If the marks scored by the students in a class test out of 50 then, Find the Median of the following data: Marks | Frequency 0-10 | 5 10-20 | 8 20-30 | 6 30-40 | 6 40-50 | 5

Correct — D. Build the cumulative frequencies, find the median class, then interpolate inside it.

Total: N = 5 + 8 + 6 + 6 + 5 = 30, so N⁄2 = 15
Cumulative frequencies: 5, 13, 19, 25, 30
15 first falls in 20–30 (13 score below 20, 19 below 30)

Median class values: L = 20, cf = 13, f = 6, h = 10
Median = L + ((N⁄2 − cf) ⁄ f) × h
= 20 + (2⁄6) × 10 = 20 + 3.33
= 23.33 ≈ 23.3 → option (d)

Why the others are wrong

• (a)
  38.9 — 38.9 lies in the 30–40 class, but 19 of the 30 students already score below 30. More than half sit below 38.9, so it is past the middle.
• (b)
  18.8 — 18.8 lies in the 10–20 class, and the cumulative count is just 13 by 20 marks. Fewer than 15 of the 30 score below it, so it falls short of the middle.
• (c)
  14.1 — 14.1 sits even lower in the 10–20 class. Only 13 students score below 20, under half of 30, so the median cannot lie anywhere below 20.

The median of grouped data is the mark below which half the students fall. Marks inside a class are not listed, so you assume they are spread evenly across it.

Find N⁄2, locate the median class where the cumulative frequency first reaches N⁄2, then move into that class in proportion to how many of its students you still need.

Here 2 of the class's 6 students are needed, so you go 2⁄6 of the way through a class 10 marks wide.

Key facts

• Grouped median = L + ((N⁄2 − cf) ⁄ f) × h, where cf is the cumulative frequency before the median class.
• The median class is the first class whose cumulative frequency reaches N⁄2.
• The grouped-data formula uses N⁄2, not (N + 1)⁄2.

Common traps

• Taking the class with the largest frequency (10–20) as the median class. That is the modal class.
• Using the cumulative frequency at the end of the median class (19) instead of the one before it (13).

No directly related past PYQ was found.