Equivalent of z-score in exponential distribution I have an exponentially distributed set of data, but there are a number of outliers at the margin. If I want to identify samples above the mean that I am 99.9+% confident are not the result of sampling error, how do I do this? In a normal distribution I can, of course, calculate the number of standard deviations and then find the number of SDs above mean required to meet my confidence criteria. Is there a similar process for exponential distributions?
 A: 
If I want to identify samples above the mean that I am 99.9+% confident are not the result of sampling error, how do I do this?

Unless you have some statistical knowledge about the source of these outliers, you can't claim this. To simplify a little bit, outliers filtering just getting rid of the most extreme data points. 
In your case you want to get rid of the 0.1% most extreme data points. To answer this question:


*

*Find $X_0$ such as $P(X>X_0)=0.1%$

*Discard every data point superior to $X_0$. 


You do not need any table to compute $X_0$ as the cumulative distribution function of the exponential distribution is computable.
Let be λ the parameter of your exponential distribution
$P(x>X_0)=e^{−λX_0}=0.1$ implies that $X_0=\frac{-ln(0.1)}{λ}$
To get back to your misconception (or just vocabulary slippery, who knows). Let's say you have N=10,000. If there are no real outliers, statistically speaking you will still get rid of something around 10 data points (which are 100% not outliers).
