Q.For a distribution the median $Q_2 = 40$, the third quartile $Q_3 = 50$ and Bowley's coefficient of skewness is $-0.2$. Find the first quartile $Q_1$.

Substitute the known values into $Sk_b = (Q_3 + Q_1 - 2Q_2)/(Q_3 - Q_1)$ and solve for $Q_1$.

Given: $Q_2 = 40$, $Q_3 = 50$, $Sk_b = -0.2$; find $Q_1$.

Substitute into Bowley's formula: $Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{50 + Q_1 - 80}{50 - Q_1} = \frac{Q_1 - 30}{50 - Q_1}.$ Setting this equal to $-0.2$: $-0.2(50 - Q_1) = Q_1 - 30 \quad\Rightarrow\quad -10 + 0.2Q_1 = Q_1 - 30.$ Collecting terms, $30 - 10 = Q_1 - 0.2Q_1$, so $20 = 0.8Q_1$ and $Q_1 = \frac{20}{0.8} = 25.$ …

Sign slips when multiplying $-0.2$ through the bracket $(50 - Q_1)$: it gives $-10 + 0.2Q_1$, not $-10 - 0.2Q_1$. Substituting the final val …