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

Step 1: Compute cumulative frequencies and locate median class.

Total $N = 3+4+5+2+4+5+4+3 = 30$.

Median position = $\frac{N}{2} = 15$th value. Cumulative frequencies: 3, 7, 12, 14, 18, 23, 27, 30.

The 15th value lies in the class with $x_i = 13$ (cumulative 18). So median $M = 13$.

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

| $x_i$ | $f_i$ | $|x_i - 13|$ | $f_i \cdot |x_i - 13|$ |

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

| 3 | 3 | 10 | 30 |

| 6 | 4 | 7 | 28 |

| 9 | 5 | 4 | 20 |

| 12 | 2 | 1 | 2 |

| 13 | 4 | 0 | 0 |

| 15 | 5 | 2 | 10 |

| 21 | 4 | 8 | 32 |

| 22 | 3 | 9 | 27 |

Sum of $f_i \cdot |x_i - 13| = 30+28+20+2+0+10+32+27 = 149$.

Step 3: Mean deviation about median

$\text{MD} = \frac{149}{30} \approx 4.967$.

Understanding Mean Deviation About the Median

For a discrete frequency distribution, we locate the median using cumulative frequencies, then take the frequency-weighted average of the absolute deviations from it.

Step-by-Step Solution

1. Build cumulative frequencies and locate the median

Total $N = 30$, so $\dfrac{N}{2} = 15$. The cumulative frequency first reaches or exceeds $15$ at $x_i = 13$ (c.f. jumps from $14$ to $18$). Hence the median is $M = 13$.

2. Frequency-weighted absolute deviations from $13$

| $x_i$ | $f_i$ | $|x_i - 13|$ | $f_i|x_i-13|$ |

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

| 3 | 3 | 10 | 30 |

| 6 | 4 | 7 | 28 |

| 9 | 5 | 4 | 20 |

| 12 | 2 | 1 | 2 |

| 13 | 4 | 0 | 0 |

| 15 | 5 | 2 | 10 |

| 21 | 4 | 8 | 32 |

| 22 | 3 | 9 | 27 |

3. Sum and divide