$P(X > \mu + \sigma) > 0$? I'm doing some data analysis. Let $X$ follow an unknown distribution, with a certain average $E[X]$ and a standard deviation $\sigma$. Both values are calculated using the (corrected) point estimators.
The data is not normal, and let's assume that the underlying distribution is continuous. Now I want to know if we can say anything about $P(X > E[X] + \sigma)$. My intuition says that this is always strictly positive. But I don't know if this is true. The name standard deviation and the formula make me think that this should be possible. So, what I actually want to know is: the fact that I found a standard deviation $\sigma$, does that imply I have a value $d \ge E[X] + \sigma$ in my dataset?
More generally, is it possible to say anything about $P(X > E[X] + k\sigma)$, with $k > 0$, being greater than some value?
(Wikipedia has a list of about 50 inequalities about probability... I checked Chebyshev, but the sign in the inequality is in the wrong way for my question.)

Suppose you have a Bernoulli random variable where $X=1$ with probability $p$ and $X=0$ with probability $1-p$.  Then $E[X]=p$ and $\sigma_X= \sqrt{p(1-p)}$. 
For any $k \gt 0$, if $p=\frac{1}{k^2+1}$ then $E[X]=\frac{1}{k^2+1}$ and $\sigma_X= \frac{k}{k^2+1}$ so $E[X]+k\sigma_X=1$ meaning that in this example $P(X\gt E[X]+k\sigma_X)=0$.  
Strictly speaking this is a discrete random variable, but any discrete distribution can be approximated arbitrarily closely by the distribution of a continuous random variable. So the answer is no, unless you have more information about the distribution of your $X$.   
