Normal, binomial and Poisson, and what swapping one for another costs
A normal distribution
A count out of a fixed number of tries
The tails here are computed straight from the complementary error function rather than as one minus the other, which is what keeps four or five deviations from collapsing into a rounded zero.
Results
| How many deviations from the mean that is | — |
| Probability of landing below it | — |
| Probability of landing above it | — |
The count, exactly and approximately
| Exact probability of that count or fewer | — |
| Exact probability of exactly that count | — |
| Normal curve, no continuity correction | — |
| How far off that is, in percent | — |
| Normal curve, with the correction | — |
| How far off that is, in percent | — |
| Poisson, with its mean set to the same value | — |
| How far off that is, in percent | — |
The rule everybody is taught is that a normal curve stands in for a count once the mean count reaches five. At exactly that boundary, with twenty tries at twenty five out of a hundred, the exact answer is 0.091260 and the uncorrected curve says 0.060668. It is wrong by a third at the very point where it was declared safe.
The continuity correction is not a refinement, it is most of the accuracy. Adding half a unit before reading the curve turns an error of minus thirty three per cent into plus seven point eight, on the same numbers, with no extra information. Anybody using the approximation without it is choosing to be wrong by more than the approximation itself costs.
Where the approximation becomes honest
| Number of tries | Mean count | Exact | Error of the corrected curve, in percent |
|---|
Each row asks the same question three counts below the mean, so the comparison is fair as the number of tries grows. The error falls from about eight per cent to about half of one, which is the real shape of the rule: not a threshold that is crossed, but a slope that is walked down.
The third row of the results is a different approximation with a different weakness. Poisson stands in for a count when the chance of each try is small, not when the number of tries is large: at twenty five out of a hundred it is off by thirty six per cent, and at one out of a hundred over a thousand tries it is off by a third of one.
So there is no single approximation that is simply better. One of them needs many tries and the other needs a small chance, and the exact answer needs neither, which is why this page computes it rather than reaching for a curve.
When can I use a normal curve instead of counting exactly?
The rule everybody is taught is that the mean count has to reach five. At exactly that boundary the uncorrected curve is wrong by a third, which is a strange thing for a threshold to allow.
The table shows what actually happens as the number of tries grows. The error falls gradually rather than switching off at a line, so the honest reading is a slope and not a threshold.
| Number of tries | Mean count | Error of the corrected curve, in percent |
|---|---|---|
| 20 | 5,0 | +7,77 |
| 50 | 12,5 | -2,05 |
| 100 | 25,0 | -1,58 |
| 200 | 50,0 | -1,23 |
| 500 | 125,0 | -0,78 |
| 1000 | 250,0 | -0,54 |
What is the continuity correction, and does it matter?
It is adding half a unit before reading the curve, because a count lands on whole numbers and a curve does not. It costs nothing and it is most of the accuracy.
On the numbers this page starts from it turns an error of minus thirty three per cent into plus seven point eight. Using the approximation without it is choosing to be wrong by more than the approximation itself costs.
When is Poisson the right approximation instead?
When the chance of each try is small, which is a different condition from having many tries. At twenty five out of a hundred it is off by thirty six per cent no matter how many tries there are.
Take the same thousand tries at one out of a hundred and the error falls to a third of one per cent. The two approximations fail in different directions, which is why neither is simply better.
Why compute the exact answer at all?
Because it is now cheap. The approximations exist because summing the terms by hand was the expensive part, and that has not been true for a long time.
This page sums them in logarithms, one term at a time, so that a thousand tries never requires a number too large to hold. The approximations are kept on screen to show what they cost, not because they are needed.
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