# If the mean is equal to the standard deviation, what is the general likelihood that the underlying distribution is normal vs exponential?

This comes from operations analysis. Often you have measurement of times, such as time between customers arriving at a store (e.g. Starbucks) or time it takes to process an order (e.g. a cappuccino).

Say you have two such variables X and Y, with X being time between customer arrival and Y being time required to process an order. Usually X is exponentially distributed (customers-per-hour would be the related Poisson distribution). In some processes, Y (time-to-process) is sometimes normal and sometimes exponential.

My question is this: if I observe a variable Y that intuitively could be normal or exponential, and I find that the mean equals the standard deviation (a characteristic of exponential distributions), can/should I assume that the distribution is exponential and not normal? Would either assumption work?

TL;DR More generally, if I have a variable for which the mean and standard deviation are equal, what is the general likelihood that the distribution is normal vs exponential? (I don't have actual data. This is more conceptual.)

[Edited: Replaced "probability" with "general likelihood" to be more specific.]

[Edited again: This question was motivated by an over-simplified textbook question that sparked my own conceptual question (being more deeply ingrained and interested in statistics than the class required). It has clarified the absurdity of the over-simplification. More precisely: any thing in operations measured as process time is not normally distributed. Exponential and lognormal distributions are often good candidates. This is true because you cannot have negative process time. However a normal approximation is often used because it is better understood and easy to teach. If I wrote the question again, I'd ask for examples where a normal approximation could result in a substantively bad analysis or recommendation. Thanks to all who helped!]

• If you have a normal variable with mean = standard deviation, then it has a 16% probability of being negative (i.e. $\Phi[-1]=0.16$, see here). So a non-negative variable can only be truncated normal at best. – GeoMatt22 Oct 14 '16 at 6:58
• (There are of course many non-negative PDFs which could work besides exponential.) – GeoMatt22 Oct 14 '16 at 7:02
• The question does not make sense, as there is no intrinsic probability distribution on the space of all probability distributions. And you mention observations in the same sentence as mean and standard deviation, which makes me wonder whether or not you mean estimates of both mean and standard deviation. – Xi'an Oct 14 '16 at 9:03
• @Xi'an You are clearly right that "probability" in the title cannot be taken literally here. I've interpreted the question is being at most "How likely is it..." but it's for the OP to reword. – Nick Cox Oct 14 '16 at 18:26
• For the title, you might include "non-negative variable", as that seems the focus. And you might re-word more like "what options besides exponential?", because as Xi'an notes it is really not possible to ask about "probability", and I would say "general likelihood" has the same issue. (It is a qualitative issue, not a quantitative one.) – GeoMatt22 Oct 14 '16 at 20:38

Any talk here of probability must at best be informal without a precise idea of what set-up you are notionally sampling from. But it's clear that a normal with mean and SD equal must have both positive and negative values, as a large fraction of data must be below mean $-$ SD, which equals zero. Distributions like that are possible but fairly unusual in my experience. Distributions of residuals from some model are examples of distributions with both positive and negative values, but their mean and SD being equal would be very unusual and -- whenever mean residuals are necessarily zero, as with some fitting procedures -- mean certainly cannot equal SD, unless trivially all residuals are zero. (Any set of values which are deviations from their mean is a case in point.)