Mathematics 9709 · Probability & Statistics 1

Quartiles from a cumulative frequency graph

A cumulative frequency curve answers: “how many are below this value?”
For quartiles, you do not usually calculate the data value directly.
You enter on the cumulative frequency axis, go across to the curve, then down to the value axis.
That is why the graph needs clear upper-class-boundary points and a smooth curve.

Read quartiles at n/4, n/2, 3n/4

For grouped data with total $n=\sum f$ , enter the count axis at $Q_1:\tfrac n4,\ \ Q_2:\tfrac n2,\ \ Q_3:\tfrac{3n}{4}$ (for a graph it is $\tfrac n4$ , not $\tfrac{n+1}{4}$ ). The $k$ th percentile sits at $\tfrac{k}{100}\,n$ .

Build cumulative frequencies; plot each against the upper class boundary.
Join with a smooth curve.
Enter the count axis at $\tfrac n4,\tfrac n2,\tfrac{3n}{4}$ ( $k$ th percentile at $\tfrac{kn}{100}$ ).
Across to the curve, down to the value; read off.

500 reaction times. Enter at 125 (=n/4) and 375 (=3n/4), across to the curve, down: Q₁ ≈ 8.3 min, Q₃ ≈ 15.7 min, so IQR ≈ 7.4 min.

Worked example

Quartiles and a percentile from the curve

$n=500$ . $Q_1$ at the 125th value reads $\approx 8.3$ ; $Q_3$ at the 375th reads $\approx 15.7$ . The 95th percentile is at $0.95\times500=475$ .

Answer

\text{IQR}\approx 15.7-8.3=7.4\text{ min}

Worked example

Interpolate quartiles from a grouped table

Times (min): $0\text{–}10$ (8), $10\text{–}20$ (20), $20\text{–}30$ (30), $30\text{–}40$ (14), $40\text{–}50$ (8). cf $8,28,58,72,80$ , $n=80$ . $Q_1$ at 20: $10+\tfrac{20-8}{20}(10)=16$ ; $Q_3$ at 60: $30+\tfrac{60-58}{14}(10)\approx31.4$ .

Answer

\text{IQR}\approx 31.4-16=15.4\text{ min}

On a graph you don't count positions — enter at n/4, n/2, 3n/4 on the COUNT axis and read across-then-down to the value.

Data form	Quartile entry point
CF graph / large grouped data	$\tfrac n4,\ \tfrac n2,\ \tfrac{3n}{4}$
Small ordered list / stem-leaf	$\tfrac{n+1}{4},\ \tfrac{n+1}{2},\ \tfrac{3(n+1)}{4}$

Examiner note

Show the dashed read-off lines from the count axis to the curve to the value — that construction is the method evidence — and quote the entry count (e.g. “at 60

=\tfrac{3n}{4}

”).

Common mistake

Plot cumulative points at the upper class boundary, never the mid-value. Entering the axis at a value and reading a count (axes swapped). And “middle 80%” means

P_{90}-P_{10}

, not

Q_3-Q_1

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Quartiles from a cumulative frequency graph

A cumulative frequency curve answers: “how many are below this value?”
For quartiles, you do not usually calculate the data value directly.
You enter on the cumulative frequency axis, go across to the curve, then down to the value axis.
That is why the graph needs clear upper-class-boundary points and a smooth curve.

Read quartiles at n/4, n/2, 3n/4

Build cumulative frequencies; plot each against the upper class boundary.
Join with a smooth curve.
Enter the count axis at $\tfrac n4,\tfrac n2,\tfrac{3n}{4}$ ( $k$ th percentile at $\tfrac{kn}{100}$ ).
Across to the curve, down to the value; read off.

500 reaction times. Enter at 125 (=n/4) and 375 (=3n/4), across to the curve, down: Q₁ ≈ 8.3 min, Q₃ ≈ 15.7 min, so IQR ≈ 7.4 min.

Worked example

Quartiles and a percentile from the curve

$n=500$ . $Q_1$ at the 125th value reads $\approx 8.3$ ; $Q_3$ at the 375th reads $\approx 15.7$ . The 95th percentile is at $0.95\times500=475$ .

Answer

\text{IQR}\approx 15.7-8.3=7.4\text{ min}

Worked example

Interpolate quartiles from a grouped table

Answer

\text{IQR}\approx 31.4-16=15.4\text{ min}

On a graph you don't count positions — enter at n/4, n/2, 3n/4 on the COUNT axis and read across-then-down to the value.

Data form	Quartile entry point
CF graph / large grouped data	$\tfrac n4,\ \tfrac n2,\ \tfrac{3n}{4}$
Small ordered list / stem-leaf	$\tfrac{n+1}{4},\ \tfrac{n+1}{2},\ \tfrac{3(n+1)}{4}$

Examiner note

Show the dashed read-off lines from the count axis to the curve to the value — that construction is the method evidence — and quote the entry count (e.g. “at 60

=\tfrac{3n}{4}

”).

Common mistake

Plot cumulative points at the upper class boundary, never the mid-value. Entering the axis at a value and reading a count (axes swapped). And “middle 80%” means

P_{90}-P_{10}

, not

Q_3-Q_1