Can a finite data set have all its values within $n$ standard deviations from the mean? Aside from the trivial $(x,x,x,x,...)$ data set, is it possible to have all the elements of a data set within some $n$ standard deviations from the mean? What is the minimum possible value for $n$ such that there exists a set with this property?
I'm also wondering if $n$ has any significance on the interpretation of the data itself. Are these distributions special in some way (other than simply "the numbers are all close together")?
Sorry if the answer is trivial, and thank you in advance!
EDIT: I just realized that in a finite data set, of course there exists some $n$ so that every data point is within $n$ standard deviations from the mean (because there are only finitely many Z-scores...)
The main question is the actual lower bound for this value of $n$ and its interpretation.
 A: This depends on how you define your standard deviation.
If you have a random experiment with a given standard deviation and you draw values then it is entirely possible but increasingly unlikely that all the values fall within n standard deviations.
If you calculate the standard deviation via the values you draw then it depends on the distribution of the values but there is almost$^1$ always a number n so that all values fall within n standard deviations.
I think the minimum n will be for a distribution with two extrema like {0,1} the mean being $\frac{1}{2}$ and the standard deviation being $\sqrt{\frac{1}{8}}$, so all values fall within $\sqrt{2}$ times the standard deviation. I am having a hard time to come up with a proof.
[1] The set with values that do not differ can be an exception depending on the definition of falling within, wether it is inclusive or exclusive.
A: The upper bound on how many standard deviations a data point can be in terms of distance from the mean is given by Samuelson's inequality. Assume we define the sample standard deviation for a random sample $x_1, x_2, \ldots , x_n$ to be $$s = \sqrt{{{1} \over {n-1}} \sum_{i=1}^n \left( x_i - \bar x \right) ^2}$$
Then Samuelson's inequality, adjusted for the $n-1$ in the denominator, states that $$\bar x - {{n-1} \over {\sqrt{n}}}s \leq x_i \leq \bar x + {{n-1} \over {\sqrt{n}}}s $$ for all $i.$  
More can be found on the Wikipedia entry, although they use a different definition (with $n$ in the denominator) for the standard deviation. 
A: Chebyshev's inequality says
$$P(|X-m| \geq k\sigma) \leq \frac{1}{k^2}$$
where $m$ is the mean of $X$ and $\sigma$ is the standard deviation of $X$.
This inequality is tight. This can be seen by considering random variables $X_x$ which are $0$ with probability $1-1/x^2$, $1$ with probability $1/2x^2$, and $-1$ with probability $1/2x^2$, where $x\geq 1$. So there is no universal lower bound on your number $n$. A similar argument should work for sample standard deviations.
Edit: A version for samples would be $-1,0,0,\dots,0,1$, with $N-2$ copies of $0$ (so $N$ points in total). Then the sample standard deviation is $\sqrt{\frac{2}{N-1}}$, so all the data points are within at most $\sqrt{\frac{N-1}{2}}$ standard deviations of the mean. But this is not quite optimal (Samuelson's inequality is optimal).
