2022 Paper 1, Q18
A percentile is a measure in statistics showing the value below which a given percentage of observations occur.
The continuous random variable $X$ has the probability density function
$$f(x)=\begin{cases}2x-2, & 1\le x\le 2\\ 0, & \text{otherwise}\end{cases}$$
Determine the 36th percentile of $X$.
Fill the density until 36% of it is behind you
A percentile is a position on the $x$-axis, found by accumulating area. The density starts at zero and climbs, so the early area comes slowly.
The percentile is the unknown limit
“The value below which 36% of observations occur” means $\int_1^a f(x)\,dx=0.36$. The unknown is a limit of integration, not a value of the function.
$\left[x^2-2x\right]_1^a=a^2-2a+1=(a-1)^2$, so the equation is $(a-1)^2=0.36$, solvable in one line without the quadratic formula.
$a=1.6$ or $a=0.4$. The support is $1\le x\le 2$, so 0.4 is impossible. And saying so is the fourth mark.
Area on the left, percentile on the right
Drag the limit. The shaded area is $(a-1)^2$ exactly, which is why the answer is a clean 1.6. And why the median sits at $1+\frac{1}{\sqrt2}\approx 1.707$, not halfway.
What does “36th percentile” ask you to find?
QCAA marking guide · 4 marks
$$\int_1^a 2x-2\,dx=0.36$$
Marker: correctly determines the definite integral. Any variable label is accepted for the unknown limit.
$$\left[x^2-2x\right]_1^a=0.36\ \Rightarrow\ \left(a^2-2a\right)-(1-2)=0.36\ \Rightarrow\ a^2-2a+0.64=0$$
Marker: determines the quadratic equation. Equivalent forms accepted, e.g. $x^2-2x-(-1)=0.36$ or $(a-1)^2=0.36$. FT marks allowed.
$$a=\frac{2\pm\sqrt{4-4\times 0.64}}{2}=\frac{2\pm\sqrt{1.44}}{2}=\frac{2\pm 1.2}{2}\ \Rightarrow\ a=1.6\ \text{or}\ 0.4$$
Marker: determines values of $a$. The perfect-square route reaches the same pair faster, because $(a-1)^2=0.36$ gives $a-1=\pm 0.6$.
Given $1\le x\le 2$, the value 0.4 is outside the domain of $X$ and is rejected.
The 36th percentile of $X$ is 1.6
Marker: evaluates the reasonableness of solutions. “$1\le x\le 2$ is used to make a judgment about the solution obtained”. A quarter of this question is for one sentence about the domain.
Why the answer is 1.6 and not 1.36
$f$ starts at 0 and climbs to 2, so the left of the interval is thinly populated. 36% of the area needs 60% of the width. The percentile is pushed right.
$(a-1)^2=p$ gives $a=1+\sqrt p$. The median is $1+\frac{1}{\sqrt2}\approx 1.707$, the lower quartile is 1.5, the upper quartile is about 1.866.
$(a-1)^2=0.36$ is symmetric about $a=1$, so the algebra always returns a mirror-image root. Whenever a percentile question gives you a quadratic, expect to reject one answer.
This is four marks from two lines of question, and nothing tells you to integrate. The QCAA define “percentile” in ordinary English, as “the value below which a given percentage of observations occur”, and the whole difficulty is hearing “below which” as an integral with the unknown in the limit. If you have only ever been asked for something like $P(X<1.6)$, you have never met the question backwards. This is an inversion. The last mark rewards a habit I want you to have on every question. Solve, then ask which root the context allows.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.