How do you calculate the cut-off of any rank-size distribution, which splits the distribution so that the top x% categories contain 1-x% of observations? The classical example is the Pareto principle, whereby the top 20% have 80% of observations. A Pareto distribution puts x at 20%, but I want to calculate x for any given distribution.

  • $\begingroup$ Welcome to Cross Validated! Please take a moment to view our tour. It looks like you answered you question unless you have a question beyond the 80-20 rule. $\endgroup$
    – Tavrock
    Mar 11, 2017 at 5:37
  • $\begingroup$ @Tavrock I want to be able to find the x for any rank-size distribution, if that wasn't clear. It could be anything between 0% and 50%, depending on the shape of the distribution. $\endgroup$
    – syre
    Mar 11, 2017 at 8:00
  • $\begingroup$ @Tavrock Some distributions could be thick-headed, so that the cut-off would be low, e.g. "the top 2% categories contain 98% of observations". (A Pareto distribution is thick-headed.) Or a distribution could be thick-tailed, e.g. "the top 49% categories contain 51% of observations" means the distribution is almost flat. $\endgroup$
    – syre
    Mar 11, 2017 at 8:06
  • $\begingroup$ It gets confusing because you are asking about the Pareto distribution, but it sounds like you really want inverse distribution functions where you are able to find the $x$ value for a given set of parameter values (such as $\alpha$ and $\beta$ for a two-parameter Weibull or $\sigma$ and $\mu$ for a Gaussian distribution) and a desired percentile ranking value. $\endgroup$
    – Tavrock
    Mar 11, 2017 at 8:25
  • 1
    $\begingroup$ Here's a solution to get this in R. $\endgroup$
    – syre
    Mar 25, 2017 at 10:20


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