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?
 A: 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). 
A: A probability density function (PDF) is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value. This probability is given by the integral of this variable’s PDF over that range — that is, it is given by the area under the density function but above the horizontal axis and between the lowest and greatest values of the range. The graph below is a graph of your PDF for the normal distribution (You simplified it by assuming you have a mean (μ) of 0 and a standard deviation (σ) of 1). Notice that p(0) is about .40. 

The graph above does not show you the probability of events but their probability density. To get the probability of an event within a given range you will need to integrate. You have an example that the probability that a standard normal random variable takes on its mean (0) is  0. This can be shown with the integral below. 

You can also get the probability of something falling in a given range. For example, if you want to find the probability of something falling within 1 standard deviation (from -1 to 1) you can integrate from -1 to 1. 

You can learn more here if you are curious. 
A: I would argue that if you are after intuition, you should change your perspective as there is a more natural way of looking at continuous probability distributions. 
I hope you will agree that in any practical situation, one is more interested in obtaining a reasonably precise estimate of the outcome of an experiment than an exact value. For example, I don't really care if someone was on hold for $15.004$ minutes- you can just tell me that they were waiting for $15$ minutes and I understand this to mean that they were waiting for a time within some interval about $15$ minutes.
Hence we are more interested in assigning a probability to the event that the RV fell within a specific interval rather than an exact point. Formally, we need to define a probability measure on the reals which is defined for all intervals. If you examine such probability measures you will find that they all take the form $[a,b] \mapsto \int_a^b f $ (on intervals) for some function $f$ which is your probability density. 
To summarise, there are more fundamental reasons why continuous probabilities on the reals should be given by integrals. By the fundamental theorem of calculus, you can differentiate integrals so we may as well give a name to the derivative of the probability measure.
I doubt this is the last word on the subject but this is a viewpoint which generalises to the measure theoretic approach to probability,
