Help reconcile probability density functions and "probability" I have some confusion stemming from the typical stumbling block that 

"the probability that a continuous random variable takes on a specific
  value is 0.".

If we take, say, the standard normal distribution, then the pdf is $p(x) = e^{-x^2/2} / \sqrt{2 \pi}$. Now, by the above quote, the probability that a standard normal random variable takes on its mean ($0$) is $0$. Yet $p(0) = 1.$ So what exactly does this number represent?
Is there an intuitive way of understand the values that a pdf takes on?

You should think of the probability $P$ that a continuous random variable $X$ takes on a specific value $a$ as finding 
$$
P(X=a)= P(a\leq X\leq a) = \int_a^a p(x)dx, 
$$
which always equals $0$ for any probability density function $p$. 
On the other hand, a probability density function, through the integral, specifies the probability of the continuous random variable falling within a particular $\textit{range}$ of values (along the $x$-axis in the one variable setting). 
