Question

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

Dear Student,

(i) The data can be written as

Class Interval	Class Mark $X_i$	Frequency $F_i$	Deviation $d_i=X_i-A$	$u_i=\frac{X_i-A}{h}$	$F_i{u}_i$
50-55	52.5	5	-20	-4	-20
55-60	57.5	20	-15	-3	-60
60-65	62.5	10	-10	-2	-20
65-70	67.5	10	-5	-1	-10
70-75	72.5	9	0	0	0
75-80	77.5	6	5	1	6
80-85	82.5	12	10	2	24
85-90	87.5	8	15	3	24
		$\sum F_i=80$			$\sum F_i{u}_i=-56$

Assumed mean, $A=72.5$
Class size, $h=5$
Mean,
$$\begin{gathered} \overline{X}=A+h\left[\frac{\sum F_i{u}_i}{\sum F_i}\right] \\ =72.5+5\left[\frac{-56}{80}\right] \\ =72.5-\frac{56}{16} \\ =72.5-3.5 \\ =69 \end{gathered}$$
(ii) Modal class is the class interval having maximum frequency. So, modal class for the given data is 55-60.
Regards,