Formula for the "Normal Curve"

Factual Questions
1

This is, admittedly, not the most fascinating question in the world.

You know the famous “bell-shaped curve” we read about–the one that so-called “normal” statistical distributions fall under?
On the cartesian plane, what is the formula that generates that curve?..you know, using “X” and “Y” and “squared” and all that stuff.

Is this (idealized) curve infinite in both its limbs, or does it have definite limits? (I’m guessing infinite.)

What is the relation of the curve, and formula, to the “standard deviation”?

2

Ta-da! http://mathworld.wolfram.com/GaussianDistribution.html

Check out equation (1), the formula for y = P(x) in terms of x, the standard deviation (sigma) and the mean (mu).

3

Achernar beat me to it. It may be of interest that although it goes out to infinity, a standard rule of thumb is that (five x standard deviation) is where the distribution is effectively zero.

4

Good God and great day in the morning! Thanks.

5

Six sigma, not five.

6

One standard deviation from the center of the distribution (in both directions) nets you about 66% of the area. Two standard deviations out gets you 95% of the area, and three standard deviations gets you 99% of the enclosed area. So three standard deviations on each side times two sides gives you six standard deviations, or six sigma. Going six standard deviations out from the center on each side would be twelve sigma, and would be overkill.

7

Oops. I was thinking of something else: exponential decay (e.g. drug dosage level) where 5 half-lives - around 3% - is taken as a practical end-point.

8

I heard somewhere that, in particle physics, 5 standard deviations above chance level is thought the be the bar for total acceptance of a theory.

9

I think those numbers are 68.3%, 95.5%, and 99.7%. The first two I wouldn’t have quibbled with, but it seems like there is a big difference between the last ones. You only need +/- 2.58 sd for a 99%.

10

There’s an infinite series that converges fairly rapidly for large values of x that gives you the integrated value:

1 - sqrt(2 / pi) × exp(-x[sup]2[/sup] / 2) × (x[sup]-1[/sup] - x[sup]-3[/sup] + 3x[sup]-5[/sup] - 15x[sup]-7[/sup] … ) [Adapted from Equation (21)]

For x = 3, then, this has a value of 99.7%. For x = 5, it’s 99.99994%. For x = 6, it’s 99.9999998%. Is 0.0000002% “effectively zero”? It depends on how accurate you want your results, of course. Probably.