With $N=36$, $L=300$, $n_b=32$, $n_w=4$ and $c=50$, the 90th percentile of the distribution is:

1. Use $P_k=L+\left(\dfrac{k\frac{N}{100}-n_b}{n_w}\right)c$ with $k=90$. 2. Position: $90\times\dfrac{36}{100}=32.4$. 3. Numerator: $32.4-32=0.4$. 4. $P_{90}=300+\dfrac{0.4}{4}\times 50=300+5=305$. 5. Reporting $32.4$ stops at the position; adding rather than subtracting $n_b$ pushes the answer past $310$. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 4: Statistics_