This is grouped data, so each class is represented by its class mark before the deviations from the (already known) mean of 27 are found.

$\sum f|x-27| = 376$, $N=40$.

$\text{MD}(\bar{x}) = 376/40 = 9.4$

Step 1 — Recall the mean. From the direct-method calculation, $\bar{x}=27$ for this exact distribution.

Step 2 — Find the class marks and the absolute deviations from 27.

| Class | $f$ | Class mark $x$ | $|x-27|$ | $f|x-27|$ |

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

| 0–10 | 4 | 5 | 22 | 88 |

| 10–20 | 6 | 15 | 12 | 72 |

| 20–30 | 14 | 25 | 2 | 28 |

| 30–40 | 10 | 35 | 8 | 80 |

| 40–50 | 6 | 45 | 18 | 108 |

| Total | 40 | | | 376 |

Step 3 — Divide.

Independent check. Re-adding the $f|x-27|$ column in a different grouping, $(88+108)+(72+80)+28 = 196+152+28 = 376$ — confirms the total.

Using the class LIMITS (e.g. 0 and 10) instead of the class mark (5) when finding $|x-\text{mean}|$ repeats the most common grouped-data error seen throughout this subject. Also make sure the mean used is the mean of THIS grouped distribution (27), not recomputed incorrectly from scratch.