Q.For the distribution $x = 10, 20, 30, 40, 50$ with $f = 4, 7, 12, 9, 3$, find

Reuse the same $cf$ column ($N = 35$, $N+1 = 36$); locate $\frac{7(N+1)}{10}$ and $\frac{25(N+1)}{100}$.

Using the cumulative frequencies from the previous example ($cf$: $4, 11, 23, 32, 35$), with $N = 35$ and $N + 1 = 36$:

(i) $D_7$: position $= \dfrac{7(36)}{10} = \dfrac{252}{10} = 25.2$. First $cf \ge 25.2$ is $32$ (against $x = 40$), so $D_7 = 40$.

(ii) $P_{25}$: position $= \dfrac{25(36)}{100} = \dfrac{900}{100} = 9$. First $cf \ge 9$ is $11$ (against $x = 20$), so $P_{25} = 20$. …

The equality is structural, not a coincidence: $\frac{25(N+1)}{100} = \frac{N+1}{4} = \frac{1(N+1)}{4}$, so the $P_{25}$ and $Q_1$ positions are identical for every d …

Rounding the decile position $25.2$ down to $25$ and then, more seriously, reading the wrong $cf$ — the position $25.2$ still needs the first $cf \ge 25.2$, which is $32$ (not …