Q.For a distribution $Q_1 = 24$, $Q_2 = 30$ and $Q_3 = 40$. Compute Bowley's coefficient of skewness and comment on the type of skewness.

Bowley's coefficient uses only the three quartiles: $Sk_b = (Q_3 + Q_1 - 2Q_2)/(Q_3 - Q_1)$.

Here the numerator is $40 + 24 - 60 = 4$ and the denominator is $40 - 24 = 16$, so $Sk_b = 4/16 = 0.25$.

Given: $Q_1 = 24$, $Q_2 = 30$ (median), $Q_3 = 40$.

Bowley's coefficient of skewness is $Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{40 + 24 - 2(30)}{40 - 24} = \frac{64 - 60}{16} = \frac{4}{16} = 0.25.$ The positive value shows the distribution is positively skewed.

Verification (dual solve). Compare the two quartile gaps directly. Upper gap $Q_3 - Q_2 = 40 - 30 = 10$; lower gap $Q_2 - Q_1 = 30 - 24 = 6$. The upper gap is the larger, which independently confirms positive skewness, and the gap form gives $Sk_b = \frac{(Q_3 - Q_2) - (Q_2 - Q_1)}{(Q_3 - Q_2) + (Q_2 - Q_1)} = \frac{10 - 6}{10 + 6} = \frac{4}{16} = 0.25,$ agreeing exactly with the compact form.

The gap form $Sk_b = \dfrac{(Q_3 - Q_2) - (Q_2 - Q_1)}{(Q_3 - Q_2) + (Q_2 - Q_1)}$ gives the same value and makes the direction of skewness visible at a glance by comparing the two quartile gaps.

A frequent error is dividing by $Q_3 + Q_1$ instead of $Q_3 - Q_1$. The denominator is the inter-quartile range, a difference, not a sum. Also remember to subtract TWICE the median in the numerator, not once.