With $N=36$, $L=300$, $n_b=32$, $n_w=4$ and $c=50$, the 90th percentil…

Question

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

Accepted Answer

C. 305

1. Use $P_k=L+\left(\dfrac{k\frac{N}{100}-n_b}{n_w}ight)c$ with $k=90$.
2. Position: $90	imes\dfrac{36}{100}=32.4$.
3. Numerator: $32.4-32=0.4$.
4. $P_{90}=300+\dfrac{0.4}{4}	imes 50=300+5=305$.
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_

Answer

300

Answer

310

Answer

32.4

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

Keep practising Statistics and Probability