Darren New wrote:
> 
> That's what I understood. If he has a linear set of data, wouldn't it
>  theoretically be equal quantities of each value, and hence wouldn't
> the statistics work to calculate the value from just one set of
> numbers per k?

Ah yes, I guess that would be like trying to numerically find the value
of the integrals instead of finding the closed form...


>> Standard deviation = sqrt(k^2/12)
> 
> (I'm kind of curious where that 1/12'th comes from.)

Well... It's 1/3 - 1/2 + 1/4 :-D

standard deviation = sqrt(\int_0^k f(x) (x-m) dx), where f is the
probability density and m the mean of the density.

Here f(x) = 1/k and m=k/2, so
\int_0^k f(x) (x-m)^2 dx = \int_0^k (x-k/2)^2/k dx
  = 1/k \int_0^k (x^2 - k x + k^2/4) dx
  = 1/k (k^3/3 - k/2 k^2 + k^2/4 k)
  = k^2/3 - k^2/2 + k^2/4
  = k^2 (1/3 - 1/2 + 1/4)
  = k^2/12

(There's probably a shorter way around this one but it works)

-- 
Vincent

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