How to equate hit probabilties on 2 different surfaces

Question

I'm looking for help in determining how to equate the following:

1. We have a surface of 13 cm X 10 cm.
2. We have a 95% probability of hit on a 16 cm diameter surface.

I would like to equate $B$, probability of hit to the 13cm X 10 cm surface. 95% probability of hit on a circle of 16 cm represent what probability of hit on the rectangle of 13cm X 10cm.

What is the mathematical steps to arrive to the solution? It is a job related problem not a school one.

Answer 1

If we assume the distribution of hits is binormal, then the distribution of the distance to the center of the target is a scaled Chi distribution with two degrees of freedom. From this (applying its inverse CDF), we find that its 95th percentile coincides with the radius of 16/2 = 8 centimeters when the scaling factor equals 1.33523. This factor is also the standard deviation of the components of the binormal distribution. Integrating the PDF of that binormal distribution over a 13 by 10 cm rectangle centered at the point of aim gives 0.83311, the desired value of $B$.

Here's a picture showing a shaded contour plot of the PDF restricted to that rectangle, with the circular target behind it for reference:

The value of 0.83311 was found with Mathematica:

With[{s = 8 / InverseCDF[ChiDistribution[2], 0.95]}, 
 NIntegrate[PDF[BinormalDistribution[{0,0},{s,s},0],{x,y}], {x,-13/2,13/2}, {y,-10/2,10/2}]
]

It was checked by simulating 100,000 independent shots and reporting the proportions that (a) fell within the 16 cm circular target and (b) fell within the rectangular target:

With[{s = 8 / InverseCDF[g, 0.95], n = 100000}, 
 data = RandomReal[BinormalDistribution[{0, 0}, {s, s}, 0], n];
 old = Length[Select[data, Norm[#] <= 8 &]] / n;
 new = Length[Select[data, Abs[#[[1]]] <= 13/2 && Abs[#[[2]]] <= 10/2 &]] / n;
 {old, new} // N
]

The output of {0.94919, 0.83331} is close enough to the intended values of {0.95, 0.83311} to confirm the correctness of the calculations.