Why is it that the normal distribution contains a factor of 1/2 in 
the exponent? (Alternately, why does the coefficient contain a factor 
of sq. rt. 1/2?) 

The two factors, 1/2 in the exponent and sq. rt. 1/2 in the 
coeffiecient, come as a set (i.e., once 1/2 is put into the exponent, 
a correction of sq. rt. 1/2 is needed in the coeffiecient to preserve 
unity in the distribution's integral). Each is necessary when the 
other is present but, why include either in the distribution at all? 
What functionality do they add? It seems that all the essential 
properties of the distribution would be present absent these factors. 

My two speculations are: One, these factors help the distribution 
better fit 'normal' data (i.e., the distribution is designed to 
accommodate a pattern seen in nature). This implies there is a 
hitherto undiscovered explanation that rests on a principle of 
natural phenomena. Two, the normal distribution was originally 
derived from other distributions (as an approximation to them) and 
the factor of 1/2 in the exponent is a vestigial remnant, an emblem 
of lineage. What's the real reason?

The 1/2 in the exponent is used to give the distribution a variance of
1. This is an arbitrary convention, but a convenient one.

The sqrt of 2 in the denominator and the 1/2 in the numerator are not
arbitrary. The central limit theorem of statistics says that averages
of simple random samples converge to the normal distribution. That's
where the numbers come from altnough it isn't trivial to show this is
true.

Chocolate: It is still a normal distribution with any number replacing
the 2. The general form is:

(1/sqrt(2*pi*s))exp(-x^2/(2s))

Where s is the variance of the distribution. The variance just
measures the width of the peak of the normal distribution. Clearly
this width will be different if you were talking about the average
size of a mosquito or the average size of a star, measuring both in
meters.

The Central Limit Theorem just tells us that the distribution of the
sum of a large number of independent random variable will obey the
general form given above. It doesn't tell us what the variance is.

Thank you, Kottekoe (and Berkeleychocolate).

To be clear, the 1/2 in the exponent is to ensure that the variance
can be succinctly described by the parameter 's'. Without the 1/2,
finding the variance from the mgf would result in a product of 's' and
some constant, presumably 2*s--I haven't completed the algebra myself.

That said, the value of 's' itself is not necessarily 1. 's' may take any value.

Correct?

Yes, that is correct.

I think everyone missed the mathematical background. I even think
everyone is using analogies to explain that the 1/2 is conventionally
inserted - which is TOTALLY WRONG. They are suggested due to the
nature of probability - never can be over 1.

Normal distribution curves or general distribution curves are
Probability Density Function (PDF). The horizontal axis covers all the
possible outcomes X. Suppose each outcome is x. PDF of a particular
outcome is thus defined by PDF(x), when x is continuous varialbe
Statisticians say P(X=x)=PDF(x) but the fact is P(X=x)=PDF(x)*dx. By
calculus, dx is the smallest possible interval thus concluding the
probability to be approximately 0, which is good enough for them.
Anyway, the probability that is always 1 is the cumulative probability
of all the outcome. We call it Cumulative Density Function (CDF). As
PDf suggests, CDF is the summation of the probability of all possible
outcomes. Geometrically it is the area under the distribution curve.
Therefore the area under distribution curve needs to be ALWAYS 1 to be
reasonable.

By the way, the general form using normal scores is:
1/{sigma*sqrt(2*pi)} * exp{-(1/2)*z^2}
where z=(x-average)/sigma, 
The general form of any level of variables is:
1/{sigma*sqrt(2*pi)} * exp{-(1/2)*(x/s)^2}, then the distribution
crosses vertical axis (average=0)

And the function we know very well satisfies both property of
symmetric bell curve (that outcomes nearer to the average occur more
likely than extremities) and the cumulative probability is always 1.
In conclusion, it is no sort of inserting 1/2 conveniently together with sqrt(1/2)

Active Alex:

I beg to differ. In his question, the original poster made it very
clear that he understood that the factor in the exponent and in the
normalizing factor were connected by the fact that the integral of the
PDF must be unity.

Your general form and mine are identical and, as you can plainly see
from either one, the factor in the exponent can take on any positive
value.