Is there a measure of 'evenness' of dispersion? I looked up on the web, but couldn't find anything helpful.
I'm basically looking for a way to measure how 'evenly' a value is distributed. As in, an 'evenly' distributed distribution like X:

and an 'unevenly' distributed distribution Y of roughly the same mean and standard deviation:

But is there any evenness measure m, such that m(X) > m(Y)? If there isn't, what would be the best way to create a measure like that?
(Images screenshot from Khan Academy)
 A: I don't know if there is a widely used method to measure this, but this idea may be useful:
First, assuming that you have the values of $X$ in increasing order, compute $X'$: $x'_k=x_{k+1}-x_k$, $k=1,\ldots,n-1$. Then compute the standard deviation of $X'$.
If the data are perfectly evenly distributed, this measure will be $0$.
To avoid ambiguity, I'll write the formula:
$$\sqrt{\frac{\sum\limits_{k=1}^{n-1}(x_{k+1}-x_k)^2}{n-1}-\bar x'^2}$$
where $\bar x'$ is the mean of $X'$, that is:
$$\bar x'=\frac1{n-1}\sum_{k=1}^{n-1}(x_{k+1}-x_k)=\frac{x_n-x_1}{n-1}$$
A: I actually think standard deviation is exactly what you're looking for and you have made an error in calculating them to be the same in your example.
The mean of $X$ is $25/6$ and the standard deviation is approximately $2.3$.
The meany of $Y$ is $23/6$ and the standard deviation is approximately $3.1$.
As you can see, as $Y$ is more spread out, it has a higher standard deviation.
EDIT - since you feel that even-ness is not adequately captured with just the standard deviation, I might recommend the kurtosis, which is the fourth standardized moment. Here are what distributions that differ on kurtosis look like

