Comment on anyway, i started blastin'
yetAnotherUser@discuss.tchncs.de 5 weeks agoI’ve found a proper approximation after some time and some searching.
Since the binomial distribution has a very large n, we can use the central limit theorem and treat it as a normal distribution. The mean would be obviously 500 billion, the standard deviation is √(n * p * (1-p)) which results in 500,000.
You still cannot plug that into WA unfortunately so we have to use a workaround.
Since WA would calculate it through:
Φ(b) - Φ(a), with b = (510 billion - mean) / (standard deviation) = 20,000 and a = (490 billion - mean) / (standard deviation) = -20,000 and Φ(x) = 0.5 * (1 + erf(x/√2))
erf(x) is the error function which has one good property: erf(-x) = -erf(x)
Therefore:
Φ(20,000) - Φ(-20,000) = 0.5 * [ erf(20,000/√2) - erf(-20,000/√2) ] = erf(20,000/√2) ≈ erf(14,142)
WolframAlpha will unfortunately not calculate this either.
However, according to Wikipedia an approximation exists which shows that:
1 - erf(x) = [(1 - e^(-Ax))e^(-x²)] / (Bx√π)
And apparently A = 1.98 and B = 1.135 give good approximations for all x≥0.
After failing to get a proper approximation from WA again and having to calculate every part by itself, the result is very roughly around 1 - 10^(-86,857,234).
So it is very safe to assume you will lose between 49% and 51% of your gut bacteria. For a more realistic 10 trillion you should replace a and b above with around ±63,200 but I don’t want to bother calculating the rest and having WolframAlpha tell me my intermediary steps are equal to zero.
Enkers@sh.itjust.works 5 weeks ago
Whoa, good work! I think I’m going to have to go over this a few times to grock how it works, especially the Φ(b) - Φ(a) bit. My stats textbook has a bit too much dust on it. ;)