> And the standard deviation is a number, just a number, just as I defined it as σ = E[(X - E[X])^2]^(1/2) which clearly is just a number and not an area.

Yes, and if I graph a function in terms of X and Y, where X is a function's argument and Y is the value returned by a particular function, one can argue that X and Y are just placeholders for numbers without any intrinsic meaning by themselves.

If I then say that Y represents the sine of X, surely someone will say, as you have said, "No, not at all, X is just a placeholder for a number, it's not what you say. And Y is just a placeholder for a number, it's not tied to any particular function."

In point of fact, the standard deviation is more than a particular number, it's an idea, and its statistical purpose is met when it's associated with a context in which that idea is expressed. That context is the normal distribution. In the present context, a particular sigma value refers to a specific area under a normal distribution, and in turn, to the probability that a particular result might have arisen by chance.

Without reference to a normal distribution, a standard deviation (a sigma) loses its conventional meaning. Variances are acquired by statistical tests of data sets, standard deviations (sigmas) are acquired from variances, and conclusions are drawn from sigmas only to the degree that they are applied to normal distributions.

Link: http://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

Quote: "In statistics, the 68–95–99.7 rule, also known as the three-sigma rule or empirical rule, states that nearly all values lie within three standard deviations of the mean in a normal distribution.
About 68.27% of the values lie within one standard deviation of the mean. Similarly, about 95.45% of the values lie within two standard deviations of the mean. Nearly all (99.73%) of the values lie within three standard deviations of the mean."

According to your thesis, the above claim is obvious nonsense, because in point of fact, sigma has no association with the normal distribution. But you know what? Wikipedia can be edited by anyone, and you can correct this egregious error today, if you like. Correct this widespread erroneous thinking -- fix these errors, everywhere you find them.

Let's perform a little test. Let's submit "5 sigma" to Wolfram Alpha and see whether it makes the same mistake you say I have been making, and that the above linked article is making:

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Link: https://www.wolframalpha.com/input/?i=5+sigma

Quote:

Input Interpretation: 5 sigma (standard deviations)
z-score : 5
Probabilities:
zɝ (left-tailed p-value) | 1-2.867×10^-7
zɱ (right-tailed p-value) | 2.867×10^-7
abs(z)ɱ (two-tailed p-value) | 5.733×10^-7
abs(z)ɝ (confidence level) | 1-5.733×10^-7
Associated two-sided confidence level: 100-5.733×10^-5%

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So it seems that, without prompting, Wolfram Alpha draws the same conclusion that Wikipedia does, and every other online reference does when confronted by terms such as "standard deviation" or "sigma" -- that, without providing a specific context, the default context is that sigmas refer to positions on a standard distribution and have statistical meanings associated with that default assumption.

> Of course, even with a Gaussian assumption, standard deviation does not have a particular numerical value.

When a scientist uses the term "5 sigma", she is in most cases using a context of a normalized normal distribution, one that follows the 68-95-99.7 rule described above. Therefore yes, 5 sigma does lead to a a particular value, by being applied to a definite integral of a normalized normal distribution that uses it as an argument. By this reasoning, 5 sigma means:

(Sage notation) integrate(e^(-x^2/2),x,a,b) / sqrt(2 * pi)

Picture: http://i.imgur.com/YhC302b.gif

If I try to compute the value used in physics, the result for 5 sigma, I can use this:

(Sage notation) N(integrate(e^(-x^2/2),x,5,infinity) / sqrt(2 * pi))

With this result: 2.86651571870318e-7

Which is correct, and is the expected one-tailed value for 5 sigma, even though according to your argument, 5 sigma has no connection to normal distributions.

> Sorry 'bout that.

What, for your specious argument? No problem, it comes with the territory. But surely you realize I have more than answered your objection.