Example Find the arithmetic mean for the following :

Marks below : 10 20 30 40 50 60 70 80

No. of students : 15 35 60 84 96 127 198 250

First, we have to convert the cumulative frequencies into frequencies of the respective classes.

Marks	Mid-values xi	Frequencies c.f. f.	U = X -A A = 45	fiui
0 - 10	5	15          15	- 40	- 600
10 -20	15	35          20	- 30	- 600
20 - 30	25	60          25	- 20	- 500
30 - 40	35	84          24	- 10	- 240
40 - 50	45 Þ A	96          12	0	0
50 - 60	55	127        31	+10	+310
60 - 70	65	198 71	+20	+1420
70 - 80	75	250 52	+30	+1560
Total

Here all class intervals are of the same width say 'c'. This method is employed in place of the Short-cut method. We measure all the class-marks (mid values) from some convenient value, say 'A', which generally should be taken as the class-mark of a class of maximum frequency or of a class which is the middle one. All the class marks happen to be multiples of c, since all class intervals are equal. We consider class frequencies as if they are centered at the corresponding class-marks.

Theorem If x₁, x₂ , x₃, ......, x_n are n values of the class marks with frequencies f₁, f₂ , f₃, ......f_n respectively and if each xi is expressed in terms of the new variable ui by the relation x_i = A + cui then, with the usual notation, we have

This method is also known as the "Coding method."

Index

4.1 Introduction
4.2 Arithmetic Mean
4.3 Properties of Arithmetic Mean
4.4 Median
4.5 Mode
4.6 Empirical relation between mean, median & mode

Chapter 5