RE: st: Estimate 95th percentile of a highly skewed distribution from summary stats?

["Feiveson, Alan H. (JSC-SK311)" <alan.h.feiveson@nasa.gov>]

Jordan - 

I suggest separating the distribution into the discrete part (zeros) plus a continuous part (the rest). If the fraction of zeros is p, then the 95th percentile of the whole distribution would be the 100(.95 - p)-th percentile of the continuous part. Perhaps you can model the continuous part with a gamma or other well-known skewed distribution.


Al Feiveson


-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Nick Cox
Sent: Wednesday, August 07, 2013 6:42 AM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: Estimate 95th percentile of a highly skewed distribution from summary stats?

Evidently the point of this is that _you know nothing except what you have shown us_.

Otherwise -summarize, detail- would do this for you, without even needing to look up the more specialised commands it cross-references.

The flip answer is somewhere in [7, 50]. I don't know how to improve on that flip answer without making very strong assumptions about the distribution. You don't even know what the fraction of zeros is.

If this were my problem I would use the maximum, knowing that it is biased.
Nick
njcoxstata@gmail.com


On 7 August 2013 12:23, Jordan Silberman <silberman.stata@gmail.com> wrote:
> Hi Stata folks,
>
> I have some (but not all) descriptive stats for a highly skewed, 
> zero-inflated distribution. I need to estimate the 95th percentile 
> score for this distribution.
>
> Here's the info I have:
>
> min value: 0
> max: 50
> mean: 5
> median: 0
> 75th percentile: 7
>
> Obviously the distribution is highly skewed to the right. Does anyone 
> know of a method through which I can estimate the 95th percentile 
> score for this distribution?
>
> The background is that this is a distribution of estimation errors for 
> an algorithm, and I need an estimate of a value that the error is 
> equal to or less than in 95% of cases.
>
> Thanks,
> Jordan
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