You cannot use the distance of an observation from a classical fit of your data to reliably detect outliers because the fitting procedure you use is itself liable to being pulled towards the outliers (this is called the masking effect). One simple way to reliably detect outliers is to use the general idea you suggested
(distance from fit) but replacing the classical estimators by robust ones much less susceptible to be swayed by outliers. Below I present a general illustration of the idea and then discuss the solution for your specific problem.
An illustration: consider the following 20 observations
drawn from a $\mathcal{N}(0,1)$ (rounded to the second
digit):
x<-c(-2.21,-1.84,-.95,-.91,-.36,-.19,-.11,-.1,.18,
.3,.31,.43,.51,.64,.67,.72,1.22,1.35,8.1,17.6)
(the last two really ought to be .81 and 1.76 but have
been accidentally misstyped).
Using a outlier detection rule based on comparing the statistic
$$\frac{|x_i-\text{ave}(x_i)|}{\text{sd}(x_i)}$$
to the quantiles of a normal distribution would never
lead you to suspect that 8.1 is an outlier, leading you
to estimate the $\text{sd}$ of the 'trimmed' series to be
2 (for comparison the raw, e.g. untrimmed, estimate of
$\text{sd}$ is 4.35).
Had you used a robust statistic instead:
$$\frac{|x_i-\text{med}(x_i)|}{\text{mad}(x_i)}$$
and comparing the resulting robust $z$-scores to the
quantiles of a normal, you would have correctly flagged
the last two observations as outliers (and correctly
estimated the $\text{sd}$ of the trimmed series to be
0.96).
(in the interest of completeness I should point out that
some people, even in this age and day, prefer to cling the raw --untrimmed--
estimate of 4.35 rather than using the more precise estimate based on trimming but this is unintelligible to me)
For other distributions the situation is not that different,
merely that you will have to pre-transform your data first.
For example, in your case:
Suppose $X$ is your original count data.
One trick is to use the transformation:
$$Y=2\sqrt{X}$$
and to exclude an observation as outlier
if $Y>\text{med}(Y)+3$ (this rule is not symmetric
and I for one would be very cautions
about excluding observations from the left 'tail' of
a count variable according to a data based threshold.
Negative observations, Obviously, should
be pretty safe to remove)
This is based on the idea that if $X$ is poisson,
then
$$Y\approx \mathcal{N}(\text{med}(Y),1)$$
This approximation works reasonably well for poisson
distributed data when $\lambda$ (the parameter of the
poisson distribution) is larger than 3.
When $\lambda$ is smaller than 3 (or when the model
governing the distribution of the majority of the data
has a mode closer to 0 than a poisson $\lambda=3$, as
in i.e. ZINB r.v.'s) the approximation tends to err on
the conservative side (reject fewer data as outliers).
To see why this is considered 'conservative' consider
that at the limit (when the data is binomial with very
small $p$) no observation would ever be flagged as outlier
by this rule and this is precisely the behaviour we want:
to cause masking, outliers have to be able to drive the
estimated parameters arbitrarily far away from their true
values. When the data is drawn from a distribution with
bounded support (such as the binomial), this can simply
not happen...
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.)
In contrast it's always true for an exponential that the mean and the SD are equal, although the converse doesn't follow. (For example, a Poisson with mean 1 also has SD 1 but is not an exponential.)
I can't say that a decision on normal or exponential is ever one that statistical people face seriously. A normal distribution is always symmetric and an exponential distribution always skewed and bounded. It should be apparent from simple graphs that one or other distribution may be plausible, but not both. Perhaps their simplest meeting place is the gamma family of distributions as the exponential and normal are both members, the latter as a limiting case.
Say what should be said about Wikipedia, but I've found its articles on individual probability distributions informative and well illustrated.
See e.g. the entry on the exponential
Your example is one where time-to-process is always positive, so normal distributions are strictly impossible. I'd expect marked skewness in many situations but approximate symmetry is also possible, e.g. when serving up burgers or coffee. But then necessarily the mean must be larger than the SD for symmetry to be possible, exactly or approximately.
Best Answer
If you have data from a Poisson process with parameter $\lambda$, then the inter-arrival times $$ X_1, \ldots, X_n \overset{iid}{\sim} \text{Exponential}(\lambda). $$ You get these data points by just taking the difference of the column of time stamps. Then you can just look at a qqplot (check this thread for R code.)
Another idea: bin the data into buckets. Then run a Poisson regression on a few polynomials in time. If any of the significance tests reject the null, you can't assume that it's a (homogeneous) Poisson process.