HELP: Probability Question

mrsimperless · 04-29-2003, 02:00 PM

Here is a probability question for those of you blessed with more schooling than I have had. (Everyone)

I am trying to locate some sort of probability/statistics library for C#/.NET that will generate numbers for me that follow a pattern of normal distribution. Ideally this would be some sort of class where I pass in the mean and standard deviation as parameters to some method and I am then returned a random number. I have fortunately been able to find a random number generator that does normal distribution, but it does not have the degreee of flexibility that I need. All numbers seem to be generated based on a mean of 0 and a standard deviation of 1. (They are of course, however, up to something like 13 or 15 decimals of precision.)

My question is this: Is there a way for me to manipulate these numbers so that I can specify the mean and standard deviation for a random set? I think I have determined that if I add any amount to the generated numbers I get that this changes the mean of that set. And I was wondering if maybe multiplying the numbers (before adding the mean) would also change the standard deviation. I have tried this out and it SEEMS to generate expected results, but I am too clueless on this probability stuff to know for sure. Any smart guys (or girls) know for sure if I am actually getting what I think I am here?

mrsimperless · 04-29-2003, 02:10 PM

Dolaquestion:

Something else just came to mind that I was wondering as well. Say for example I determine that a given running back has a certain mean (let's say 80) and standard deviation (let's say 5) for rushing yards per game using normal distribution. Knowing this, how do I determine the mean and standard deviation for his yards per QUARTER?

I think that I would just divide the YPG mean by four to get 20, but how do I handle the standard deviation?

Brillig · 04-29-2003, 02:31 PM

Short answer... yes

By multiplying the initial distribution by X, you go from a standard deviation of 1 to X.

By subsequently adding Y, you get a normal distribution with a mean of Y with standard deviation X.

04-29-2003, 02:10 PM	#2
mrsimperless College Prospect Join Date: Apr 2003	Dolaquestion: Something else just came to mind that I was wondering as well. Say for example I determine that a given running back has a certain mean (let's say 80) and standard deviation (let's say 5) for rushing yards per game using normal distribution. Knowing this, how do I determine the mean and standard deviation for his yards per QUARTER? I think that I would just divide the YPG mean by four to get 20, but how do I handle the standard deviation? __________________ "All I know is that smart women are hot. Susan Polgar beat me in 24 moves in a simultaneous exhbition. I slept with the scoresheet under my pillow." Off some dude's web site. Last edited by mrsimperless : 04-29-2003 at 02:11 PM.

04-29-2003, 02:31 PM	#3
Brillig College Prospect Join Date: Oct 2000 Location: Mountain View, California	Short answer... yes By multiplying the initial distribution by X, you go from a standard deviation of 1 to X. By subsequently adding Y, you get a normal distribution with a mean of Y with standard deviation X.

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