Average number of comparisons in sorted insertion
Well, that would strongly depend on the algorithm that you use to find out where to insert the new element, and on the distribution of new elements. 

Assuming that you do a linear search, and the new element could go to any position with the same probability, your result is close, but not quite exact. The new element can go into one of (n + 1) positions, not n. There may be up to n comparisons needed; n comparisons are needed both if the new element goes into the first array position, and if it goes into the second array position. So the result is 

(1 + 2 + 3 + ... + (n-1) + n + n) / (n + 1) = 

= (n (n+1) / 2 + n) / (n + 1)

= n (n + 3) / 2 / (n + 1)

= (n + 2 - 2 / (n + 1)) / 2 ≈ (n + 2) / 2,


which is just a tiny bit more than your answer (n + 1) / 2.