Thread Subject:

Random Number Generation

I am trying to create a random number array given a minimum, maximum, and mean value. Is there any way to do this?

Thanks,
Corey

"Corey Adrian" <uryguitarist777@yahoo.com> wrote in message
news:i1fhko$rsc$1@fred.mathworks.com...
>I am trying to create a random number array given a minimum, maximum, and
>mean value. Is there any way to do this?

Sure, just write a function that errors if the specified mean value is less
than the specified minimum or greater than the specified maximum value and
otherwise returns your mean value.

http://dilbert.com/strips/comic/2001-10-25/

If you want your numbers distributed differently in the interval between
your minimum and maximum than "centered at the mean value", you need to
provide a description of your desired distribution.

--
Steve Lord
slord@mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
To contact Technical Support use the Contact Us link on
http://www.mathworks.com

"Corey Adrian" <uryguitarist777@yahoo.com> wrote in message <i1fhko$rsc$1@fred.mathworks.com>...
> I am trying to create a random number array given a minimum, maximum, and mean value. Is there any way to do this?
>
> Thanks,
> Corey
- - - - - - - -
  How about using the beta distribution? Let a = minimum, b = maximum, and c = mean. Of course, it is assumed that a < c < b here. Then do:

 x = a+(b-a)*betarnd((c-a)/(b-a),(b-c)*(b-a),m,n);

to get an m by n array x of such random numbers.

  You can multiply the first two arguments of betarnd by any common factor to change the distribution's variance at your pleasure without affecting the mean. By making this factor very large you can approach Steven's interesting suggestion.

Roger Stafford

  Oops! I meant

 x = a+(b-a)*betarnd((c-a)/(b-a),(b-c)/(b-a),m,n);

Roger Stafford

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in message <i1g240$rho$1@fred.mathworks.com>...
> Oops! I meant
>
> x = a+(b-a)*betarnd((c-a)/(b-a),(b-c)/(b-a),m,n);
>
> Roger Stafford

Thanks Roger that is exactly what I needed!

Corey Adrian