how to generate a normal distribution with mean=1, min=0.2, max=0.8
On Fri, 29 Apr 2011, David Winsemius wrote:
On Apr 29, 2011, at 1:29 PM, Mike Miller wrote:
On Fri, 29 Apr 2011, Giovanni Petris wrote:
Well, but the original poster also refers to 0.2 and 0.8 as "expected min and max", in which case we are back to a joke...
Well, he is a lot better with English than I am with Mandarin. He seemed to like the truncated normal answers, so we'll let those be his answers. It is possible to choose parameters for a normal distribution with 500 observations such that the expected value of the maximum is .8 and the expected value of the minimum is .2. Obviously, the mean would be .5, not 1, but what would the variance then have to be to provide the correct expected max and min? That's another legitimate question.
You would need to specify an N since the expected first and last order statistic would decrease/increase with increasing N.
Right -- I chose N=500, as did the OP. I think the order statistics for the normal are pretty complex, but it wouldn't be hard to use the density for order statistics for the uniform to compute the appropriate values for a standard normal, then rescale. http://en.wikipedia.org/wiki/Order_statistic#The_order_statistics_of_the_uniform_distribution You'd have to multiply the beta density times the inverse normal cdf and get the weighted average for a set of points. It doesn't sound terribly difficult but I don't want to do it! ;-) Mike