A beta probability distribution has two parameters, *a* and *b*. You can think of these as the number of successes and failures out of *a*+*b* trials. The PDF of a beta distribution is approximately normal if *a* and *b* are approximately equal and *a* + *b* is large.

If *a* and *b* are close, they don’t have to be very large for the beta PDF to be approximately normal. (In all the plots below, the solid blue line is the beta distribution and the dashed orange line is the normal distribution with the same mean and variance.)

On the other hand, when *a* and *b* are very different, the beta distribution can be skewed and far from normal. Note that *a* + *b* is the same in the example above and below.

Why the sharp corner above? The beta distribution is only defined on the interval [0, 1] and so the PDF is zero for negative values.

An application came up today that raised an interesting question: What if *a* + *b* is very large, but *a* and *b* are very different? The former works in favor of the normal approximation but the latter works against it.

The application had a low probability of success but a very large number of trials. Specifically, *a* + *b* would be on the order of a million, but *a* would be less than 500. Does the normal approximation hold for these kinds of numbers? Here are some plots to see.

When *a* = 500 the normal approximation is very good. It’s still pretty good when *a* = 50, but not good at all when *a* = 5.

**Update**: Mike Anderson suggested using the skewness to quantify how far a beta is from normal. Good idea.

The skewness of a beta(*a*, *b*) distribution is

2(*b* – *a*)√(*a* +* b* + 1) / (*a* + *b* + 2) √(*ab*)

Let *N* = *a* + *b* and assume *N* is large and *a* is small, so that *N*, *N* + 2, *b – **a*, and *N* – *a* are all approximately equal in ratios. Then the skewness is approximately 2 /√*a*. In other words, once the number of trials is sufficiently large, sknewness hardly depends on the number of trials and only depends on the number of successes.

Nice graphs. You could also devine this by examining the skewness of the three distributions, which are 0.089, 0.283, and 0.894, respectively. What a nice homework problem for my junior-level probability course; thanks!