Q.Find $Q_1$, $Q_2$ and $Q_3$ for the discrete frequency distribution: values $x = 10, 20, 30, 40, 50$ with frequencies $f = 4, 7, 12, 9, 3$.

Build the less-than $cf$ column, get $N = 35$, and locate each quartile position $\frac{i(N+1)}{4}$ in that column.

$Q_1 = 20$, $Q_2 = 30$, $Q_3 = 40$.

Form the less-than cumulative frequency column:

So $N = \sum f = 35$ and $N + 1 = 36$.

$Q_1$: position $= \dfrac{1(36)}{4} = 9$. First $cf \ge 9$ is $11$ (against $x = 20$), so $Q_1 = 20$.

$Q_2$: position $= \dfrac{2(36)}{4} = 18$. First $cf \ge 18$ is $23$ (against $x = 30$), so $Q_2 = 30$.

$Q_3$: position $= \dfrac{3(36)}{4} = 27$. First $cf \ge 27$ is $32$ (against $x = 40$), so $Q_3 = 40$.

Independent check. $Q_2$ is the median: the median position is $\tfrac{N+1}{2} = 18$, whose $cf$ falls in $x = 30$ ✓. The quartiles are in order $20 \le 30 \le 40$ ✓.

Each quartile can be cross-read from the cumulative counts directly: $9$ of the $35$ observations (about a quarter) are reached by the end of $x = 20$, half by the end of $x = 30$, and three-quarters by the end of $x = 40$ — matching $Q_1, Q_2, Q_3$.

Using $N$ instead of $(N+1)$ for a discrete distribution, or reading the value of $x$ against a $cf$ that is still below the position. The rule is 'first $cf$ that reaches or exceeds the position' — do not stop at the $cf$ just short of it.