Q.For a distribution $Q_1 = 20$, $Q_2 = 26$ and $Q_3 = 36$. Find Bowley's coefficient of skewness.

Use $Sk_b = (Q_3 + Q_1 - 2Q_2)/(Q_3 - Q_1)$.

So $Sk_b = 4/16 = 0.25$, a positive value.

Given: $Q_1 = 20$, $Q_2 = 26$, $Q_3 = 36$.

$Sk_b = \frac{Q_3 + Q_1 - 2Q_2}{Q_3 - Q_1} = \frac{36 + 20 - 2(26)}{36 - 20} = \frac{56 - 52}{16} = \frac{4}{16} = 0.25.$ The distribution is positively skewed.

Verification (dual solve). Quartile gaps: upper $Q_3 - Q_2 = 36 - 26 = 10$; lower $Q_2 - Q_1 = 26 - 20 = 6$. The upper gap is larger, confirming positive skewness, and the gap form gives $(10 - 6)/(10 + 6) = 4/16 = 0.25$, matching the compact form.

Computing $2Q_2$ as $2 \times Q_1$ or forgetting to double the median at all. The middle term of the numerator is always twice the median $Q_2$.