Firstly, the Daigian issue is a sound one. The fact that Daigian clearly delineates the limit between the bottom 37.5% of channelers and the top 62.5% is very useful.

It tells us a lot about the standard deviation. We know that about 68% of channelers fall within 1 standard deviation from the Mean. At the same time, we know that only 25% of channelers lie within Daigian's distance from the Mean (above and below).

Meaning that Daigian lies well within 1 standard deviation from the Mean. In fact, she lies about 0.32 standard deviations from the Mean, based on RJ's stats.

Right, so Daigian is Marker A, at 0.32 standard deviations from the Mean. Remember this, as I'll come back to it in the concluding formula.

Next, we also know that Moiraine is at least 3 times as strong as Daigian, and that Egwene at full strength is just shy of twice Moiraine's strength. Meaning that Egwene is likely around 5 times Daigian's strength.

Based on the evidence (the known channeling sample, which is around 5000 female channelers), there seems to be about a 1 in 500 chance of a channeler of Egwene's strength occurring. I calculate this as follows:

Out of the approximately 5000 (roughly) known female channelers, about 10 modern day channelers are as strong or stronger than Egwene. Off the top of my head, they are

Sharina
Someryn
Tamela
Viendre,
Talaan
Aviendha
Elayne
and one or two novices with Egwene level potential.

About 10 people in total, give or take a few.

That means there will be about 20 channelers out of 5000 that are further than Egwene from the Mean, on both the lower and upper side. From a population of 5000, that's 1 in 250, approximately.

For the purposes of this discussion, I am happy to assign to her the 1 in 370 rarity which equates to 3 standard deviations from the Mean (using Sidious's quick SD quide in the post above).

So Egwene is Marker B, which is 3 Standard Deviations from the Mean.

Marker A and B allow us to unravel the entire strength model.

If we assign a random value to Daigian, we can determine the strength of Egwene, the magnitude of the standard deviation, and the Mean itself, all by simple extrapolation.

That means we've cracked the entire code! Here it is, in all its beautiful simplicity.

I'll just present the numbers that were the end result of lots of iterations to get it to fit a 100 point scale. But you can use any number in this model and come to the same relative strengths on any scale you choose. Here it is:

If Daigian is a hypothetical 12, then Egwene is approximately a 60. (5 times stronger than Daigian)

Now, since we know Daigian lies 0.32 Standard Deviations below the Mean, and Egwene lies 3 standard deviations above the Mean, that means the absolute distance between them, which is 60-12 = 48, equates to 3.32 standard deviations.

This means that the Standard Deviation of the One Power strength Curve, is 48/3.32 = 14.46 (on a 100 point scale).

Since Daigian is 0.32 Standard Deviations below the Mean, we can therefore determine that the Mean is at 16.6, using Daigians assumed starting point of 12.

Now that we have the Mean and the Standard Deviation, everything else falls into place.

Since Egwene is 3 Standard Deviations above the Mean, Egwene's strength is 16.6 + 3 Standard Deviations, which ties back to the 60 strength we had before.

Then it is simple to calculate the strengths of persons 4, 5 and 6 standard deviations above the Mean.

3 SD (Egwene) = 60.

4 SD (Moghedien) = 60+14.5 = 74.5
5 SD (Graendal) = 74.5+14.5 = 89
6 SD (Lanfear) = 89+14.5 = 103.

Meaning Lanfear is just short of 6 Standard Deviations from the Mean.

All of the above are statistically irrefutable based on 3 simple assumptions:

1 - 37.5% of channelers lie below Daigian on the Curve. (Based on RJ's own quote)

2. Egwene is approximately 5 times as strong as Daigian. (Based on evidence from the books)

3. Egwene's rarity among the modern day channeler population is approximately 1 in 370, equating to 3 Standard Deviations. (Based on the known modern day channelers in the Books).

There. The code has been cracked.

Small permutations are of course a given, but the overall methodology seems entirely sound in my view.

Please feel free to critique it, because this may just be the solution to it all, if we've got it right.