Mean deviation for discrete frequencies

A discrete frequency distribution lists distinct values x₁, x₂, …, xₙ with frequencies f₁, f₂, …, fₙ. Write N = Σfᵢ for the total frequency.

Mean: x̄ = (1/N) Σ fᵢxᵢ. Then M.D.(x̄) = (1/N) Σ fᵢ|xᵢ − x̄|: each distance is counted as many times as its value occurs.

Median of a discrete distribution: arrange the values in ascending order, write the cumulative frequencies, and take the value whose cumulative frequency first equals or exceeds N/2. Then M.D.(M) = (1/N) Σ fᵢ|xᵢ − M|.

Bowling-machine test (made up): 40 trialists each face 10 balls and the clean hits are counted. Hits 3, 4, 5, 6, 7, 8, 9, 10 occur 1, 4, 6, 7, 12, 6, 3, 1 times.

Σfᵢxᵢ = 260, so x̄ = 260/40 = 6.5 hits. The weighted distances Σfᵢ|xᵢ − 6.5| add to 52, giving M.D.(x̄) = 52/40 = 1.3 hits.

Cumulative frequencies 1, 5, 11, 18, 30, 36, 39, 40. N = 40 is even, so the median is the mean of the 20th and 21st values; both fall in the class with cumulative frequency 30, so M = 7. Σfᵢ|xᵢ − 7| = 50 gives M.D.(M) = 50/40 = 1.25 hits.

NCERT's worked results: a distribution with N = 40 and Σfᵢxᵢ = 300 has mean 7.5 and M.D.(x̄) = 92/40 = 2.3; another with N = 30 has median 13 and M.D.(M) = 149/30 ≈ 4.97.