Answer 1

Let A be the assumed mean and d be the deviation of x from the assumed mean.

Let A = 45.5.

d = x – A

Marks (C.I.)	No. of students (Frequency f)	Mid-point of C.I. (x)	d = x – A	f × d
11-20	2	15.5	-30	-60
21-30	6	25.5	-20	-120
31-40	10	35.5	-10	-100
41-50	12	45.5	0	0
51-60	9	55.5	10	90
61-70	7	65.5	20	140
71-80	4	75.5	30	120
				`sumfd = 70`

Mean  = `A + (sumfd)/(sum f)`

`=> Mean = 45.5 + 70/50`

`=> Mean  = 45.5 + 1.4`

`:. Mean = 45.9`