Understanding variance and standard deviation I get that the variance of a random variable X is the expected value of the squared deviation from its mean, and the standard deviation is just the square root of that. Can you interpret the standard deviation as the absolute average deviation from the mean? Is the variance defined as such because if we defined the standard deviation as the expected value of X - E(X), we would get 0?
 A: The average deviation from the mean is $0$, that is $E[X-\mu]=0$, and so, taken literally, the absolute average deviation is also $0$.  Changing your question slightly to

Can you interpret the standard deviation as the average absolute deviation from the mean?

No, $E[|X-\mu|]$ is not the standard deviation, that is,
$$\sigma = \sqrt{E[(X-\mu)^2]} \neq E[|X-\mu|].$$
A: The standard deviation is at least as large as the mean absolute deviation: The function $\varphi(z) = z^2$ is strictly convex over non-negative argument values; from Jensen's inequality it follows that:
$$\mathbb{V}(X)=\mathbb{E}\Big((X-\mu)^2\Big) = \mathbb{E}\Big(\varphi(|X-\mu|)\Big) \geqslant \varphi\Big(\mathbb{E}(|X-\mu|)\Big) = \mathbb{E}(|X-\mu|)^2,$$
with the inequality being strict unless $\mathbb{V}(|X-\mu|)=0$.  It follows that $\mathbb{S}(X) \geqslant \mathbb{V}(|X-\mu|)$, with the inequality being strict unless $\mathbb{V}(|X-\mu|)=0$.  So as you can see, unless it is equal to zero (i.e., for a random variable with a point mass distribution), the standard deviation is strictly greater than the mean absolute deviation.
