Q.For the marks $15, 20, 22, 25, 28, 30, 35, 40, 45, 50$ of $10$ students, find

Data are ascending with $n = 10$, so $n + 1 = 11$; use $\frac{j(n+1)}{10}$ for the decile and $\frac{k(n+1)}{100}$ for the percentile.

(i) $D_4 = 26.2$; (ii) $P_{70} = 38.5$.

The marks are already ascending, $n = 10$, $n + 1 = 11$.

(i) $D_4$: position $= \dfrac{4(11)}{10} = 4.4$ → between the 4th ($25$) and 5th ($28$) marks.

(ii) $P_{70}$: position $= \dfrac{70(11)}{100} = 7.7$ → between the 7th ($35$) and 8th ($40$).

Independent check. $D_4$ should have about $40\%$ of the data below it: four marks ($15, 20, 22, 25$) lie below $26.2$, i.e. $40\%$ of $10$ ✓. $P_{70}$ should have about $70\%$ below: seven marks ($15$ through $35$) lie below $38.5$, i.e. $70\%$ ✓.

Since $D_4 = P_{40}$, the fourth decile could equally be found from the percentile formula: position $= \frac{40(11)}{100} = 4.4$ — the same $4.4$th position, giving the same $26.2$. This confirms the relationship $D_j = P_{10j}$.

Mixing up the divisors — using $/4$ (quartiles) or $/100$ (percentiles) when a decile ($/10$) is required. Read which family is asked for first, then pick $4$, $10$ or $100$ in the denominator.