Mathematics 9709 · Probability & Statistics 2

Median, quartiles & percentiles

The median is the balance point of area: walk in from the left and stop the instant you have covered half the total. Quartiles and percentiles are the same move with a different fraction (0.25, 0.75, $p/100$ ).

Set the running area equal to the fraction

Median: $\int_{\text{low}}^{m}f(x)\,dx=0.5$ ; lower quartile $=0.25$ ; $p$ th percentile $=p/100$ .

Median solves (area from the left) = 0.5. Quartiles use 0.25 / 0.75; the pth percentile uses p/100.

f(x) = x²/72 on [0, 6]. The median is where the shaded area first reaches 0.5: m³/216 = 0.5 → m = ∛108 ≈ 4.76.

Worked example

Easiest first: median and IQR of one curve

Bus arrival time $f(x)=x^2/72$ on $[0,6]$ . Set the area from 0 equal to each fraction:

Median: $\int_0^m\tfrac{x^2}{72}dx=\tfrac{m^3}{216}=0.5\Rightarrow m^3=108,\ m=\sqrt[3]{108}\approx4.76$ .
Quartiles: $t^3=54\Rightarrow t\approx3.78$ and $v^3=162\Rightarrow v\approx5.45$ .
$\text{IQR}=v-t=5.45-3.78=1.67$ .

Answer

m\approx4.76,\ \text{IQR}=1.67

Worked example

A “show that” median equation, then solve it

$f(x)=\tfrac{2}{21}(6-x)$ on $[1,4]$ . (a) Show the median satisfies $2m^2-24m+43=0$ ; (b) find $m$ .

Set $\int_1^m\tfrac{2}{21}(6-x)dx=0.5$ : $\tfrac{2}{21}\big[6x-\tfrac{x^2}{2}\big]_1^m=0.5$ .
Clear denominators: $12m-m^2-11=10.5\Rightarrow 2m^2-24m+43=0$ (shown).
Complete the square: $(m-6)^2=14.5\Rightarrow m=6\pm\sqrt{14.5}$ .
Reject $6+\sqrt{14.5}=9.81\notin[1,4]$ ; take $m=6-\sqrt{14.5}$ .

Answer

m=6-\sqrt{14.5}\approx2.19

A “show that” needs full algebra — the calculator earns no marks here. Always reject the root outside the domain.

“Show that” median: set ∫ = 0.5, clear denominators to reach the given quadratic, solve, then reject the out-of-domain root.

Worked example

Percentiles in surd form

$f(x)=\tfrac{2}{25}(5-x)$ on $[0,5]$ . The median in exact (surd) form, and the IQR.

Median: $\int_0^m\tfrac{2}{25}(5-x)dx=0.5\Rightarrow m^2-10m+12.5=0\Rightarrow m=5-\tfrac{\sqrt{50}}{2}\approx1.46$ .
Lower quartile $(0.25)\Rightarrow t\approx0.67$ ; upper quartile $(0.75)\Rightarrow v=2.5$ .
$\text{IQR}=2.5-0.67=1.83$ .

Answer

m=5-\tfrac{\sqrt{50}}{2}\approx1.46,\ \text{IQR}\approx1.83

Worked example

Piecewise: which piece holds the cutoff?

$f=x^2/27$ on $[0,3]$ , then $\tfrac13(5-x)$ on $[3,5]$ . Locate the piece first.

Area of piece 1: $\int_0^3\tfrac{x^2}{27}dx=\tfrac13<0.5$ , so the median is in piece 2.
Solve in piece 2: $\int_m^5\tfrac13(5-x)dx=0.5\Rightarrow m^2-10m+22=0$ .
$(m-5)^2=3\Rightarrow m=5-\sqrt3$ (reject $5+\sqrt3$ ).
The lower quartile $(0.25<\tfrac13)$ sits in piece 1: $t^3/81=0.25\Rightarrow t\approx2.73$ .

Answer

m=5-\sqrt3\approx3.27,\ \text{LQ}\approx2.73

Pick the fraction: median 0.5, quartiles 0.25 / 0.75, percentile $p/100$ .
Piecewise? integrate to each breakpoint to find which piece holds the cutoff.
Set the area from the low end = the fraction; solve inside that piece.
Reject any root outside the domain.

On a piecewise f, locate the piece FIRST (compare the running area with the fraction), then solve inside it — and reject the out-of-range root.

Examiner note

To show a median lies in an interval, check the running area straddles 0.5; then solve algebraically and take the in-range root. (9709/62/M/J/24 Q7) (9709/62/M/J/23 Q7)

Common mistake

On a piecewise pdf, solving in the first piece without checking the cutoff is even there; keeping the root outside the domain; and doing a “show that” on the calculator (it needs full algebra).

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Median, quartiles & percentiles

Set the running area equal to the fraction

Median: $\int_{\text{low}}^{m}f(x)\,dx=0.5$ ; lower quartile $=0.25$ ; $p$ th percentile $=p/100$ .

Median solves (area from the left) = 0.5. Quartiles use 0.25 / 0.75; the pth percentile uses p/100.

f(x) = x²/72 on [0, 6]. The median is where the shaded area first reaches 0.5: m³/216 = 0.5 → m = ∛108 ≈ 4.76.

Worked example

Easiest first: median and IQR of one curve

Bus arrival time $f(x)=x^2/72$ on $[0,6]$ . Set the area from 0 equal to each fraction:

Median: $\int_0^m\tfrac{x^2}{72}dx=\tfrac{m^3}{216}=0.5\Rightarrow m^3=108,\ m=\sqrt[3]{108}\approx4.76$ .
Quartiles: $t^3=54\Rightarrow t\approx3.78$ and $v^3=162\Rightarrow v\approx5.45$ .
$\text{IQR}=v-t=5.45-3.78=1.67$ .

Answer

m\approx4.76,\ \text{IQR}=1.67

Worked example

A “show that” median equation, then solve it

$f(x)=\tfrac{2}{21}(6-x)$ on $[1,4]$ . (a) Show the median satisfies $2m^2-24m+43=0$ ; (b) find $m$ .

Set $\int_1^m\tfrac{2}{21}(6-x)dx=0.5$ : $\tfrac{2}{21}\big[6x-\tfrac{x^2}{2}\big]_1^m=0.5$ .
Clear denominators: $12m-m^2-11=10.5\Rightarrow 2m^2-24m+43=0$ (shown).
Complete the square: $(m-6)^2=14.5\Rightarrow m=6\pm\sqrt{14.5}$ .
Reject $6+\sqrt{14.5}=9.81\notin[1,4]$ ; take $m=6-\sqrt{14.5}$ .

Answer

m=6-\sqrt{14.5}\approx2.19

A “show that” needs full algebra — the calculator earns no marks here. Always reject the root outside the domain.

“Show that” median: set ∫ = 0.5, clear denominators to reach the given quadratic, solve, then reject the out-of-domain root.

Worked example

Percentiles in surd form

$f(x)=\tfrac{2}{25}(5-x)$ on $[0,5]$ . The median in exact (surd) form, and the IQR.

Median: $\int_0^m\tfrac{2}{25}(5-x)dx=0.5\Rightarrow m^2-10m+12.5=0\Rightarrow m=5-\tfrac{\sqrt{50}}{2}\approx1.46$ .
Lower quartile $(0.25)\Rightarrow t\approx0.67$ ; upper quartile $(0.75)\Rightarrow v=2.5$ .
$\text{IQR}=2.5-0.67=1.83$ .

Answer

m=5-\tfrac{\sqrt{50}}{2}\approx1.46,\ \text{IQR}\approx1.83

Worked example

Piecewise: which piece holds the cutoff?

$f=x^2/27$ on $[0,3]$ , then $\tfrac13(5-x)$ on $[3,5]$ . Locate the piece first.

Area of piece 1: $\int_0^3\tfrac{x^2}{27}dx=\tfrac13<0.5$ , so the median is in piece 2.
Solve in piece 2: $\int_m^5\tfrac13(5-x)dx=0.5\Rightarrow m^2-10m+22=0$ .
$(m-5)^2=3\Rightarrow m=5-\sqrt3$ (reject $5+\sqrt3$ ).
The lower quartile $(0.25<\tfrac13)$ sits in piece 1: $t^3/81=0.25\Rightarrow t\approx2.73$ .

Answer

m=5-\sqrt3\approx3.27,\ \text{LQ}\approx2.73

Pick the fraction: median 0.5, quartiles 0.25 / 0.75, percentile $p/100$ .
Piecewise? integrate to each breakpoint to find which piece holds the cutoff.
Set the area from the low end = the fraction; solve inside that piece.
Reject any root outside the domain.

On a piecewise f, locate the piece FIRST (compare the running area with the fraction), then solve inside it — and reject the out-of-range root.

Examiner note

To show a median lies in an interval, check the running area straddles 0.5; then solve algebraically and take the in-range root. (9709/62/M/J/24 Q7) (9709/62/M/J/23 Q7)

Common mistake

On a piecewise pdf, solving in the first piece without checking the cutoff is even there; keeping the root outside the domain; and doing a “show that” on the calculator (it needs full algebra).