Median, quartiles and percentiles | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Translate the percentile into a left-hand area

A percentile is an area boundary

50% area

median m ≈ 1.59, not the interval midpoint 1

The median splits probability area in half; it does not usually split the x-axis width in half.

\int_L^{q_p}f(x)\,dx=\frac{p}{100}

Symbols

$L$ = lower end of the support (same as X)
$q_p$ = p-th percentile (same as X)

The median is the 50th percentile: left area 0.5.
The lower quartile has left area 0.25.
The upper quartile has left area 0.75.
$\mathrm{IQR}=Q_3-Q_1$ ; do not subtract their areas.

Solve the area equation, then check the support

Worked example

Current-paper median

Earlier parts give $0\le X\le\tfrac43$ and $f(x)=\tfrac{81}{64}x^2$ . Find the median m.

\int_0^m\frac{81}{64}x^2\,dx=0.5

\frac{27m^3}{64}=\frac12

m^3=\frac{32}{27}

Answer

m=\sqrt[3]{\frac{32}{27}}=1.06

(9709/62/M/J/25 Q7(c))

Examiner note

If an equation gives several roots, keep only the root inside the support. State the exact form first when the question asks for it.

Equal areas do not usually have equal widths

Worked example

Quartiles of an increasing density

Let $f(x)=\tfrac38x^2$ for $0\le x\le2$ . For any left-hand area r,

\int_0^q\frac38x^2\,dx=r

\frac{q^3}{8}=r,\qquad q=2\sqrt[3]{r}

Q_1=1.2599

Q_3=1.8171

Q_3-Q_1=0.557

Answer

\mathrm{IQR}=0.557

Your turnUnlabelled percentile decision [4 marks]

The PDF is $f(x)=\tfrac29x$ for $0\le x\le3$ . Find the 80th percentile.

Show worked answer

\int_0^q\frac29x\,dx=0.8

\frac{q^2}{9}=0.8

Answer

q=\sqrt{7.2}=2.68

Common mistake

Do not place the median halfway across the support unless symmetry justifies it. Percentiles divide area, not horizontal width.

Translate the percentile into a left-hand area

A percentile is an area boundary

50% area

median m ≈ 1.59, not the interval midpoint 1

The median splits probability area in half; it does not usually split the x-axis width in half.

\int_L^{q_p}f(x)\,dx=\frac{p}{100}

Symbols

$L$ = lower end of the support (same as X)
$q_p$ = p-th percentile (same as X)

The median is the 50th percentile: left area 0.5.
The lower quartile has left area 0.25.
The upper quartile has left area 0.75.
$\mathrm{IQR}=Q_3-Q_1$ ; do not subtract their areas.

Solve the area equation, then check the support

Worked example

Current-paper median

Earlier parts give $0\le X\le\tfrac43$ and $f(x)=\tfrac{81}{64}x^2$ . Find the median m.

\int_0^m\frac{81}{64}x^2\,dx=0.5

\frac{27m^3}{64}=\frac12

m^3=\frac{32}{27}

Answer

m=\sqrt[3]{\frac{32}{27}}=1.06

(9709/62/M/J/25 Q7(c))

Examiner note

If an equation gives several roots, keep only the root inside the support. State the exact form first when the question asks for it.

Equal areas do not usually have equal widths

Worked example

Quartiles of an increasing density

Let $f(x)=\tfrac38x^2$ for $0\le x\le2$ . For any left-hand area r,

\int_0^q\frac38x^2\,dx=r

\frac{q^3}{8}=r,\qquad q=2\sqrt[3]{r}

Q_1=1.2599

Q_3=1.8171

Q_3-Q_1=0.557

Answer

\mathrm{IQR}=0.557

Your turnUnlabelled percentile decision [4 marks]

The PDF is $f(x)=\tfrac29x$ for $0\le x\le3$ . Find the 80th percentile.

Show worked answer

\int_0^q\frac29x\,dx=0.8

\frac{q^2}{9}=0.8

Answer

q=\sqrt{7.2}=2.68

Common mistake

Do not place the median halfway across the support unless symmetry justifies it. Percentiles divide area, not horizontal width.