)$, I think my question relates to the expected variance of $z = y^{-1} = U^{-1}(0,100)$. This Wikipedia page highlights the inverse uniform distribution (for the continuous case, assuming the lower bound is strictly positive). The variance given there is
$$ \frac{1}{a*b} - \left( \frac{ln(b) - ln(a)}{b-a} \right)^2 $$
where $a = 1$ and $b = 100$ are the bounds.
Where my mind is blank is here:
- Am I even right here? Since I want to know the 'to be expected' differences in the number of people in each group, I think I am asking for the variance of the pdf values of a uniform distribution, though I could have gotten this wrong.
- How do I scale this back to 'number of people' from the frequencies? Is this a simple multiplication with 10000, or am I overlooking something?