# Formula for the "Normal Curve" **URL:** \ **Category:** Factual Questions\ **Created:** [January 19, 2003, 9:44pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684 "2003-01-19T21:44:14Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![Scott\_Dickerson](https://avatars.discourse-cdn.com/v4/letter/s/f04885/32.png) [@Scott\_Dickerson](https://boards.straightdope.com/u/Scott_Dickerson)\ **Post date:** [January 19, 2003, 9:44pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/1 "2003-01-19T21:44:14Z")
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”? ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 19, 2003, 10:00pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/2 "2003-01-19T22:00:51Z")
Ta-da! [http://mathworld.wolfram.com/GaussianDistribution.html](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). ---
**Author:** ![raygirvan](https://avatars.discourse-cdn.com/v4/letter/r/ac91a4/32.png) [@raygirvan](https://boards.straightdope.com/u/raygirvan)\ **Post date:** [January 20, 2003, 5:15am UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/3 "2003-01-20T05:15:34Z")
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. ---
**Author:** ![Scott\_Dickerson](https://avatars.discourse-cdn.com/v4/letter/s/f04885/32.png) [@Scott\_Dickerson](https://boards.straightdope.com/u/Scott_Dickerson)\ **Post date:** [January 21, 2003, 6:06am UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/4 "2003-01-21T06:06:51Z")
Good God and great day in the morning! Thanks. ---
**Author:** ![Urban\_Ranger](https://avatars.discourse-cdn.com/v4/letter/u/e9c0ed/32.png) [@Urban\_Ranger](https://boards.straightdope.com/u/Urban_Ranger)\ **Post date:** [January 21, 2003, 11:39am UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/5 "2003-01-21T11:39:36Z")
> [@](#): > > \*Originally posted by raygirvan \* > 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. Six sigma, not five. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [January 21, 2003, 12:04pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/6 "2003-01-21T12:04:09Z")
> [@](#): > > ## quote: > > ## Originally posted by raygirvan 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. > > Six sigma, not five. 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. ---
**Author:** ![raygirvan](https://avatars.discourse-cdn.com/v4/letter/r/ac91a4/32.png) [@raygirvan](https://boards.straightdope.com/u/raygirvan)\ **Post date:** [January 21, 2003, 1:46pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/7 "2003-01-21T13:46:17Z")
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. ---
**Author:** ![Shalmanese](https://avatars.discourse-cdn.com/v4/letter/s/45deac/32.png) [@Shalmanese](https://boards.straightdope.com/u/Shalmanese)\ **Post date:** [January 21, 2003, 2:20pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/8 "2003-01-21T14:20:55Z")
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. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [January 21, 2003, 5:48pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/9 "2003-01-21T17:48:34Z")
> [@](#): > > \*Originally posted by CalMeacham \* > **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.** 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%. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [January 21, 2003, 6:23pm UTC](https://boards.straightdope.com/t/formula-for-the-normal-curve/149684/10 "2003-01-21T18:23:17Z")
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)](http://mathworld.wolfram.com/Erf.html)] 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.