Q.For the distribution classes $0\text{–}10, 10\text{–}20, 20\text{–}30, 30\text{–}40, 40\text{–}50$ with frequencies $5, 8, 15, 12, 10$, find

Reuse the $cf$ column ($N = 50$); the locating figures are $\frac{3N}{10}$ and $\frac{90N}{100}$.

Using the cumulative frequencies $5, 13, 28, 40, 50$ ($N = 50$):

(i) $D_3$: locating figure $= \dfrac{3N}{10} = \dfrac{150}{10} = 15$. First $cf \ge 15$ is $28$ → class $20\text{–}30$: $L = 20, cf = 13, f = 15, h = 10$.

(ii) $P_{90}$: locating figure $= \dfrac{90N}{100} = \dfrac{4500}{100} = 45$. First $cf \ge 45$ is $50$ → class $40\text{–}50$: $L = 40, cf = 40, f = 10, h = 10$.

$P_{90} = 40 + \frac{45 - 40}{10}\times 10 = 40 + \frac{5}{10}\times 10 = 40 + 5 = 45.$ …

Because $P_{90}$'s locating figure ($45$) coincides exactly with a class boundary's cumulative frequency ($cf = 50$ at the top, with $40$ just before), the interpolation lands neatly at $45$; whenever the locating figure hits …

For $P_{90}$, choosing the class where $cf = 40$ (i.e. $30\text{–}40$) instead of the first class whose $cf$ reaches $45$. The locating figure $45$ exceeds …