Normal Distribution Curve

[pianoman]

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!

[leafy]

It's a Gaussian distribution of the form f(x)=ae^something. I forget.

[pianoman]

I found it here:
http://en.wikipedia.org/wiki/Normal_distribution
So would this be a rational function?

[leafy]

I do believe so.

[pianoman]

Got it.
Thank you!

[AngelFish]

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...

[ruler501]

how would you make one that goes up to one at 100 then goes down to .01 at 140 and 60?

[pianoman]

How exactly do you find the area under the curve?
PS-Please dumb it down a bit if its really complicated.

[ruler501]

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

The factor 1/sqrt{2*pi} in this expression ensures that the total area under the curve ϕ(x) is equal to one

[pianoman]

But to find the area under parts of the curve, you would need calc?

[ruler501]

yeah. It confuses me so I will try this again later when i understand integrals more.

[AngelFish]

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

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 ).

[ruler501]

I like the idea of softtware. I don't even understand elementary much less sophisticated methods

[leafy]

There are commands normalcdf() and tcdf() but I think they use calculus.

[AngelFish]

I'll try to go through one method of finding the definite integral. No guarantees that I'll do it correctly

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 All | Copy To Clipboard

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

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.