But it isn't that hard to prove for yourself for the simple case of the binomial curve if you're willing to supply some elbow grease. Here is an outline of how I did it when I was an undergrad.
1. Prove that log(n!) = n log(n) - log(n)/2 + k + o(1) where k is some constant. (Trick: approximate the integral of log(x) with the trapezoidal rule, that gives you n log(n). You wind up with log(n)/2 + log(1)/2 left over. Prove that the sum of the errors to infinity converges to a constant, which gives you, to finite n, k + o(1) where the o(1) term is the sum of the rest of the errors that you haven't reached yet.)
2. Suppose that X_1, X_2, etc are an identically distributed independent set of random variables that have probability 0.5 of being +1 or -1. Prove that each X_i has mean 0 and variance 1.
3. Consider the random variable Y_n = (X_1 + X_2 + ... + X_n)/sqrt(n). Prove that it has mean 0 and variance 1.
4. What is the "density" (in quotes because it is discrete) of Y_n around x? Well you can calculate an exact formula in terms of the binomial formula, and stick in your approximation from step 1, then show that the point weight of the nearest point to x is going to be e^(-x^2)(c + o(1))/sqrt(n) where c is a constant that can be derived from k in step 1. Remember that that's the only point for a distance of sqrt(n), so the density is proportional to e^(-x^2).
5. Look up the standard argument that the integral from -infinity to infinity of e^(-x^2) is sqrt(pi). Note that Y has to sum up to a total weight of 1 (because it is a probability density), substitute that fact in then you get both the standard normal distribution out AND as an unexpected bonus, Stirling's formula!
(If you wish you can do the same for a biased coin, the derivation is slightly more complex but follows the same lines. However to get arbitrary probability distributions you need to use different approaches. People usually don't encounter that proof until graduate school.)
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The derivation does require one trick that is usually not taught in high school. You know how you can maximize a function by setting its derivative to zero? Well, there is a generalization of that to functions of multiple variables: you set the gradient to zero. But there is yet another generalization to that for functions of continuously infinite number of variables! That's called the calculus of variations. If we want to maximize the entropy over all possible probability distributions over the real numbers, then we have a continuously infinite number of variables, since for each real number x, we need to assign a probability mass p(x). Here is the derivation for those interested:
Let m be the mean we want and v be the variance we want, and let p(x) be the probability density function.
We want to maximize the entropy S = int(p(x)log(p(x))) given the constraints that:
int(p(x)) = 1 // total probability should sum to 1
E(X) = int(x*p(x)) = m // mean should be m
E((X-m)^2) = int((x-m)^2*p(x)) = v // variance should be v
This problem can be solved with calculus of variations with Lagrange multipliers. The Lagrangian is all the terms inside the integrals summed up: L = p*log(p) + a*p + b*x*p + c*(x-m)^2*p
Calculus of variations tells us that for the optimal p, we have: dL/dp = 0
This is the analogue of setting the derivative to zero for maximizing a function.Calculating that we get:
a + bx + c(x-m)^2 + log p + 1 = 0
rewriting: p = e^-(a + bx + c(x-m)^2 + 1)
We have the general form of the Bell curve. Now we just need to find the right values for a,b,c to satisfy the 3 constraints.As special cases of this we have:
1) The probability distribution with maximum entropy without constraints is the uniform distribution.
2) The probability distribution with maximum entropy with only the constraint that the mean is m, is the exponential distribution.
In general if you want to find maximum entropy distribution with known expectation values E(f(X)) and E(g(X)), then that is of this form:
p(x) = C*e^(a*f(x) + b*g(x))
This gives an indication why so many distributions have this form. For example the entire exponential family can also be derived as maximum entropy distributions with certain known expectation values.Of course this derivation is only a good motivation for the bell curve if you already cared about entropy, mean and variance.
The reason that _so many_ things have normal distributions is because of the Central Limit Theorem. If there's any reason to study math stats (other than understand the underpinnings of statistics at a deeper level than most of the population), it's that you get to become really familiar with how amazing this Theorem is.
The sum of die rolls shows a good example of why you get the bell shape: there are more ways (combinations) with which to sum up to the center values than the extremal values.
For the majority of the people I had in the intro statistics courses I TA'ed for, this was enough explanation. In undergrad, we didn't prove Lindeberg-Levy CLT until my third quarter of math stats, and the versions with weaker assumptions we didn't prove until graduate math stats.
When you're struggling to understand the fundamentals of hypothesis testing and simple combinatorics, going through the rigorous process of proving the CLT is daunting, mind-boggling, and largely unnecessary :).
It is slightly easier if you start off assuming Stirling's formula. However there is no need to assume that - you can derive it at the same time. Elsewhere in this thread I gave an outline of how to derive this simple case.
http://intromath.ca/aakkozzll/
The trick is to keep in mind that in this model, each ball has a 50/50 chance of either going left or right when it crosses each peg.
This is analogous to a game where you flip a coin 9 times in a row and win a dollar for every heads, lose a dollar for every tails. On average, you come out even, but the outcome of any particular game is like one of the balls falling down that Pachinko board. You might get really lucky and get 9 heads in a row, but usually there's a mix of heads and tails, and if you look at all the possible results, it would stack up like those balls in the pachinko board.
To give an easy to visualize example, a teacher that grades on the curve, does so by normalizing the data (taking the top 10% of the grades for a test and making those A's regardless of previous score, then making the top 10%-20% B's regardless of score and so on. There are different methods to calculate the percentiles (and what those percentiles are could vary by grading system), but that's just the most common one.
The bell curve and Gaussian curve (and normal curve) are almost always synonymous. Bell Curve describes its shape, while the Gaussian curve describes the guy who derived that nasty-looking pdf. While some other distributions may look bell-shaped (especially ones that converge to a Normal in some asymptotic sense, like the t-distribution), people will look at you funny if you refer to anything other than the normal (or in rare cases, the t) as "The Bell Curve".
The point about "the data is laid out so the sum of the data adds up to 1" doesn't make much sense since the Normal takes values on the entire real line. Could you clarify?
The procedure you describe is not what most people would call "normalizing", so much as breaking the data up into quantiles/percentiles. Normalizing the data would be for the instructor will subtract the average from everyone's raw score, then divide by the standard deviation. If the raw scores themselves were distributed normally, then these new "normalized" scores are now standard normal, so letter grades are given based on how far away from 0 a student's scaled score is. This procedure is pretty flawed since scores are usually not super close to normality (many exams are not symmetric in scores, the distribution may have several modes, etc). The procedure you describe makes no assumptions about the underlying data, and does not require normality, so I like it more.
Thanks for the correction. My method would be more fair, but is incorrect when applied to standard normalization. It should be percentiles. In a strict bell curve, if the top 20% get A's, then the bottom 20% would also have to fail (even if they didn't really get an F) so a pretty unfair system as you mentioned as grades are already predetermined :(
Your method is incorrect as a method of normalization, but it seems we are in agreement that perhaps normalization itself is a silly method for this. Most instructors I've known that claimed they were "curving" were in fact using the percentile approach. Grading based on percentiles (top 10% get A's, next 20% get B's, 40% C's, 10% D, bottom 20% F, for example) allows you far more flexibility, as the percentiles don't even need to be symmetric. My grad courses usually only had three grades: A+, A, and "Please see me to discuss your future in the department".
Your background (being just shy of a PhD in Math), should have lots of useful information you can share.