(i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.
(i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) 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

Solution:

(i) Class mark (xi) = (upper limit + lower limit)/2

Let assumed mean (A) = 67.5

Class size (h) = 5

Class Interval Frequency (fi) Class mark (xi) di = xi – A u= di/h fiui
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 10 0 0
70-75 9 72.5 15 1 9
75-80 6 77.5 20 2 12
80-85 12 82.5 25 3 36
85-90 8 87.5 30 4 32
Total ∑fi = 80 ∑fiui = 24

By step deviation method, Mean = x̄ = A+h∑fiui /∑fi

 = 67.5+5(24/80)

= 67.5+5×0.3

= 67.5+1.5

= 69

Hence the mean of the distribution is 69.

(ii) Modal class is the class with highest frequency.

Here the modal class is 55-60.