Your first question seems to have nothing to do with your second question. (Maybe that is what you intended.)
Note: Corrected response thanks to Mauro Benjamin Mistretta.
For a sample of size
$n$, the sample variances you describe with the denominator being
$n-1$ and with the denominator of
$n$ are found in the following manner:
(* Sample *)
x = {1, 2, 4, 5, 1, 2};
n = Length[x]
(* 6 *)
(* Sample variance with n-1 in the denominator *)
s2 = Variance[x]
(* 27/10 *)
(* Sample variance with n in the denominator *)
variance = Variance[x] (n-1)/n
(* 9/4 *)
The sample mean is found with
Mean[x]
