Asymmetric Normal Probability Distribution

Solution 1:

I suggest Rayleigh distribution as it is quite similar to your figure, however it starts from zero. But one can shift it how he/she wants to. It is the distribution of the amplitude of the complex gaussian random variable.

http://en.wikipedia.org/wiki/Rayleigh_distribution

Solution 2:

If $X$ is normal, consider $Y=e^X$. It will be distributed as $$ \frac{1}{x}e^{-\ln(x)^2} = \frac{1}{x^{1+\ln(x)}}$$

This is quite similar to your picture.

As Sasha said, it's the log-normal distribution.

Solution 3:

If your asymmetric random variable is defined on $\mathbb{R}$, as opposed to $\mathbb{R}^+$, you may want to look into Azzallini's skew-normal distribution. It is implemented in R (package sn), and in Mathematica (ref-page).

Solution 4:

Try the log-normal distribution. It is probably the simpliest distribution that mimic the behavior you are searching for. It is easily implementable and has been successfully applied to many applied mathematics fields.