On Thu, 24 Apr 2008, Keith Goodman wrote: > It's random.normal(loc, scale). But that will give you random x values > drawn from the normal distribution, not y values at your x. So you'll have > to code your own normal, which will probably look something like
Keith, I overlooked that, thanks. > norm = 1 / (scale * sqrt(2 * pi)) > y = norm * exp(-power((x - loc), 2) / (2 * scale**2)) Can do. So, scale would equate to width and loc to center, yes? It's been so many years since I've dealt intimately with statistics and mathematics that I've gotten quite rusty. (However, I still retain what I need to accomplish a given objective.) Much appreciated, Rich -- Richard B. Shepard, Ph.D. | Integrity Credibility Applied Ecosystem Services, Inc. | Innovation <http://www.appl-ecosys.com> Voice: 503-667-4517 Fax: 503-667-8863 _______________________________________________ Numpy-discussion mailing list [email protected] http://projects.scipy.org/mailman/listinfo/numpy-discussion
