The less-than ogive's cumulative frequency reaches N/2 = 25 partway through the 20–30 class, so the median is found by locating that point on the segment joining (20,14) and (30,30).

Step 1 — Build the less-than cumulative frequency table.

Check: $4+10+16+14+6=50$ matches the total cumulative frequency of $50$ at the last point.

Step 2 — Draw the less-than ogive. Plot $(0,0), (10,4), (20,14), (30,30), (40,44), (50,50)$ and join consecutively with straight line segments.

Step 3 — Locate $N/2$ on the curve. $N=50$, so $N/2=25$, which falls between the plotted points $(20,14)$ and $(30,30)$.

Step 4 — Interpolate along that segment. The segment rises by $30-14=16$ in cumulative frequency as marks rise by $30-20=10$. To rise from $14$ to $25$ (an increase of $11$) requires $\dfrac{11}{16}\times10=6.875$ extra marks beyond $20$. So the median is at $20+6.875=26.875$.

Independent check (interpolation-formula form).

matching Step 4 exactly. …

The median can also be verified by locating where a plotted more-than ogive (points (0,50),(10,46),(20,36),(30,20),(40,6),(50,0)) intersects the less-than ogive — both metho …

Using the class mark or the lower boundary instead of the correct upper boundary as the X-value for a less-than ogive point is a common error. When comparing to a more-than ogive, forgetting that its cumulative frequencies are …