Informally, a probability distribution defines the relative frequency of outcomes of a random variable - the expected value can be thought of as a weighted average of those outcomes (weighted by the relative frequency). Similarly, the expected value can be thought of as the arithmetic mean of a set of numbers generated in exact proportion to their probability of occurring (in the case of a continuous random variable this isn't exactly true since specific values have probability $0$).
The connection between the expected value the arithmetic mean is most clear with a discrete random variable, where the expected value is
$$ E(X) = \sum_{S} x P(X=x) $$
where $S$ is the sample space. As an example, suppose you have a discrete random variable $X$ such that:
$$ X = \begin{cases} 1 & \mbox{with probability } 1/8 \\
2 & \mbox{with probability } 3/8 \\
3 & \mbox{with probability } 1/2
\end{cases} $$
That is, the probability mass function is $P(X=1)=1/8$, $P(X=2)=3/8$, and $P(X=3)=1/2$. Using the formula above, the expected value is
$$ E(X) = 1\cdot (1/8) + 2 \cdot (3/8) + 3 \cdot (1/2) = 2.375 $$
Now consider numbers generated with frequencies exactly proportional to the probability mass function - for example, the set of numbers $\{1,1,2,2,2,2,2,2,3,3,3,3,3,3,3,3\}$ - two $1$s, six $2$s and eight $3$s. Now take the arithmetic mean of these numbers:
$$ \frac{1+1+2+2+2+2+2+2+3+3+3+3+3+3+3+3}{16} = 2.375 $$
and you can see it's exactly equal to the expected value.