Q.For a distribution $Q_1 = 15$, the median $Q_2 = 20$ and Bowley's coefficient of skewness is $0.2$. Find the third quartile $Q_3$.

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

Given: $Q_1 = 15$, $Q_2 = 20$, $Sk_b = 0.2$; find $Q_3$.

Substitute into Bowley's formula: $Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{Q_3 + 15 - 40}{Q_3 - 15} = \frac{Q_3 - 25}{Q_3 - 15}.$ Setting this equal to $0.2$: $0.2 = \frac{Q_3 - 25}{Q_3 - 15} \quad\Rightarrow\quad 0.2(Q_3 - 15) = Q_3 - 25.$ Expanding, $0.2Q_3 - 3 = Q_3 - 25$, so $25 - 3 = Q_3 - 0.2Q_3$, giving $22 = 0.8Q_3$ and $Q_3 = \frac{22}{0.8} = 27.5.$ …

Errors in clearing the fraction: every term on the left must be multiplied by $(Q_3 - 15)$, and the $Q_3$ terms must then be collected on one side. Substitu …