Could someone explain explain to what correspond the moments of a random variable? Could someone explain what are (at least the four first) moments ? (normalized moment to be more precise) Let $X$ a r.v. 


*

*So the first moment is the expectation. This will correspond to $\mathbb E[X]$ and is going to be the "barycenter" (in the sense that if $X$ is supported on $[a,b]$ and $\rho(t)$ is the density of the line $[a,b]$ (i.e. $\rho(dt)=\mathbb P_X(dt)$), then the barycenter of the line is $\mathbb E[X]$ (I like to see this in physic term).

*So the second moment is the variance. So, it's going to represent the dispersion of the mass around $\mathbb E[X]$, i.e. $66\%$ of the total mass is going to be concentrate on $[\mathbb E[X]-\sqrt{Var(X)},\mathbb E[X]+\sqrt{Var(X)}]$, but I don't really understand why... 
Is my intuition for $\mathbb E[X]$ and $\text{Var}(X)$ correct ? What about the the third and the fourth moments ? (I think of higher dimension it's much more abstract).
 A: Actually, the geometric interpretation of kurtosis is quite simple: Following J.G.'s notation, let $V = Z^4$, and let $p_V(v)$ denote the pdf of $V$. Then the kurtosis of $X$ is the center of mass (point of balance) of $p_V(v)$. 
Now, place a fulcrum on the horizontal axis of the graph of $p_V(v)$ at 3.0: If $p_V(v)$ falls to the right, then kurtosis of $X$ is >3, and the distribution of $X$ is heavier-tailed than the normal distribution.  If it falls to the left, then the kurtosis is <3, and the distribution of $X$ is lighter-tailed than the normal distribution.  
Yes, "flatness" as a descriptor of kurtosis is quite silly - kurtosis has nothing to do with that. Just look at the beta(.5,1) distribution: It has kurtosis less than 3.0, and hence is classified as "platykurtic." But it is nowhere near flat-topped.  In fact, it is infinitely peaked. 
Nor does large kurtosis have anything to do with "peakedness," or "concentration toward the mean."  Think about it: With kurtosis > 3, why does the pdf of $Z^4$ ($p_V(v)$) fall to the right when the fulcrum is placed at 3.0: Is it because of the "peakedness," or concentration near the mean (which translates to large mass where $X$ is near to $\mu$, and hence $Z^4$ is near $0$), or is it due to tail extremity (where $X$ is far from $\mu$, and hence $Z^4$ is large)? The answer is obviously that $p_V(v)$ falls to the right due to tail extremity.
You can make a similar geometric argument for skewness: Consider the pdf of $Z^3$: Its center of mass is the skewness.  Place a fulcrum at $0.0$: If the pdf of $Z^3$ falls to the right, then the distribution of $X$ is positively skewed, meaning that the right tail of the distribution of $X$ is heavier (as measured by combined leverage of cubed deviations from the mean) than the left tail of the distribution of $X$.  The converse description applies for negative skewness.
A: The first moment is the (arithmetic) mean, $\mu:=\Bbb E[X]$. If this is finite, any $a$ satisfies $\Bbb E[X-a]=\mu-a$. So if we want to measure spread around $\mu$, we can't use $\Bbb E[X-\mu]=0$, but we can use the variance $\sigma^2=\Bbb E[(X-\mu)^2]$, or better still (so we're working with a quantity that has the same dimensions as $X$ and $\mu$, should $X$ not be dimensionless) the standard deviation $\sigma=\sqrt{\Bbb E[(X-\mu)^2]}\ge 0$. (You could also look at the absolute deviation $\Bbb E[|X-\mu|]$, but that's usually neither as useful nor as easy to analyse.)
If $\mu,\,\sigma$ exist and are finite, the $z$-score of $X$ is defined as $Z:=\frac{X-\mu}{\sigma}$. This has mean $0$ and variance $1$. We can prove $P(|Z|\ge k)\le k^{-2}$, but there are theoretical reasons to have an especial interest in a "Normal distribution", for which a much tighter bound can be obtained, which your numerical example considers.
While $Z,\,Z^2$ must have respective means $0,\,1$, $\Bbb E[Z^3],\,\Bbb E[Z^4]$ are another story. Their values are the skewness and kurtosis of $X$, and indeed of $aX+b$ for $a>0$ (for the kurtosis we can relax this to $a\ne 0$), including $Z$ itself. These quantities provide additional information about $Z$'s distribution. In particular, they provide ways a distribution can notably fail to be Normal. For a Normal distribution, their respective values are $0$ and $3$, and $\Bbb E[Z^4-3]$ is often called the excess kurtosis of $X$. We classify distributions by whether this is positive, zero or negative. Note that $\Bbb E[Z^4-1]$ is the variance of $Z^2$, so the excess kurtosis is at least $-2$. This can be saturated.
A symmetric distribution (i.e. one for which $P(X\le x)=P(X\ge 2\mu-x)$) has either zero or undefined skewness. When the skewness of an asymmetric distribution is defined but non-zero, the sign of the skewness indicates whether the distribution "leans" to the left or right. But despite a common misconception, it doesn't tell you how the mean compares to the median.
A geometric interpretation of the kurtosis is a little harder. You'll sometimes read people saying it describes how flat the distribution's peak is, but it's better to think of it in terms of the behaviour of the distribution's tails.
