You're stumped because most of the "statistics primer" section in that post doesn't make any sense. The connection between the Gaussian density and the sqrt(pi) heuristic is mostly imaginary. The original heuristic (pi, sqrt(pi), pi^2) works pretty much the same with 3 instead of pi, so you can view the pi versions as numerology or charitably a nice mnemonic.
Could you elaborate? Assume we convert all our happy path estimates to minutes. What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution than a uniform standard distribution,because normal distribution is more likely to occur in nature.
while I can understand that this is a controversial assumption, I'm not the first one to make it. Referring to numerology seems a bit odd?
I'm really looking for a proper way here, I'm quite a rational person, so numerology is not really my cup of tea TBH.
Your blog post has so many errors that I don't even know where to start. As another poster mentioned, areas under non-degenerate probability density functions are 1 by definition, whether they're uniform, Gaussian, standardized or not. What you described as a "standard uniform distribution" is really a degenerate distribution[1], meaning that you assume no uncertainty at all (stdev=0). There's nothing "uniform" about that, you might just as well start with a Gaussian with stdev=0.
"converting from a standard uniform distribution to a Gaussian distribution" as you described does not make any sense at all. If you replace an initial assumption of a degenerate distribution with a Gaussian, as you seem to be doing, you replace a no-uncertainty (stdev=0) assumption with some uncertainty (so the uncertainty blow-up is infinite), but it doesn't affect point estimates such as the mean or median, unless you make separate assumptions about that. There is nothing in your story that leads to multiplying some initial time estimate by sqrt(pi). The only tenuous connection with sqrt(pi) in the whole story is that the Gaussian integral happens to be sqrt(pi). There are some deep mathematical reasons for that, which has to do with polar coordinates transformations. But it has nothing to do with adjusting uncertainties or best estimates.
Thank you for your valuable feedback; it will take some time to process, and I will adjust the blog post as my insights grow (potentially discarding the whole idea, but for me it's a learning process.)
> What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution
If that were your actual assumption, you should measure the variance of the difference between your estimate and the actual time taken, use that to determine a confidence interval (e.g. 95% of the time, the additional delay is less than x) and then add it to your estimate.
