Concept: Mean Deviation About Median for grouped data.

Step 1 – Convert to continuous classes and find the median class.

Adjusting boundaries (subtract 0.5 from lower, add 0.5 to upper), the classes become 15.5–20.5, 20.5–25.5, …, 50.5–55.5.

Cumulative frequencies: 5, 11, 23, 37, 63, 75, 91, 100.

$N/2 = 50$, so the median class is 35.5–40.5 (cumulative 63).

Median $M = L + \frac{N/2 - cf}{f} \times h = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + 2.5 = 38$ years.

Step 2 – Compute deviations $|x_i - M|$ and their weighted sum.

Take class marks $x_i$: 18, 23, 28, 33, 38, 43, 48, 53.

Deviations from $M=38$: 20, 15, 10, 5, 0, 5, 10, 15.

Multiply by frequencies: …

Mean Deviation About the Median (Continuous Grouped Data)

Following the hint, subtract $0.5$ from each lower limit and add $0.5$ to each upper limit to get continuous classes, then apply

Step-by-Step Solution

1. Continuous classes, mid-points and cumulative frequencies.

Total $N = 100$.

2. Median class.

$\dfrac{N}{2} = 50$; the cumulative frequency first exceeds $50$ in class 35.5–40.5 (where $cf = 63$).

$M = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + \frac{13}{26}\times 5 = 35.5 + 2.5 = 38$ …