I don't understand normal probability distribution... So I'm on the Wiki page for normal distribution.
Let's say you have a standard normal distribution $X \sim \mathcal{N}(0,1)$ and you do the probability density plot so that when $X = 0$, $y$ is 0.4.
What does that 0.4 mean?  Does that mean there's a 40% chance that the value is exactly 0?  That doesn't seem right.  I don't get how to interpret the $Y$-axis. 
 A: Here is your plot:

Correct to four places, the exact value of the density function at $z = 0$ is $\varphi(0) = \frac{1}{\sqrt{2\pi}} = 0.3989.$ (Exact computation in R statistical software, in case you're interested.)
dnorm(0)
## 0.3989423
1/sqrt(2*pi)
## 0.3989423

Areas under the density curve, correspond to probabilities. Suppose I take
a narrow interval around $0,$ say $(-0.1, 0.1),$ which is of width $0.2.$
Then $P(-.1 < Z < .1) \approx (0.2)(0.4) = 0.08,$ which is the area of a 
rectangle with base $.2$ and height $.4.$ (The random variable $Z$ is standard normal.) To four places, the exact probability is a little
smaller: 0.0797. (Smaller because the density curve is a little below $0.4$
throughout the interval.)
diff(pnorm(c(-.1,.1)))
## 0.07965567

See the red rectangle in the figure below. Can you use printed tables of the
standard normal CDF to get this answer?

Maybe you can use the height of the density function at $z = 1$ to estimate
the probability $P(.9 < Z < 1.1) = 0.4839.$ (Such approximations are better
for very short intervals than for longer ones. Recall that @ThomasAndrews' $\epsilon$ is intended to be a very small number.)
dnorm(1)*.2
## 0.04839414            # approx
diff(pnorm(c(.9,1.1)))
## 0.04839406            # exact

Note: If you're looking at the R code (certainly not necessary), it might
help to know that in R dnorm is a normal density function and pnorm is a
normal CDF. (Without extra arguments, the PDF and CDF for standard normal, respectively.)
