Distribution with all data further than one standard deviation from the mean Chebyshev inequality states that the lower bound for the percentage of data in the interval $[\mu-\sigma , \mu+\sigma]$ is $0\%$. Can someone provide a distribution satisfying this bound or a sequence of distribution that gets arbitrary close?
We tried coming up with such distribution but everytime we try putting more data outside the interval, the standard deviation rises and renders our attempt vain.
EDIT : Following Marc-Élie's answer, can we find, for every $100\%>\epsilon>0\%$ a distribution such that at most $\epsilon$ of the values are in the interval?
 A: We want $\dfrac{|x_i-\mu|}{\sigma}>1$ for all $x_i$. From this inequation follows that $(x_i-\mu)^2>\sigma^2$ for all $x_i$. 
We also have (for the discrete case) that $\sigma^2=\sum (x_i-\mu)^2/n\ge n (\min\{(x_i-\mu)^2\})/n=\min\{(x_i-\mu)^2\}$, where the equality is valid only when every $x_i$ are at the same distance from the mean.
Joining those two inequations we obtain :
$$(x_i-\mu)^2>\sigma^2\ge \min\{(x_i-\mu)^2\}$$
Where we want the left inequation to be true for every $x_i$. We obtain from the right inequation that this cannot be possible since $\sigma^2$ is at least bigger or equal than one $(x_i-\mu)^2$.
So it seems that it is impossible to get 0% of the distribution in $[\mu-\sigma,\mu+\sigma]$ in the discrete case. Note that it would be possible to get 0% in $]\mu-\sigma,\mu+\sigma[$. 
For the continuous case, I am not sure it is possible to have 0% in the wanted interval, but we can get arbitrarily close to it. 
If we take the beta distribution and we set the two parameter equal ($\alpha=\beta$) to each other (for symmetry). I'm pretty sure that we have : $P(\mu-\sigma<X<\mu+\sigma)\underset{\alpha\rightarrow 0}{\rightarrow}0$.
I do not have a formal proof, but if we look at $P(\mu-\sigma<X<\mu+\sigma)$ as a function of $\alpha$ we have : 

where the simulation where run in R.
