Q.Using the same daily-wages data (0–20: 6, 20–40: 10, 40–60: 18, 60–80: 15, 80–100: 11 workers), calculate the Median and Q2, and hence the quartile deviation using Q1 = ₹38 and Q3 = ₹74.67 from Example 3.

The median and Q2 must come out identical, since both locate the exact halfway point of the same distribution; the quartile deviation then follows directly from Q1 and Q3. …

Step 1 — Locate the median class. $k = N/2 = 30$. The first cf ≥ 30 is 34, in class 40–60 (L = 40, cf before = 16, f = 18, h = 20).

Step 2 — Locate Q2 by the quartile formula, with i = 2. $k = \frac{2N}{4} = \frac{2(60)}{4} = 30$ — the identical position as the median, so Q2 lands in the same 40–60 class and gives the identical value:

$Q_2 = 40 + \frac{30-16}{18}\times20 = 55.56$ …

This IS the cross-check. Computing the median via the Measures of Central Tendency formula and Q2 via the quartile formula independently, and getting the SAME ₹55.56 both times, is a direct numerical confirmation of th …

Assuming Q2 = Median without actually working it out, and therefore missing an arithmetic slip in one of the two independent c …