Re: [Maxima-discuss] "distrib" package

[Robert D. <rob...@gm...>]

On 2014-09-03, Henry Baker <hb...@pi...> wrote:

> On a particular Olympic sport, the abilities of any individual
> in a population is normally distributed.
>
> Thus, for country A, with population Ap, mean Am, variance Av,
> the abilities of the individuals follow a normal distribution.
>
> Clearly, each country will choose from the upper tail of the
> distribution -- its highest 1%, for example.
>
> What is the probability that a 1% athlete from A will beat a 1%
> athlete from B, given the population, mean and variance of each country ?

We are looking for P(ability(A) - ability(B) > 0) where ability(A)
and ability(B) are Gaussian tail distributions. That is, not the
entire bump (i.e. national population) but just the 1% who
are sent to the games. I think Maxima can figure out the density of
ability(A) - ability(B) exactly, but one must be careful with the
limits of integration in the convolution since we are working with
just the tails. Note sure if Maxima can then compute P(diff > 0)
since that requires another integration.

I worked on a related problem recently. For the record:
http://maxima-solved.blogspot.com/2014/06/probability-density-of-sum-of-gamma-and.html

> Now, suppose that we have a _group sport_ which requires a small
> number (k) athletes, and that the performance of each country is
> additive -- e.g., the gymnastics "team" medal, where the team is
> of size k.
>
> Now, what is the probability that country A will beat country B ?

Well, for k larger than a few, the group ability is nearly Gaussian
even though the individual abilities are very non-Gaussian (via
central limit theorem). So one can come up with an approximation
more easily in the group case.

Sounds like fun -- good luck.

Robert Dodier