This is a discrete frequency distribution; find the mean, then the frequency-weighted average of the absolute deviations from it.

$\sum fx=45$, mean$=1.8$; $\sum f|x-1.8|=22$; $\text{MD}(\bar{x})=22/25=0.88$

Step 1 — Find the mean.

$\bar{x} = 45/25 = 1.8$.

Step 2 — Find $|x-1.8|$ and $f|x-1.8|$.

| $x$ | $f$ | $|x-1.8|$ | $f|x-1.8|$ |

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

| 0 | 3 | 1.8 | 5.4 |

| 1 | 7 | 0.8 | 5.6 |

| 2 | 9 | 0.2 | 1.8 |

| 3 | 4 | 1.2 | 4.8 |

| 4 | 2 | 2.2 | 4.4 |

| Total | 25 | | 22 |

Step 3 — Divide.

Independent check. Re-adding in a different pairing, $(5.4+4.4)+(5.6+4.8)+1.8 = 9.8+10.4+1.8=22$ — confirms the total.

Rounding the mean (1.8) further before computing the deviations changes the final answer slightly — keep the exact mean throughout the calculation, rounding only the final mean deviation if needed.