Fermi-Dirac distribution definition and meaning I'm studying statistical mechanics and band theory, with two different professors.
My statistical mechanical teacher defines the Fermi Occupation function this way:
\begin{equation}
F(E,T)=\frac{1}{1+e^{\beta({E-\mu})}}
\end{equation}
where $\beta=kT$. He this defines the Fermi level as $E_F=\mu(T)$ as T tends to zero. 

So, about this I have some questions because even at this point I'm very confused. What does F(E,T) represent? Is it a probability distribution? And if not, what does it represent?
I got very confused because, as $E_F$ is defined as being independent on the temperature, thus a constant in the picture (depending on the solid considered I suppose) I see from the image that there are no fermions over $E_F$, even when the temperature increases. Why is that? If, for example, I consider a semiconductor I have that the Fermi level lies in the middle of an energy gap, so at very low temperatures I have no conduction. But, as the temperature increases, I need some electron over $E_F$ for the conduction to take place. 
With the other professor is a different story. He defines
\begin{equation}
F(E,T)=\frac{1}{1+e^{\beta({E-E_F})}}
\end{equation}
and he says it's a distribution probability. He uses a picture like this one:

The two pictures obviously can't refer to the same thing. Can anyone give me a clue?
 A: 
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 $\nu = 50k_bT$. 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.
