Estimating R90 (radius of a circle expected to include 90% of impacts) I want to determine how big a target I can hit with a bow at a certain distance with 90% probability. I place some paper targets at that distance and shoot 20 arrows at them. I have a ruler and a pencil so I can measure distances on paper and do simple arithmetic, but not something complex like MLE.
Impact coordinates mostly follow bivariate normal distribution centered at the target, but sometimes there is some bias, vertical and horizontal variances might not be exactly the same, and occasionally there is an outlier (flier).
This analysis (source code) recommends L-estimators based on 9th miss radius in a 10 shot group, or sum of 6th and 9th miss radiuses in a 10 shot group. Does this approach make sense? Are there better alternatives?
Proposed L-estimators:


*

*Measure 9th miss radius R9:10 in each of the two 10 shot groups, take the average, multiply by 1.15 to get estimate of R90.

*In a 10 shot group, add 6th miss radius R6:10 and 9th miss radius R9:10. Do it again with the second group and take the average. Multiply by 0.69 to get estimate of R90.
 A: If you're satisfied that a symmetric bivariate normal is an appropriate model (as are most marksmen), then there is a closed-form solution for this using the Rayleigh distribution.
When the distribution is parameterized by σ then the Rayleigh CDF gives us $Pr(r \leq \alpha) = 1 - e^{-\alpha^2 / 2 \sigma^2}$.  This probability is more commonly known as a Circular Error Probable (CEP).
You're looking for the 90% CEP, which is thus $\sigma \sqrt{-2 \ln(90\%)}\approx 2.15\sigma$.
The estimator for the Rayleigh parameter σ is not terribly complex if you can measure your sample impact coordinates and have basic spreadsheet functions.  The unbiased estimator for n sample shots is computed as follows:


*

*Calculate the sample center $(\bar{x}, \bar{y})$.

*For each sample shot calculate the sample radius $r_i = \sqrt{(x_i - \bar{x})^2 + (y_i - \bar{y})^2}$.

*The unbiased Rayleigh estimator is $\widehat{\sigma_R^2} = \frac{n}{n-1} \frac{\sum r_i^2}{2n}$.

*Use the same Gaussian correction factor to remove the bias when taking the square root of $\widehat{\sigma_R^2}$ as when taking the square root of variance.  I.e., ${}^1/c_{G}(n) = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$.  Or, in code or spreadsheet formula, $c_{G}(n)=$ 1/EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2)).

*Finally, the unbiased parameter estimate is $\hat{\sigma} = c_G(2n-1) \sqrt{\widehat{\sigma_R^2}}$.

