# Assessing deviation from the mean without Chebyshev's inequality

I have a set of experimental data which comprises distances between successive points: And a simulation which attempts to recapture the experimental data by choosing points at random and calculating the distances between them. This is repeated many times as to create a population of simulations. The red line below shows the simulated population average: x-axis = No. of Distances (plotted in increasing size)
y-axis (logarithmic) = Distance size


Neither the experimental data nor the simulated data fits a normal distribution (as assessed by the one-sample Kolmogorov–Smirnov test). Given the nature of the data i.e. distances between successive points and visual inspection of the data, a gamma-distribution appears to best fit the data.

I need to be able to calculate by how much any given simulation can deviate from the mean such that I can place two lines either side of the population average and show whether or not the experimental data (blue line) falls within this range. This is largely because the experimental data is NOT population based and represents a single sample.

As the 68–95–99.7 rule does not apply here - is there any more precise method than using Chebyshev's inequality and -/+ 2-3SD? This produces a rather wide band that makes it difficult to assess changes in the simulation parameters.

EDIT: Starting with formula for the statistic used in KS test $$D_n = \sup_x | F_n(x) - F(x) |$$ and knowing that the goodness of fit is built using critical values of Kolmogorov distribution you have $$\sqrt{n}D_n > K_\alpha$$ Now by choosing your alpha you are able to determine $K_\alpha$ such that $$P(K<K_\alpha) = 1-\alpha$$ So you can determine some quantity $D_\alpha$ such that $P(D_n>D_\alpha)=\alpha$ and use that as a width around the graph of $F(x)$.