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

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

1 answer

Answer: Mean fund = ₹ 987.5

Take \(A = 1250\) and class width \(h = 500\).

Class	\(x_i\)	\(d_i = x_i - 1250\)	\(u_i = d_i/500\)	\(f_i\)	\(f_iu_i\)
0–500	250	-1000	-2	35	-70
500–1000	750	-500	-1	28	-28
1000–1500	1250	0	0	32	0
1500–2000	1750	500	1	15	15
2000–2500	2250	1000	2	10	20
Total				\(\sum f_i = 120\)	\(\sum f_iu_i = -63\)

\[ \bar{u} = \frac{-63}{120} \]

\[ \bar{X} = A + \bar{u}\,h = 1250 + \frac{-63}{120} \times 500 \]

\[ \bar{X} \approx 987.5 \]

Note: the answer key gives ₹ 985 or ₹ 987.5; the exact value is ₹ 987.50 (₹ 985 comes from rounding \(\bar{u}\) early).

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?