The two pictures obviously can't refer to the same thing.
Actually, they do.
You see, $FE(E,T)$, the Fermi function, is the mean number of fermions in the state of given energy E at temperature T. Hopefuly, it is bounded between 0 and 1 so only zero or one fermion in these quantum states. In this case, $\mu$ is the limit of energy at wich you have one half of a chance to found a fermion in you gas. Lower energy levels will mostly be filled and higher ones will be empty.
Obviously, in order to generalise the distribution it is common to normalize regarding $k_bT$ i.e $\beta = \frac{1}{k_bT}$. $\mu$ actually depends on your system and mostly on your system's reservoir and so is fixed with regards to N and T (else the distribution isn't valid, meaning you don't have enough particles or the temperature is too high).
I did a plot of two characterising limit $\nu$ (sorry for notation change, it's old) relatively different to the temperature of their systems as $\nu = 10k_bT$ and $\nu = 50k_bT$. 28
e is in $k_bT$ units.
As you can see, as the distribution is exponentially decaying between $\nu-5k_bT$ and $\nu+5k_bT$. This gives you the density of distribution of your fermions in the different energy states around.
If you keep an eye on a constant $\mu$ (or $\nu$ in the case of my plot) you will see the distribution sharpen as you decrease the temperature. This defines the so called "Fermi sphere" in the p space containing all the energy levels of your fermions at zero temperature. As such you define it's radius as $p_F$ giving you a total number of quantic states (for electrons): $$\frac{4}{3}\pi p_F^3\frac{V}{h^3}2=N$$ (2 is due to the 1/2 spin of electrons) As such, you can define the Fermi energy as $\epsilon_F = \frac{p_F^2}{2m}$ giving you the fermi level at $0\ K$: $\mu_F=\epsilon_F$.
Remember that $\mu$ depends on the system bath and that the distribution along the energy states varies with $T$. Hope this helps a bit. 28