How do I calculate PERT stddev if I use custom weight factors? According to (for example) http://tynerblain.com/blog/2006/04/13/foundation-series-basic-pert-estimate-tutorial/, the PERT mean estimate is 
$$\text{meanA}=(\text{optimistic}\ +\ 4 \times \text{likely}\ +\ \text{pessimistic})/6$$
I understand I can use "custom" weight factors for the basic estimates. For example, I could say I weight the optimistic estimate by a factor of 2. Then, of course, the divisor would not be 6, but 7:
$$\text{meanB}= (2 \times \text{optimistic}\ +\ 4 \times \text{likely}\ +\ \text{pessimistic})/7$$
Now a PERT estimate includes an approximation of the standard deviation. With the widely used weights of 1:4:1, the stddev is approximately 
$$\text{stdevA}=(\text{pessimistic}\ –\ \text{optimistic})/6$$
I am not sure where the 6 comes from, and I think this is my question. 
I am wondering about this because if I use the weights from $\text{meanB}$, for example 2:4:1, which divisor would I use for approximating the stddev? Still 6? Or 7?
 A: The mean and sd calculations you give are used in PERT three-point estimation.
They might be regarded as based on either assuming particular cases of a 
four-parameter beta distribution 
or a particular double triangular* distribution, such that the median 
and mode coincide.
* The double triangular has a density consisting of two triangles back-to-back (with a discontinuity at the mode). The general case has probability $p$ in the left triangle and $(1-p)$ in the right, but in the particular one discussed above, $p=0.5$
The values given for calculating the mean and sd are based only on the three specified values.  In the case of the double triangular with equal probability in both halves, the given mean is right, but I haven't checked whether the sd estimate makes sense (I believe it's based on a rough approximation).
To derive a suitable sd estimate for your differently weighted average, you'd need some basis on which the weighted average 
was a suitable way to approximate the distribution's mean and then obtain a suitable sd estimate from that same basis.
I think you might perhaps be able to choose a different beta or perhaps some other distribution so as to make your weighted mean work and then base an estimate of standard deviation off whichever distribution you ended up with. (No double triangular will produce relative weights of 2,4,1.)
