Question 456951
I will assume that you're looking at the distribution of the sample means (which follows a normal distribution) of size 50.

In this case, {{{z = (x - mu)/(sigma/sqrt(n)) = (12 - 16)/(2/sqrt(50)) = -2sqrt(50) = -14.14213562}}} is the standard score for a sample mean of 12.

Looking at the empirical rule for normal distributions, the probability is virtually 1 that a sample mean is greater than 12.