weighted averages and binomials(i think)

consider a multiple choice exam with 50 questions. a student who has gone to class and done homework has a 75% probability of answering any questions. Assume that 60% of the students have gone to class and done their home. The remaining 40% have not gone to class and have a 25% probability of answering questions correctly.

a) What is the mean and standard deviation
b) To receive and A, a student must get 43 answers correct out of 50. How many out of 100 students are likely to get As?

To find the mean and standard deviation, this is my thought process-

if this is a binomial, with success = answer correctly,the mean will be n(number of trials)x p(probability) = 50x.75= 37.5
for 60% of the group. and 50x.25=12.5 for 40%. so mean would be (.6)x37.5 + (.4)x12.5 = 22.5+5= 27.5

i dont know if you can actually to it that way. the thing is then finding the standard deviation, by finding the variance, which according to the binomial theorem is np(1-p). so for 60% of the class it would be 50x(.75)(.25) = 9.375 it would be the same for the 40% as well. thus sqrt of 9.375 = 3.062

For the 2nd part, I was thinking I use the binomial probability function which is 50C43 x (p^x)(1-p)^(n-x)

That number is small, and multiply it by 100. that would be number of students who get an A (43 or more correct).