Concept: Mean Deviation About Median — we first find the median of the grouped data, then average the absolute deviations from it.

Step 1 – Find cumulative frequencies and locate the median class.

Cumulative frequencies: 3, 8, 14, 21, 29.

$N = 29$, so $\frac{N+1}{2} = 15$th observation lies in the class with $x_i = 30$ (since cumulative frequency just exceeds 15 is 21).

Thus median $M = 30$.

Step 2 – Compute absolute deviations $|x_i - M|$ and multiply by frequencies.

| $x_i$ | $f_i$ | $|x_i - 30|$ | $f_i \cdot |x_i - 30|$ |

|---|---|---|---|

| 15 | 3 | 15 | 45 |

| 21 | 5 | 9 | 45 |

| 27 | 6 | 3 | 18 | …

Mean Deviation About the Median (Discrete Frequency Data)

Step-by-Step Solution

1. Cumulative frequencies.

Total $N = 29$.

2. Locate the median.

Here $\dfrac{N}{2} = 14.5$. The cumulative frequency first exceeds $14.5$ at $cf = 21$, which corresponds to $x_i = 30$.

3. Weighted absolute deviations.

| $x_i$ | $f_i$ | $\lvert x_i-30\rvert$ | $f_i\lvert x_i-30\rvert$ |

|---|---|---|---| …