This is a discrete frequency distribution, so the mean deviation weights each $|x-\text{mean}|$ by its frequency rather than treating all five values equally.

Mean $=100/50=2$. $\sum f|x-2| = 40$, $N=50$.

$\text{MD}(\bar{x})=40/50=0.8$

Step 1 — Find the mean.

$\bar{x} = 100/50 = 2$.

Step 2 — Find $|x-\bar{x}|$ and $f|x-\bar{x}|$ for each row.

| $x$ | $f$ | $|x-2|$ | $f|x-2|$ |

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

| 0 | 5 | 2 | 10 |

| 1 | 10 | 1 | 10 |

| 2 | 20 | 0 | 0 |

| 3 | 10 | 1 | 10 |

| 4 | 5 | 2 | 10 |

| Total | 50 | | 40 |

Step 3 — Divide.

Independent check. Re-adding the $f|x-2|$ column in a different grouping, $(10+10)+(0+10)+10 = 20+10+10=40$ — confirms the total.

Averaging only the five listed $x$-values' plain absolute deviations from their own (unweighted) average, ignoring how many employees each value represents, is the classic mistake — the mean deviation here must be frequency-weighted, exactly like the mean itself. Always confirm $\sum f$ (here 50) is used as the divisor, not the number of distinct $x$-values (5).