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Reading my forthcoming SMXR358 Summer School material, I came across the following method of estimating uncertainty when counting randomly occurring events (I quote).

*If the number of randomly occurring events counted in a given period is N, the uncertainty in this count (i.e. the likely difference between N and the mean that would be found from a long series of repeated experiments) is √(N).*

Example: if the mean is 100 counts, 68% 0f experiments will produce results within 90 and 110 counts i.e. +/- √100.

Example: if the mean is 100 counts, 68% 0f experiments will produce results within 90 and 110 counts i.e. +/- √100.

So where does that come from? I asked a tutor last year and we had an inconclusive discussion about Poisson distributions which seemed beside the point. Anyway, I reviewed it again and came to the following conclusion.

Suppose that there is an underlying binomial process with very low probability p of producing an observable event, and high probability q = (1-p)≈ 1 of not producing an observable event in a given time period. Radioactive decay would be an example.

Let T be the total number of underlying events (observable and not) which occur during the course of a single experiment and let Tp be the expected number of

*observable*events. The average of the counts N which are measured would tend to this value over a number of experiments.

The standard deviation σ of the distribution is √(Tpq) = √(N *q) ≈ √N, as q is nearly 1.

So using the normal distribution approximation, 68% of measured values of N will lie within +/- 1σ of the mean, Tp, which is +/- √N as stated. Note that we use measured N as a best estimator for Tp.

**Update**

I could have got to this more elegantly. The p ≈ 0 and q ≈ 1 condition and the large number of trials T is the condition for the Poisson distribution to apply. And in the Poisson distribution, the variance = the mean N, so the standard deviation is the square root of the mean = √N.