# The Precision Parameter

Measuring Precision showed how a single parameter σ characterizes the precision of a shooting system.

This σ is the parameter for the Rayleigh distribution with probability density function $$\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}$$. The associated Cumulative Distribution Function gives us the probability that a shot falls within a given radius of the center:

$$Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma}$$

Therefore, we expect 39% of shots to fall within a circle of radius σ, 86% within , and 99% within .

Using the characteristics of the Rayleigh distribution we can immediately compute the three most useful precision measures:

Radial Standard Deviation $$RSD = \sigma \sqrt{2}$$. The expected sample RSD of a group of size n is

$$RSD_n = \sigma \sqrt{\frac{2}{c_{G}(n)}} \approx \sigma \sqrt{2 - \frac{1}{2n} - \frac{7}{16n^2} - \frac{19}{64n^3}}$$

Mean Radius $$MR = \sigma \sqrt{\frac{\pi}{2}}$$. The expected sample MR of a group of size n is

$$MR_n = \sigma \sqrt{\frac{\pi}{2 c_{B}(n)}}\ = \sigma \sqrt{\frac{\pi (n - 1)}{2 n}}$$

## Circular Error Probable (CEP)

Circular Error Probable $$CEP = \sigma \sqrt{\ln(4)}$$. The expected sample CEP of a group of size n is

$$CEP_n = \sigma \frac{\sqrt{\ln(4)}}{c_{G}(n) c_{R}(n)}$$ The extreme spread median from the table for 5 shots is 3.0 σ. I've determined my rifle has precision σ = ½MOA. If I take five shots at 100 yards we would expect half my groups to be less than 3.0/2 = 1.5MOA $$\approx$$ 1.6". Dividing the rest of the distribution data for that row by 2 we can also say that 90% of my groups should exhibit an extreme spread in the range (1.0, 2.1) MOA — i.e., between 1.1" and 2.2" at 100 yards.