# Where did this risk exposure 'estimation-formula' come from?

I was reading a book and the authors metioned that risk exposure can be estimated scientifically using this forumula:

$risk(\$) = \frac{(a + 4m + b)}{6}$and standard deviation$\sigma = \frac{b-a}{6}$Where: a = minimum dollar exposure m = most likely dollar exposure b = maximum dollar exposure Where a, m, b are solicited from past data or expert judgement. All this is fine, but I've never seen this formula - where did this come from? Is it founded in statistics or any scientific reference? Anyone know anything about this? Pointers/suggestions/clarifications would be greatly appreciated. I'm curious as to how does the above formula provides a risk exposure metric. Why '6' in the denomiator and why the a + 4m + b in the numerator? Similarly for$\sigma$? Any ideas? EDIT: This formula seems to be that of weighted averaging but am not sure about it's inception or the reason of giving 'm' a weight of 4 and how that influences the calculation of$\sigma$? UPDATE:Book - Making the Software Business Case. However just Googling for the numerator threw up a few results - it seems to be known as the PERT formula and is used for estimation of 'time' - but my question about the 'formula' still holds irrespective of it being used for risk... DERIVATION OF FORMULA - The reason for the formula to be what it is seems to be beautifully explained in a JSTOR article for those interested! • You could help us by specifying the book and providing its definition of risk! This word "risk" means so many things that it's practically impossible to reverse-engineer a formula like this. – whuber Apr 27, 2012 at 21:23 ## 1 Answer This is called Three Point Estimation sometimes used in project management. It is a rule of thumb used to translate simple estimates from non-statisticians into helpful statistics. Rather strangely, the estimate of the mean comes from assuming a so-called double-triangular distribution, where$m$is both the median and the mode, so with density •$f(x) = \frac{x-a}{(m-a)^2} \text{ when } a \lt x \lt m$•$f(x) = \frac{b-x}{(b-m)^2} \text{ when } m \lt x \lt b\$

but the estimate of the standard deviation is rather less than this distribution might suggest.

• It has a name ;) Thanks for the pointer. However I still left wondering "why" that particular selection of weights? i.e. why 4m and not 7m why not 2a or 7b etc.? Any theoretical grounds for this formula?
– PhD
Apr 28, 2012 at 1:15