A grouped distribution of 36 stones has quartile class lower boundary $L=150$, $n_b=8$, $n_w=10$ and class width $c=50$. The first quartile is:

1. For grouped data $Q_1=L+\left(\dfrac{\frac{N}{4}-n_b}{n_w}\right)c$. 2. $\dfrac{N}{4}=\dfrac{36}{4}=9$, so the numerator is $9-8=1$. 3. $Q_1=150+\dfrac{1}{10}\times 50=150+5=155$. 4. Answering $9.0$ applies the ungrouped rule $Q_1=\tfrac14 N$, which gives a position, not a value. 5. $149.5$ uses a continuity-corrected boundary the class limits do not call for. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 4: Statistics_