Why is a fourth type of active neutrinos not allowed by the invisible width of the $Z$-boson? I don't understand the following statement from my notes:
"A fourth active neutrino is not allowed by the invisible width of the $Z$-boson to which it would contribute as much as one active neutrino, $Z \to \nu_\alpha \bar{\nu}_\alpha$."
I understand that the invisible width is the uncertainty of a particle's mass and therefore the rate at which particles decay ( larger width, faster decay). $Z$-bosons decay into a pair of neutrinos, and that neutrinos barely interact with matter, making them "invisible", hence the term.
But I don't understand how this invisible width doesn't allow a fourth active neutrino to exist.
Why does it imply that besides the three types of active neutrinos ($e, \mu, \tau$) only sterile neutrinos could potentially exist?
 A: The Z boson coupling to all fermions is predicted by the Standard Model, and does not depend on what family the fermion is in. Because the mass of all neutrinos is much, much lighter than the mass of the Z, they all have the same partial decay widths, which can be determined using precision electroweak measurements. Thus if we know the invisible width of the Z boson, we can determine the number of decays of the Z boson simply by $n=\Sigma_{\text{invisible}}/\Sigma_{\text{one decay}}$. The only integer consistent with the number computed this way is $3$. Since we already know of three generations of neutrinos, this doesn't leave room for any others.
If there is a fourth generation of neutrinos, this implies either that they do not couple to the Z boson much (so-called "sterile neutrinos") or that their mass is very, very large (close to or greater than $M_Z/2$).
A: Your question is quite obscure, but I'm starting a placeholder answer, to avoid a garland of comments in a clarificatory conversation.
The full width of the Z is about 2.5 GeV, of which the visible leptonic and hadronic (quark) decays observed add up to 1.9 GeV or so; so the invisible width is the rest, about 0.5 - 0.6 GeV, and must be due to decay into neutrinos. Don't worry about the precise numbers: the PDG will give you more than you ask for.
Now, estimating the widths to neutrinos theoretically with the peculiar telltale SM couplings of the neutral current, you find good agreement with 3 "active" neutrinos, i.e. neutrinos coupling in that precise manner.
A fourth family with neutrinos coupling in the same way  (and very heavy "visible" charged leptons and quarks, so unobservable)  would only couple to this invisible width  through a 4th active neutrino. But that would require 4/3 the invisible width observed.
So, if there are more neutrinos, they cannot be active; they must be sterile, i.e. they don't couple to the Z in the conventional SM manner, and, so they don't contribute to its invisible width.
