🚀 go-pugleaf

Jon Dodd wrote:
> ...
> From a normal distribution with mean m and standard deviation sd,  I
> pick a random sample A and then second B. If A>B then the value of A is
> assigned to group X, if B>A then value of A is assigned to group Y. What
> is the resulting mean and standard deviation of the two distributions X
> and Y? I am not mathamatically oriented so a step by step proof would be
> great.
> ...


Hi Jon,

For a normal distribution A and B are independent random variables.
Therefore, the likelihood for P(A>B) = 0.5 (i.e. 50%) and 
P(B>A) = 1 - 0.5 = 0.5 (P(u) + P(v) = 1, if v is the complementary event
of u  - think of a coin: P(head) + P(tail) = 1). 
You are just dividing a set of N random numbers into two sets of N/2
elements by random! (However, if you had specified something like A > m
+ sd this would make a different case.) Of course both sets are random
representations of the same distribution with the same mean and standard
deviation - but slightly different sample means and sample standard
deviations as those are only estimators of the "true" parameters of the
distribution. For N -> Infinity: mean(x) = mean(y) =
mean_of_normal_distribution.

I think, you should read one or two basic books on statistics, there are
even some focussing on medical/biological issues.

Regards,
Harald

(BTW, I had the same thought as Ben.)

-- 
Harald Giese
Email: giese@dkrz.de
Phone: +49 (0)40 4123 5796; Fax: +49 (0)40 5605724
Institut fuer Meereskunde der Universitaet Hamburg
(Institute of Oceanography of the University of Hamburg)
Troplowitzstrasse 7, D-22529 Hamburg