Assuming 30 days in a month and 1 year consists of 12 months. So total temperature values for 50 years would be 30 x 12 x 50 = 18000.
Now lets say we store these 18000 values in a vetor A whose dimension would be 18000 x1
A= [ ] 18000 x1
Next step is to find temperature mean for each month. This can be done using for loop and mean command as follows:
Let N = 12 x 50 = 600
y(i) = mean (A(30*(k-1)+1:30*k));