Question

Calculate the mean of the distribution given below using the shortcut method.

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
No. of students	2	6	10	12	9	7	4

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
				${\displaystyle \sum fd =70}$

Mean  = ${\displaystyle A+\frac{\sum fd}{\sum f}}$

${\displaystyle \Rightarrow Mean=45.5+\frac{70}{50}}$

${\displaystyle \Rightarrow Mean =45.5+1.4}$

${\displaystyle \therefore Mean=45.9}$