How is the distribution of random variable a random variable? I am having trouble understanding this notion. How is a distribution function of a random variable a random variable, because we already have the distribution function determined. Also, more importantly, if $X$ is a random variable, how would we find $D(X)$ where $D(a)$ is $Pr (X\leq a)$. I am unable to wrap my mind around this. 
 A: One case in which the distribution of a random variable is itself random is this: First choose a number $p$ between $0$ and $1$ randomly---say with a uniform distribution.  Then toss a biased coin with probability $p$ of heads.  The conditional probability distribution of the outcome of the coin toss is itself chosen randomly.
Your other question is quite a different thing.
Now suppose $X$ has a standard normal distribution.  It is conventional to denote the cumulative probability distribution function for this distribution by the capital Greek letter $\Phi$ (and the density by the lower-case $\varphi$).  So we have
$$
\Pr(X \le x) = \Phi(x),
$$
where lower-case $x$ can be any real number at all.  If one chooses $X$ randomly from that distribution and then evaluates $\Phi(X)$, then $\Phi(X)$ is itself a random variable.  Notice that $\Phi(X)$ is always between $0$ and $1$.  And for (lower-case) $x$ between $0$ and $1$ we have (since $\Phi$ is a strictly increasing function)
$$
\Pr(\Phi(X) \le x) = \Pr (X \le \Phi^{-1}(x)) = \Phi\Big( \Phi^{-1}(x)\Big) = x.
$$
Therefore $\Phi(X)$ is actually uniformly distributed between $0$ and $1$.
