Minor errata (nitpicking)....

I wrote:
The real problem is with your math: the average of averages
is not necessarily the same as the average of all. They are
equal only [when] the divisors of each average are the same,
which is not true in your example.

I shouldn't say that is the "only" time they are equal. Of course, there
are examples when they are equal by coincidence (e.g. when all the numbers
to be averaged are equal). And there are examples when they are "close
enough" to equal; for example, equal to 1 decimal place.