Why is the variance of the beta value among BJTs much lower in the saturation region than in the forward active region?
In active mode you give the transistor a high enough \$V_{ce}\$ such that the transistor can determine the collector current \$I_c\$ that is flowing.

Then we can apply a given base current \$I_b\$, let the transistor determine \$I_c\$ and calculate \$\beta = I_c/I_b\$.

In saturation mode the transistor does not get a high enough \$V_{ce}\$ so the transistor cannot determine the collector current \$I_c\$ that is flowing (for that a higher \$V_{ce}\$ would be needed but then we would be in active mode!).

So in saturation mode \$I_c\$ is determined outside the transistor. To make any current flow of course a base current \$I_b\$ is needed, we cannot make that \$I_b\$ small as that would result in a small \$I_c\$, maybe smaller than what we want (remember that \$I_c\$ is set outside the transistor) and then again we would end up in active mode.

So \$I_b\$ has to be fairly large. How large? Well \$I_b\$ needs to be more than \$I_c/\beta\$ because that's the current that would flow when in active mode. So if we force \$I_b &gt; I_c/\beta\$ then we would have a high enough \$I_b\$ to be sure we're in saturation mode. Many datasheets use \$I_c/I_b = 10\$ when measuring in saturation mode.

So in saturation mode you cannot measure \$\beta\$ as it is determined by the measurement itself (both \$I_b\$ and \$I_c\$ are forced from external).

You can only measure the transistor's \$\beta\$ properly when it is in active mode.

The reason why \$\beta\$ varies a lot is due to it being a very poorly controlled property of a transistor. \$\beta\$ depends on doping levels and thickness of the base region, both of these are difficult to control in manufacturing. This may sound like a huge problem but it is not, the circuits in which we use transistors do not rely on the value of \$\beta\$ to be predictable. Instead, proper circuits are designed to work with the smallest value \$\beta\$ can have, then when \$\beta\$ is larger, the circuit will still work as designed.