Quick question on sketching wavefunction in well 
Usually for an infinite well, the sketch for n=3 level is this:

Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left):

 A: Just so this doesn't slip past:

Now I think if one side of the potential barrier is higher, the particle will be more likely to spend time on the left side than the right side, so the wavefunction should have higher amplitudes on the left (skewed to the left):

This is incorrect. Between A and B the well is deeper, so the particle goes faster. Between B and C the well is shallower and the particle goes slower, so it takes a longer time to cross this region. If you take a snapshot at a random time, it will be more likely to be between B and C, and the wavefunction amplitude there is higher:

As Gert mentioned, for a finite well there is also a slight penetration (tunnelling) of the walls, giving an exponential tail there instead of strict zeros at A and C.
A: You are correct that for $n = 3$ there are $2$ non-boundary zero points.
Also, the modulus of $\psi(x)$ is lowest where $V(x)$ is lowest.
Where you are wrong is that $\psi(A)$ is not zero and $\psi(B)$ is not zero either, as you indicate in your schematic. For $x < A$ and $x > B$, $V(x)$ is not $\infty$ (your well is a finite, not infinite well) and in those regions the wave functions decay to $0$ exponentially ("tunnelling" effect).
See for example this simple case.
