Question 375569
The sampling distribution of the means would also be normally distributed with a {{{mu = 122}}} and standard deviation {{{sigma/sqrt(n) = 9.6/sqrt(144) = 9.6/12 = 0.8}}}.  Then 
{{{P(X < 124) = P((X - mu)/(sigma/sqrt(n)) < (124 - mu)/(sigma/sqrt(n))) }}}
={{{P(Z < (124 - 122)/0.8) = P(Z < 2.5)}}}.  You can easily get the probability now by looking at a usual standard normal table.