Average, standard deviation and min/max values I'm analzying a computer science paper and just found in the experimental setup the following statement:


*

*Average (standard deviation) of number of files per peer: 464 (554)

*Min - max number of files per peer: 100 - 4,774


Are these numbers possible at all? It does not say anything of a normal distribution, but how is it possible that the standard deviation is 554, but the min number of files per peer is 100?
 A: Suppose we have $n$ numbers $x_1, x_2, \ldots, x_n$ ranging from $a$ to $b$, meaning
that $x_i = a$ for at least one $i$, $1 \leq i \leq n$, and $x_j = b$ for at least
one $j$, $1 \leq j \leq n$. Duplicates are allowed, meaning that two or more of the
$n$ numbers could possibly have the same value in $[a, b]$.  Define the mean $\bar{x}$ and the variance $\sigma_x^2$ of the set of $n$ numbers as
$$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i, ~~ 
\sigma_x^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2 
= \left(\frac{1}{n}\sum_{i=1}^n x_i^2\right) - \bar{x}^2.$$
For convenience, let us change the set to $y_1, y_2, \ldots, y_n$ where
$y_i = x_i - a$, so that the new set has values ranging from $0$ to $b-a$.
The mean $\bar{y}$ is just $\bar{x}-a$, while the variance is unchanged:
$\sigma_y^2 = \sigma_x^2$.  Now, it is shown in the answers to 
this question on stats.SE 
that the ratio 
$\sigma_y/\bar{y} = \sigma_x/(\bar{x}-a)$ can be no larger than $\sqrt{n-1}$.
Note that the value does not depend on $b$ at all
You don't say what the value of $n$ is, but given the numbers in your
answer, the upper bound on the standard deviation is
$$\sigma_x = 554 \leq (464-100)\sqrt{n-1}$$
which is certainly satisfied except in the unusual circumstance
that there are only $3$ peers in the experiment and thus
only $3$ numbers $x_1$, $x_2$, and $x_3$ are being described
in terms of mean/standard-deviation/min-max: certainly overkill!
In summary, there are no obvious problems with the standard deviation
being larger than the mean.
