Question

A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of the funds by ‘step deviation’ method.

Fund (Rupees)	0-500	500-1000	1000-1500	1500-2000	2000-2500
No. of workers	35	28	32	15	10

Answer 1

Class Fund (Rupees)	Class Mark (xi)	di = xi − A	`u_i = d_i/g`	Frequency (Number of farm owners) fi	Frequency × deviation (fi × ui)
0-500	250	−1000	−2	35	−70
500-1000	750	−500	−1	28	−28
1000-1500	1250 → A	0	0	32	0
1500-2000	1750	500	1	15	15
2000-2500	2250	1000	2	10	20
				∑fi = 120	∑fiui = −63

Here, ∑f_iu_i = −63, ∑f_i = 120, g = 500

`bar u = (sum f_iu_i)/(sum f_i)`

= `(-63)/120`

= −0.525

Mean `(bar X) = A + bar u g`

= 1250 + (−0.525) × 500

= 1250 − 262.50

= 987.50

∴ The mean of the funds is ₹ 987.50.