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Offline pianoman

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Normal Distribution Curve
« on: May 25, 2011, 06:20:18 pm »
Hi everyone!
We were just learning about bell curves in class, and it just brought up a question to my mind.
Is there any specific equation that can be used to describe/graph a bell curve?
I'm not sure if I'm wording that very well.
Thanks! :P
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Offline leafy

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Re: Normal Distribution Curve
« Reply #1 on: May 25, 2011, 06:23:59 pm »
It's a Gaussian distribution of the form f(x)=ae^something. I forget.
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Offline pianoman

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Re: Normal Distribution Curve
« Reply #2 on: May 25, 2011, 06:29:32 pm »
I found it here:
http://en.wikipedia.org/wiki/Normal_distribution
So would this be a rational function?
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Offline leafy

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Re: Normal Distribution Curve
« Reply #3 on: May 25, 2011, 06:32:53 pm »
I do believe so.
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Offline pianoman

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Re: Normal Distribution Curve
« Reply #4 on: May 25, 2011, 06:48:02 pm »
Got it.
Thank you!
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Offline AngelFish

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Re: Normal Distribution Curve
« Reply #5 on: May 25, 2011, 06:56:40 pm »
f(x)=(1/sqrt(2*pi))*e^((-x^2)/2) is the standard normal Gaussian distribution. The general form for any normalized Gaussian is f(x)=(1/sqrt(2*pi*StdDev^2))*e^(-(x-mean)^2/(2*StdDev^2))

As it happens the 3d analogue of the Standrd normal Gaussian is f(x,y)=(1/sqrt(2*pi))*e^(-(x)^2-(y)^2/(2))

And yes, I have those memorized... :P
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Offline ruler501

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Re: Normal Distribution Curve
« Reply #6 on: May 25, 2011, 06:59:50 pm »
how would you make one that goes up to one at 100 then goes down to .01 at 140 and 60?
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Re: Normal Distribution Curve
« Reply #7 on: May 25, 2011, 07:01:09 pm »
How exactly do you find the area under the curve?
PS-Please dumb it down a bit if its really complicated.
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Offline ruler501

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Re: Normal Distribution Curve
« Reply #8 on: May 25, 2011, 07:03:09 pm »
thats calculus...
you have to take the integral of it.(not an easy task) I learning how to do this. If I figure it out I'll write it out for you

EDIT: what are the rules for substitution and multiplication again in integrals. I never did manage to fully understand that.

EDIT2: I just found this while brosing the wikipedia article
Quote from: wikipedia
The factor 1/sqrt{2*pi} in this expression ensures that the total area under the curve ϕ(x) is equal to one
« Last Edit: May 25, 2011, 07:15:32 pm by ruler501 »
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Re: Normal Distribution Curve
« Reply #9 on: May 25, 2011, 07:16:35 pm »
But to find the area under parts of the curve, you would need calc?
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Offline ruler501

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Re: Normal Distribution Curve
« Reply #10 on: May 25, 2011, 07:19:25 pm »
yeah. It confuses me so I will try this again later when i understand integrals more.
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Offline AngelFish

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Re: Normal Distribution Curve
« Reply #11 on: May 25, 2011, 07:48:35 pm »
Quote from: ruler501 on May 25, 2011, 07:03:09 pm
thats calculus...
you have to take the integral of it.(not an easy task) I learning how to do this. If I figure it out I'll write it out for you

EDIT: what are the rules for substitution and multiplication again in integrals. I never did manage to fully understand that.

EDIT2: I just found this while brosing the wikipedia article
Quote from: wikipedia
The factor 1/sqrt{2*pi} in this expression ensures that the total area under the curve ϕ(x) is equal to one

I wouldn't even bother with that type of integration. The e^(-x^2) factor means that the integral is non-elementary (IE, you can't write the answer down in terms of elementary functions). The area under the curve has to be found with more sophisticated methods (or using software :P).
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Offline ruler501

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Re: Normal Distribution Curve
« Reply #12 on: May 25, 2011, 07:52:17 pm »
I like the idea of softtware. I don't even understand elementary much less sophisticated methods
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Offline leafy

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Re: Normal Distribution Curve
« Reply #13 on: May 25, 2011, 09:00:38 pm »
There are commands normalcdf() and tcdf() but I think they use calculus.
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Offline AngelFish

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Re: Normal Distribution Curve
« Reply #14 on: May 25, 2011, 11:18:08 pm »
I'll try to go through one method of finding the definite integral. No guarantees that I'll do it correctly :P

First of all, there's some number representing the area under the curve that we'll call A. The integral is taken from a to b, thus...

Code: [Select]
b
∫1/√(2π)e^(-x²/2)=A
a

        b
=1/√(2π)∫e^(-x²/2)
        a

           b
A²=(1/√(2π)∫e^(-x²/2))²
           a

<skip about half a page of painful to type integrals>
      b          b
(1/2π)∫e^(-x²/2)*∫e^(-y²/2)
      a          a

      b b
(1/2π)∫∫e^(-x²/2-y²/2)
      a a

      b b
(1/2π)∫∫e^(-1/2(x²+y²))
      a a

<convert to polar with x^2+y^2=r^2>

 tan^-1(b/b) √(2b²)
(1/2π)∫dΘ∫e^(-2(r²)) dr
 tan^-1(a/a) √(2a²)

<U substitution>

<blah blah blah>

It was at this point that I realized that the integral only worked for a= -∞ and b=∞, which is still better than the normal methods taught using anti-derivatives.
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