Question

1) Using step–deviation method, calculate the mean marks of the following distribution.

2) State the modal class.

Class Interval	50 - 55	55 - 60	60 - 65	65 - 70	70 - 75	75 - 80	80 - 85	85 – 90
Frequency	5	20	10	10	9	6	12	8

Answer 1

1) 

Class interval	Frequency (f)	x	d = x – A= x – 67.5	t = d/i  i = 5	f xt
50 – 55	5	52.5	-15	-3	-15
55 – 60	20	57.5	-10	-2	-40
60 – 65	10	62.5	-5	-1	-10
65 – 70	10	67.5	0	0	0
70 – 75	9	72.5	5	1	9
75 – 80	6	77.5	10	2	12
80 – 85	12	82.5	15	3	36
85 – 90	8	87.5	20	4	32
Total	80				24

Assumed mean (A) = 67.5

Class size,  i = 5

Mean = `A + i (sum ft)/(sum f)``

`= 67.5 + 5 xx 24/80`

`= 67.5 + 1.5`

= 69

Thus, the mean of the given data is 69.

2) Modal class is the class corresponding to the greatest frequency.

So, the modal class is 55 – 60.