I am not too sure how to start this problem:

Suppose that 40%, 35% and 25% of the Stat302 students come from math, stat
and computer science, respectively.

The mean and variance of the final mark for a randomly selected student from
the math program are 72 and 25; from the stat program are 75 and 36; and from
the computer science program are 73 and 16.

Tabulate the pmf of E(X|Y)

using Conditional Expectation-math-equation.png

mu would = 73.3, and using the individual variance we can obtain the probability of individual group getting mu.
So for Math it would be for P(math) = (72-73.3)/5 = 1 - P(Z<.26) = 0.3974
P(stats) = (75 - 73.3)/6 = P(Z< 0.28) = 0.6103
p(CS) = (73 -73.3)/ = 1-P(Z<0.075) = 0.39165

For the class as a whole, we find the fraction of class getting the final mark
= 0.3974 *0.4 + 0.6103*.35 + .39165*.25 = .47048

I am not sure if I am on the right track. I would really appreciate it if someone can give me some guidance. Thanks!