Partition Values for a Continuous (Grouped) Frequency Distribution

For a continuous (class-interval) frequency distribution the exact observations are unknown — we know only how many fall in each class — so a partition value is found by interpolation inside the class in which it lies.

Method.

1. Ensure classes are in ascending order and are continuous (convert any inclusive classes such as $10\text{–}19,\ 20\text{–}29$ to boundaries $9.5\text{–}19.5,\ 19.5\text{–}29.5$ first).
2. Build the less-than $cf$ column; find $N = \sum f$.
3. For the required partition value compute the locating figure — $\tfrac{iN}{4}$ for $Q_i$, $\tfrac{jN}{10}$ for $D_j$, $\tfrac{kN}{100}$ for $P_k$ (using $N$, not $N+1$).
4. The partition class is the first class whose $cf \ge$ that locating figure. Apply:

Worked illustration. For the classes below ($N = 50$):

$Q_1$ locating figure $= \tfrac{N}{4} = 12.5$; the first $cf \ge 12.5$ is $13$ (class $10\text{–}20$), so $L = 10,\ cf = 5,\ f = 8,\ h = 10$:

$Q_1 = 10 + \frac{12.5 - 5}{8}\times 10 = 10 + 9.375 = 19.375.$ …