Mathematics 9709 · Probability & Statistics 1

Range, IQR & box plots

A mean tells you where the data sits; variation tells you how spread out it is.
Range is the full width from smallest to largest.
IQR is the width of the middle 50%.
IQR ignores the lowest quarter and highest quarter, so outliers cannot wreck it.
The exam core is: median, IQR, box plot, and one clear comparison sentence.

Quartiles by splitting in half

Sort the data, find the median, then the quartiles are the medians of the two halves. Odd $n$ : discard the median first, then split. $\text{IQR}=Q_3-Q_1$ .

Worked example

IQR of an even list (no median to discard)

2, 5, 9, 13, 29, 33, 49, 55 ( $n=8$ ). Median $=\tfrac{13+29}{2}=21$ ; split 4 / 4.

Lower 2, 5, 9, 13 → $Q_1=\tfrac{5+9}{2}=7$ . Upper 29, 33, 49, 55 → $Q_3=\tfrac{33+49}{2}=41$ .

Answer

\text{IQR}=41-7=34

Worked example

IQR of an odd list (discard the median)

Unordered: 69, 17, 43, 6, 73, 77, 39. Sort → 6, 17, 39, 43, 69, 73, 77 ( $n=7$ ). Median = 43 (discard it).

Lower half 6, 17, 39 → $Q_1=17$ . Upper half 69, 73, 77 → $Q_3=73$ .

Answer

\text{IQR}=73-17=56

IQR ignores the extreme quarter at each end, so one extreme value cannot move it, which is why it is the standard measure for comparing spread.

Median & IQR from an ordered display (stem-and-leaf):

Decode the key first — fix the true place value before reading anything.
Count $n$ ; locate $Q_2$ at the $\tfrac{n+1}{2}$ th value, then read $Q_1,\,Q_3$ by inspection as the medians of the lower and upper halves.
Read the data values at those positions (not the positions).
$\text{IQR}=Q_3-Q_1$ ; range = max − min.
State each with its unit.

Worked example

Read median + IQR off a stem-and-leaf

Lengths (mm), key $14\mid2=142$ :

14 | 2 2 4 8 9
15 | 1 3 5 6 7 9
16 | 5 8

That is 13 values: 142, 142, 144, 148, 149, 151, 153, 155, 156, 157, 159, 165, 168. $Q_2$ is the $\tfrac{13+1}{2}=7$ th value, 153 — discard it. Lower six → $Q_1=\tfrac{144+148}{2}=146$ ; upper six → $Q_3=\tfrac{157+159}{2}=158$ .

Answer

\text{median }153,\ \text{IQR }=158-146=12\text{ mm}

Worked example

Read median + IQR off 15 ordered values

Team A times (min), ordered: 23, 27, 29, 31, 34, 36, 38, 41, 43, 45, 48, 52, 55, 58, 61 ( $n=15$ ). Positions 8th, 4th, 12th: $Q_2=41$ , $Q_1=31$ , $Q_3=52$ . (9709/52/O/N/23 Q4) (9709/52/M/J/24 Q4)

Answer

\text{median }41,\ \text{IQR }=52-31=21\text{ min}

Drawing the box plot the way the mark scheme wants:

One linear scale across the page (≥1 cm per stated unit); label the axis with quantity + unit.
Mark the five-number summary; box from $Q_1$ to $Q_3$ with the median line inside.
Whiskers to min and max — not through the box, not from its corners.
Label each plot (team / company name).

A box-and-whisker plot: the box spans the IQR (Q₁ to Q₃), the line inside is the median, the whiskers reach the min and max.

Read skew straight off the box: if $Q_3-Q_2>Q_2-Q_1$ (the right side of the box is longer) the data are positively skewed; the reverse is negative. Drag the quartiles and watch the IQR and the skew change:

Q₁Q₃

IQR = Q₃ − Q₁ = 20·Q₃−Q₂ > Q₂−Q₁ → positive skew (right)

Worked example

Two box plots + a spread comparison

Team A (above): five-number summary 23, 31, 41, 52, 61 → IQR 21. Team B: 20, 38, 44, 50, 66 → IQR 12. Drawn on one scale, A's box is wider. (9709/52/M/J/23 Q3)

Answer

\text{A: IQR }21\ >\ \text{B: IQR }12

So Team A has the greater spread.

The question / data	Use	Why
Extreme value / skew present	IQR (and median)	robust — an outlier can't move it
Clean, symmetric data	range / SD (and mean)	uses every value
'Compare the spread' of two groups	IQR or range, in context	central-tendency comments score 0
'Mean or median better here?'	median if an outlier exists	name the measure + cite the anomaly

Examiner note

The method mark is stating both quartile values then $\text{IQR}=Q_3-Q_1=\dots$ . For a comparison, write it in context with a spread word (“Team A's heights are more spread out, IQR 21 vs 12”); a comment about who is taller on average scores zero when the question asked about spread.

Common mistake

Two killers: not ordering first, and forgetting to discard the median when

n

is odd. For data rounded “to the nearest cm”, the range uses class boundaries: heights 150 and 169 cm give a range between

168.5-150.5=18

and

169.5-149.5=20

cm.

Common mistake

Decoding the key wrong (writing 14.2 instead of 142), or stating the position (“the 7th value”) instead of the data value sitting there. And a box plot with whiskers through the box, no axis unit, or only one of the two plots labelled.

Why (n+1)/4 here, but n/4 on a CF graph?

On a discrete ordered list you are counting to a specific person, so positions go as $\tfrac{n+1}{4}$ . On a continuous cumulative-frequency graph (next note) you read a smooth curve and enter at $\tfrac n4$ . Keep the two apart — an accept-range hangs on it.

PreviousThe median & choosing an average NextQuartiles from a cumulative frequency graph

Range, IQR & box plots

A mean tells you where the data sits; variation tells you how spread out it is.
Range is the full width from smallest to largest.
IQR is the width of the middle 50%.
IQR ignores the lowest quarter and highest quarter, so outliers cannot wreck it.
The exam core is: median, IQR, box plot, and one clear comparison sentence.

Quartiles by splitting in half

Sort the data, find the median, then the quartiles are the medians of the two halves. Odd $n$ : discard the median first, then split. $\text{IQR}=Q_3-Q_1$ .

Worked example

IQR of an even list (no median to discard)

2, 5, 9, 13, 29, 33, 49, 55 ( $n=8$ ). Median $=\tfrac{13+29}{2}=21$ ; split 4 / 4.

Lower 2, 5, 9, 13 → $Q_1=\tfrac{5+9}{2}=7$ . Upper 29, 33, 49, 55 → $Q_3=\tfrac{33+49}{2}=41$ .

Answer

\text{IQR}=41-7=34

Worked example

IQR of an odd list (discard the median)

Unordered: 69, 17, 43, 6, 73, 77, 39. Sort → 6, 17, 39, 43, 69, 73, 77 ( $n=7$ ). Median = 43 (discard it).

Lower half 6, 17, 39 → $Q_1=17$ . Upper half 69, 73, 77 → $Q_3=73$ .

Answer

\text{IQR}=73-17=56

IQR ignores the extreme quarter at each end, so one extreme value cannot move it, which is why it is the standard measure for comparing spread.

Median & IQR from an ordered display (stem-and-leaf):

Decode the key first — fix the true place value before reading anything.
Count $n$ ; locate $Q_2$ at the $\tfrac{n+1}{2}$ th value, then read $Q_1,\,Q_3$ by inspection as the medians of the lower and upper halves.
Read the data values at those positions (not the positions).
$\text{IQR}=Q_3-Q_1$ ; range = max − min.
State each with its unit.

Worked example

Read median + IQR off a stem-and-leaf

Lengths (mm), key $14\mid2=142$ :

14 | 2 2 4 8 9
15 | 1 3 5 6 7 9
16 | 5 8

Answer

\text{median }153,\ \text{IQR }=158-146=12\text{ mm}

Worked example

Read median + IQR off 15 ordered values

Team A times (min), ordered: 23, 27, 29, 31, 34, 36, 38, 41, 43, 45, 48, 52, 55, 58, 61 ( $n=15$ ). Positions 8th, 4th, 12th: $Q_2=41$ , $Q_1=31$ , $Q_3=52$ . (9709/52/O/N/23 Q4) (9709/52/M/J/24 Q4)

Answer

\text{median }41,\ \text{IQR }=52-31=21\text{ min}

Drawing the box plot the way the mark scheme wants:

One linear scale across the page (≥1 cm per stated unit); label the axis with quantity + unit.
Mark the five-number summary; box from $Q_1$ to $Q_3$ with the median line inside.
Whiskers to min and max — not through the box, not from its corners.
Label each plot (team / company name).

A box-and-whisker plot: the box spans the IQR (Q₁ to Q₃), the line inside is the median, the whiskers reach the min and max.

Q₁Q₃

IQR = Q₃ − Q₁ = 20·Q₃−Q₂ > Q₂−Q₁ → positive skew (right)

Worked example

Two box plots + a spread comparison

Team A (above): five-number summary 23, 31, 41, 52, 61 → IQR 21. Team B: 20, 38, 44, 50, 66 → IQR 12. Drawn on one scale, A's box is wider. (9709/52/M/J/23 Q3)

Answer

\text{A: IQR }21\ >\ \text{B: IQR }12

So Team A has the greater spread.

The question / data	Use	Why
Extreme value / skew present	IQR (and median)	robust — an outlier can't move it
Clean, symmetric data	range / SD (and mean)	uses every value
'Compare the spread' of two groups	IQR or range, in context	central-tendency comments score 0
'Mean or median better here?'	median if an outlier exists	name the measure + cite the anomaly

Examiner note

Common mistake

Two killers: not ordering first, and forgetting to discard the median when

n

is odd. For data rounded “to the nearest cm”, the range uses class boundaries: heights 150 and 169 cm give a range between

168.5-150.5=18

and

169.5-149.5=20

cm.

Common mistake

Why (n+1)/4 here, but n/4 on a CF graph?