Mean deviation for class intervals

A continuous frequency distribution groups data into class intervals with no gaps, each with its frequency. Every class is taken to be centred at its mid-point xᵢ, and the discrete method is then applied to the mid-points.

Practice match (made up): runs of 50 trialists in classes 0-10, 10-20, …, 60-70 with frequencies 1, 12, 15, 8, 7, 6, 1. Mid-points 5, 15, …, 65 give Σfᵢxᵢ = 1550, so x̄ = 31 runs, and Σfᵢ|xᵢ − 31| = 616 gives M.D.(x̄) = 12.32 runs.

Step-deviation shortcut for the mean: pick an assumed mean a near the middle and the common class width h, and use dᵢ = (xᵢ − a)/h. Then x̄ = a + h × (Σfᵢdᵢ)/N. It shifts the origin to a and changes the scale by h, which keeps the arithmetic small. With a = 35, h = 10: Σfᵢdᵢ = −20, so x̄ = 35 + 10 × (−20/50) = 31.

Median of a continuous distribution: the median class is the one whose cumulative frequency first equals or exceeds N/2. Then Median = l + ((N/2 − C)/f) × h, where l is its lower limit, f its frequency, h its width and C the cumulative frequency of the class before it.

Practice match: cumulative frequencies 1, 13, 28, …; N/2 = 25 falls in 20-30, so l = 20, C = 13, f = 15, h = 10 and M = 20 + (12/15) × 10 = 28 runs. Deviations of mid-points from 28 give Σfᵢ|xᵢ − 28| = 598 and M.D.(M) = 11.96 runs.

NCERT's worked results: a marks distribution with mean 45 has M.D.(x̄) = 10; another with N = 50 has median 28 and M.D.(M) = 508/50 = 10.16.

If classes are written with gaps, such as 16-20, 21-25, make them continuous first by moving each lower limit down by 0.5 and each upper limit up by 0.5.