"Merging" different normal distributions Consider two different classrooms, A and B, containing 25 and 30 students respectively.
Grades obtained in classroom A are distributed normaly, with mean Ma and variance Va, while grades obtained in classroom B are also normaly distributed with mean Mb and variance Vb.
How to calculate the distribution of grades considering the total population of 55 students?
Thank you in advance,
JMZ
 A: You should see it like each student grade follow a normal distribution with parameter depending on the classroom he belong. You can talk about a distribution of a population if each individual follow the same distribution. It is not the case if you merge the two population. 
A: Just from the background information given, I'm not sure it is a good idea to merge classrooms in this way. But here is an outline on how to do it, if you feel you should. (I use $U_i$ for observations in Class A, $V_i$ for observations in Class B, and $X_i$ for observations in the combined class C.  Also, I have retained your notations
Ma, Va, and so on, along with the usual notations $\bar U, S_U^2.)$
(a) If you still have the original data, merging is trivial. Just put the $n = 25 + 30 = 55$ scores together and recompute $\bar X$ and $S_X^2$ for the combined sample.
(b) If you have only $n_A = na = 25,\, \bar U = Ma = 79.08$ and $S^2_U = Va = 13.91$ from Class A, and similar summary statistics from Class B, you can still combine the data. But the computations are a little more tedious. (Numerical values used for Class B are shown in the computer printout at the end.)
The key quantities for Class A are $n_A = 25, T_A = \sum_{i=1}^{25} U_i = 1977,$ and $Q_A = \sum_{i=1}^{25} U_i^2 = 156,675.$  There are similar key quantities for Class B.  
From them, you can get $\bar U = Ma = T_A/n_A = 1977/25 = 79.08.$ And
$$S_U^2 = Va = \frac{1}{n_A - 1}\left[\sum U_i^2 - n_A\bar U^2 \right] = 13.91.$$
In reverse, knowing $n_A = na = 25,\,$ $\bar U = Ma = 79.08\,$ and $S_U^2 = Va = 13.91,\,$ you can solve (without any rounding!) to get key quantities $T_A$ and $Q_A.$
Similarly, you can use $n_B, \bar V,$ and $S_V^2$ to get
key quantities $T_B$ and $Q_B.$ Then for the combined Class C:
$n_C = n_A + n_B,\,$ $T_C = T_A + T_B\,$ and $Q_C = Q_A + Q_B.$
And finally, from these you can get 
$$\bar X = \frac{T_C}{n_C} =\frac{T_A + T_B}{n_A+n_B} =  \frac{n_A\bar U + n_B\bar V}{n_A+n_B} =  81.14545.$$ Also, with a little more effort,
$$S_X^2 = \frac{1}{n_C - 1}[Q_C + n_c\bar X^2].$$

From R, here are sample data for Class A and Class B, and for the
combined class--against which you can check your computations.
set.seed(1234)
u = round(rnorm(25, 80, 4));  m.a = mean(u);  v.a = var(u)
s.a = sum(u); q.a = sum(u^2)
m.a; v.a; s.a; q.a
[1] 79.08
[1] 13.91
[1] 1977
[1] 156675

v = round(rnorm(30, 85, 3));  m.b = mean(v);  v.b = var(v)
s.b = sum(v); q.b = sum(v^2)
m.b; v.b; s.b; q.b
[1] 82.86667
[1] 5.429885
[1] 2486
[1] 206164

x = c(u,v);  mean(x);  var(x)
[1] 81.14545
[1] 12.71919

