The runs scored by 50 players in a local tournament are grouped with inclusive classes:
0-9 -> 6, 10-19 -> 11, 20-29 -> 15, 30-39 -> 12, 40-49 -> 6.
Find the median runs, using the true (continuous) boundaries.

Cumulative: 6, 17, 32, 44, 50, so $n=50$ and $\frac{n}{2}=25$. The first cf to reach 25 is 32, so the median class is 20-29. The gap between classes is $20-19=1$, so subtract 0.5: the continuous class is 19.5-29.5, giving $l=19.5$ and $h=29.5-19.5=10$. Median = $19.5+\left(\frac{25-17}{15}\right)\times 10=19.5+5.33=24.83$. 25.33 skips the boundary correction and uses $l=20$, landing exactly 0.5 too high. 24.3 uses $h=9$ (from $29-20$), forgetting that the corrected class really spans 10 units. 14.83 uses cf = 32, the median class's own cumulative frequency.