# Re: Formula Needed

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Posted by jerry crick on October 21, 2002 at 14:02:32:

In Reply to: Re: Formula Needed posted by Brad Paul on October 20, 2002 at 15:06:57:

: : : Hi - can anyone help?

: : : I am looking for a formula that will calculate values for a curve whose end points defined by a and b would typically be the values 0 to 1 and 1 to 4 repectively - these to be chosen by the user and represent the x axis. In my example these have been multiplied by a cost estimate.

: : : The y access will run between 0 and 100 representing the probability. The probability of 0 must occur where x = 1, 400 in my example since my cost estimate is 400.

: : : In addition to this I need to be able to alter the shape of the curve using a single value. This would represent a level of confidence where very confident leads to a very narrow and spiky curve and not confident leads to a flatter less spiky curve.

: : : Well that's the description. I have placed an image of the resultant graph I would like to see in the link below. This one just used a log scale to get an approximation!
: : I have a function for you but you have to promise that you do not use
: : it to do any real statistics. This is just a made up function that
: : fits what you have described and has no statistical value!

: : f(x,a,b,n)=22n{(x-a)n(x-b)n}/{(b-a)n(a-b)n}

: : where n=1,2,3....

: : f(a,a,b,n)=f(b,a,b,n)=0

: : f({a+b}/2,a,b,n)=1
: : df({a+b}/2,a,b,n)=0

: : As n gets larger the curve gets thiner.

: : PS I could not find you link to your curve.

: To make your shape the only constrant on n is
: n>0 It does not have to be an int.

Paul

This is I know not for statisitcal results at this stage though at some point I hope I can move the process on a bit more.

Thanks again

Jerry

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