Definition of symmetric random variable [in terms of distribution function ] This is what i know about symmetric distributions:
The distribution of rv (random variable) $X$ is symmetric about $a$ iff
$$
 P ( X \le a - x ) = P ( X \ge a + x )  \qquad \forall x  \in \mathbb{R}
$$
I just want to confirm ,.,is this " iff " correct ? i mean it is clear to me that if $x$ is symmetric about $a$ it will satisfy above equation, is the converse also true ?
 A: A (real) random variable $X$ is symmetric about zero iff $X$ and $-X$ have the same distribution, often written as $\newcommand{\eqD}{\stackrel{\small{D}}{=}} X \eqD -X$.  Now if $X$ is symmetric about $a$, $X-a$ will be symmetric about 0, so we have $X-a \eqD a-X$.  If $X$ has a density, in terms of density functions this becomes $f(x) = f(-x)$, or, for symmetry about $a$, $f(a+x)=f(a-x)$ so the density function is symmetric about $a$.
Now, we can use this definition to calculate the cumulative distribution function $F(x)=P(X \le x)$ in two different ways:
\begin{align}
  P(X-a \le x) &= P(-X+a \le x) \\
  P(X \le x+a) &= P(-X \le x-a) \\
  P(X \le x+a) &= 1-P(X \le a-x)  \\
  F(x+a)  &=  1-F(a-x)
\end{align}
and this is almost what you have given, but you need to switch your second inequality sign!  And, yes, this is an equivalence.
A: Just check the skewness of the distribution. If it is 0, then the distribution will be symmetric around expected value of that random variable.
Checking with probability is not a good idea. In ideal cases, it may be equal but it totally depends over the distribution. In case of normal distribution probability checking may be correct, but it all depends. 
Skewness is the only way to check symmetry.
