For grouped data with $L=39.5$, $N=180$, $f_b=37$, $n_w=73$ and $c=10$, the lower quartile is:

1. Use $Q_1=L+\left(\dfrac{\frac{N}{4}-f_b}{n_w}\right)c$. 2. First find the quartile position: $\dfrac{N}{4}=\dfrac{180}{4}=45$. 3. Subtract the cumulative frequency below the class: $45-37=8$. 4. Then $\dfrac{8}{73}\times10=1.096$. 5. Adding to the lower boundary: $39.5+1.096=40.596$. Reporting 45 stops at the position rather than converting it to a value. _Source: NECTA ACSEE 2022 Advanced Mathematics 142/1, Question 4: Statistics_