Grouped data needs the interpolation formula for the median first, then the same $\sum f|x-M|/N$ pattern as every mean-deviation calculation.

c.f.: $8,20,40,54,60$; $M=60+\left(\dfrac{30-20}{20}\right)\times10=65$; $\sum f|x-65|=540$; $\text{MD}(M)=540/60=9$

Step 1 — Build the cumulative frequency table and locate the median class.

$N=60$, $N/2=30$. The c.f. first reaches or exceeds 30 at the class 60–70 (c.f.$=40$), so $L=60$, $cf=20$, $f=20$, $h=10$.

Step 2 — Apply the interpolation formula.

Step 3 — Find $|x-M|$ and $f|x-M|$ using class marks.

| Class | $f$ | Class mark $x$ | $|x-65|$ | $f|x-65|$ |

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

| 40–50 | 8 | 45 | 20 | 160 |

| 50–60 | 12 | 55 | 10 | 120 |

| 60–70 | 20 | 65 | 0 | 0 |

| 70–80 | 14 | 75 | 10 | 140 |

| 80–90 | 6 | 85 | 20 | 120 |

| Total | 60 | | | 540 |

Step 4 — Divide.

Independent check. Re-adding, $(160+120)+(0+140+120) = 280+260=540$ — confirms the total.

Picking the wrong median class — e.g. stopping at the first class whose c.f. exceeds 20 instead of the correct threshold $N/2=30$ — is the most common interpolation-formula error. Always compute $N/2$ first, then scan the c.f. column for where it FIRST reaches or exceeds that value.