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 (Production in Thousand rupees)	Class Mark xi	di = xi − A	\[u_i = \frac{d_i}{h}\]	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
				\[\sum f_i = 120_{}\]	\[\sum_{} f_i u_i = - 63\]

Required Mean = \[A + h\frac{\sum_{} f_i u_i}{\sum_{} f_i}\]
\[= 1250 - \left( \frac{63}{120} \right)500\]
= 1250 − 262.5
​= Rs 987.5
Hence, the mean of the funds is Rs 987.5.