Find the median class, then the median.

Histogram of a grouped distribution with its cumulative curve; a line at N/2 finds the median class and the interpolation formula l + ((N/2 − C)/f)h is drawn as a ramp inside it.

For the practice-match data, place the probe at the median (28 runs).

Taking the median class as the one with the largest frequency.

• 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.

Teaching this idea? Put the simulation on your own page or LMS with this code.