Thanks, John. I'll give this a try. On 1/16/07, John Randall <[EMAIL PROTECTED]> wrote:
Devon McCormick wrote: > What I'd like to do is to construct an equivalent 5-element > distribution > with the same mean and standard deviation but (more or less) normally > distributed. How about fitting a binomial distribution to the data? If a frequency table on i.(n+1) has mean m, the binomial distribution with generating function (q+px)^n has the same mean if np=m. mf =:(+/ .* [EMAIL PROTECTED]) % +/ NB. mean of frequency table n =:<:@# NB. binomial n from frequency table p =:mf % <:@# NB. binomial p from frequency table b =:[: p. (p^n);(n # [EMAIL PROTECTED]) NB. binomial dist from freqency table d =:0 0.13333333 0.4 0.46666667 0 NB. data b d NB. binomial with same mean 0.0301408 0.168789 0.354456 0.330826 0.115789 mf d 2.33333 mf b d 2.33333 Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
-- Devon McCormick ^me^ at acm. org is my preferred e-mail ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
