Question

Calculate the mean of the distribution, given below using the short cut 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

Marks	f	x	d = x – A = x – 45.5	fd
11 – 20	2	15.5	–30	–60
21 – 30	6	25.5	–20	–120
31 – 40	10	35..5	–10	–100
41 – 50	12	A = 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
	∑f = 50			∑fd = 70

∴ Mean = `A + (sumfd)/(sumf)`

= `45.5  + 70/50`

= `45.5 + 7/5`

=`(227.5 +7)/5`

= `234.5/5`

= 46.9

Answer 2

Class Interval (Inclusive form)	Class Interval (Exclusive form)	No. of Students `bb((f_i))`	`bb(x_i)`	`bb(A = 45.5)` `bb(d_i = x - 45.5)`	`bb(f_i d_i)`
11 – 20	10.5 – 20.5	2	15.5	–30	`{:(-60),(-120),(-100):}} - 280`
21 – 30	20.5 – 30.5	6	25.5	–20
31 – 40	30.5 – 40.5	10	35.5	–10
41 – 50	40.5 – 50.5	12	45.5	0	0
51 – 60	50.5 – 60.5	9	55.5	10	`{:(90),(140),(120):}} 350`
61 – 70	60.5 – 70.5	7	65.5	20
71 – 80	70.5 – 80.5	4	75.5	30
		`sumf_i = 50`			`sumf_i d_i = 70`

Assumed mean (A) = 45.5

`sumf_i = 50, sumf_i d_i = 70`

Mean = `A + (sumf_i d_i)/(sumf_i)`

= `45.5 + (70)/(50)`

= 45.5 + 1.4

= 46.9