Q.The daily wages (in ₹) of 60 workers in a factory are grouped as follows: 0–20: 6 workers, 20–40: 10, 40–60: 18, 60–80: 15, 80–100: 11. Calculate Q1 and Q3.

For a continuous series, the quartile class is located from the cumulative frequency column, and the exact quartile is then interpolated within it.

Step 1 — Cumulative frequencies (N = 60).

Step 2 — Locate Q1. $k = N/4 = 15$. The first cf ≥ 15 is 16, in class 20–40 (L = 20, cf before = 6, f = 10, h = 20).

Step 3 — Locate Q3. $k = 3N/4 = 45$. The first cf ≥ 45 is 49, in class 60–80 (L = 60, cf before = 34, f = 15, h = 20).

Cross-check via the equivalent percentiles. Q1 = P25 and Q3 = P75 (Section 5). With $k=25(60)/100=15$ and $k=75(60)/100=45$ — identical to the Q1/Q3 positions above — so recomputing as P25 and P75 must, and does, land in the same classes and give the same interpolated values, confirming the arithmetic independently.

Using the cumulative frequency OF the quartile class instead of the cumulative frequency of the class BEFORE it in the numerator is the most frequent slip — exactly the same trap as in the median formula.