Median, Quartiles and Interquartile Range

AS · S1 · 14 min

The median is the middle value when the data are put in order, and the quartiles cut the data into quarters. Together they describe a data set in a way that ignores a few extreme values: the median is the typical value, and the interquartile range measures how spread out the middle half is. The syllabus lists median, range and interquartile range among the measures you must "understand and use", and they appear in nearly every Paper 5 data question: from raw lists, stem-and-leaf diagrams, frequency tables and cumulative frequency graphs.

The median

Definition

The median is the middle value of the data when arranged in order of size. For $n$ ordered values it is the $\tfrac{n+1}{2}$ th value. When $n$ is even this position is a half, and the median is the mean of the two middle values.

$n = 9$ : the $5$ th value.
$n = 10$ : halfway between the $5$ th and $6$ th values.

The position formula gives where the median is, not what it is. A common slip is to write "median $= 5.5$ " when $5.5$ was the position.

Quartiles

The median splits the data into a lower half and an upper half. The quartiles are the medians of those halves.

Definition

The lower quartile $Q_1$ is the median of the lower half of the ordered data.
The upper quartile $Q_3$ is the median of the upper half.
When $n$ is odd, the median itself belongs to neither half.
The interquartile range is $\text{IQR} = Q_3 - Q_1$ .
The range is the largest value minus the smallest value.

Key result

$n$	Median	Lower half	$Q_1$	$Q_3$
$11$	$6$ th	values $1$ – $5$	$3$ rd	$9$ th
$12$	mean of $6$ th, $7$ th	values $1$ – $6$	mean of $3$ rd, $4$ th	mean of $9$ th, $10$ th
$15$	$8$ th	values $1$ – $7$	$4$ th	$12$ th
$16$	mean of $8$ th, $9$ th	values $1$ – $8$	mean of $4$ th, $5$ th	mean of $12$ th, $13$ th
$20$	mean of $10$ th, $11$ th	values $1$ – $10$	mean of $5$ th, $6$ th	mean of $15$ th, $16$ th

Tip

Textbooks differ slightly on quartiles for small data sets (some use the $\tfrac{n+1}{4}$ th value with interpolation). The halves method above is clear and widely accepted for Paper 5; whatever method you use, state the positions so the examiner can follow you. For large grouped data sets the question will expect a cumulative frequency graph, where these small differences vanish.

Why the IQR and not just the range

The range depends on only two values, the two most extreme, so a single unusual value can make it enormous. The IQR describes the spread of the middle $50\%$ of the data, so it ignores the top and bottom quarters, where outliers live. The IQR goes naturally with the median; the standard deviation goes naturally with the mean.

Percentiles

The $k$ th percentile is the value below which $k\%$ of the data lie. The lower quartile is the $25$ th percentile, the median the $50$ th and the upper quartile the $75$ th. Percentiles are almost always estimated from a cumulative frequency graph, at cumulative frequency $\tfrac{k}{100}n$ . The interval between the $10$ th and $90$ th percentiles (the $10$ to $90$ interpercentile range) contains the middle $80\%$ of the data.

The median of a frequency table

When discrete data are given in a frequency table, you cannot see the ordered list, but you can find any position using cumulative frequencies.

Method

Add a running total (cumulative frequency) column.
Work out the position(s) needed: $\tfrac{n+1}{2}$ for the median; the middle of each half for the quartiles.
The value at a position is the first value whose cumulative frequency reaches that position.

The median of grouped data

For grouped data, the median is estimated from a cumulative frequency graph at $\tfrac{n}{2}$ , or calculated by linear interpolation:

\text{median} \approx L + \frac{\tfrac{n}{2} - F}{f} \times w,

where $L$ is the lower boundary of the class containing the median, $F$ the cumulative frequency before that class, $f$ its frequency and $w$ its width. Full method and examples are in Cumulative frequency graphs.

Properties of the median

It is not affected by extreme values: making the largest value ten times larger leaves the median unchanged.
It is a good average for skewed data or data with outliers.
It does not use the actual size of every value, only their order, so it uses less information than the mean.
Two medians cannot be combined to give the median of the combined data (unlike means, which combine through totals).

Worked examples

Median and quartiles of raw data (routine)

Find the median, quartiles and interquartile range of:

12,\ 15,\ 9,\ 22,\ 18,\ 11,\ 25,\ 14,\ 19,\ 16

Solution

Ordered: $9, 11, 12, 14, 15, 16, 18, 19, 22, 25$ . $n = 10$ .

Median: halfway between the $5$ th and $6$ th values, $\tfrac{15 + 16}{2} = 15.5$ .

Lower half: $9, 11, 12, 14, 15$ , so $Q_1 = 12$ . Upper half: $16, 18, 19, 22, 25$ , so $Q_3 = 19$ .

$\text{IQR} = 19 - 12 = 7$ . (Range $= 25 - 9 = 16$ .)

Median from a frequency table

The numbers of children in $50$ families:

Number of children	$0$	$1$	$2$	$3$	$4$	$5$
Frequency	$6$	$11$	$14$	$9$	$7$	$3$

Find the median and the interquartile range.

Solution

Cumulative frequencies: $6, 17, 31, 40, 47, 50$ .

$n = 50$ : the median is halfway between the $25$ th and $26$ th values. Values $18$ to $31$ are all $2$ , so both are $2$ : median $= 2$ .

Lower half: values $1$ to $25$ . Its median is the $13$ th value. Values $7$ to $17$ are $1$ , so $Q_1 = 1$ .

Upper half: values $26$ to $50$ . Its median is the $13$ th of these, i.e. the $38$ th value overall. Values $32$ to $40$ are $3$ , so $Q_3 = 3$ .

$\text{IQR} = 3 - 1 = 2$ children.

Effect of an extreme value

The times, in seconds, for $13$ swimmers are:

31,\ 28,\ 35,\ 40,\ 26,\ 33,\ 29,\ 38,\ 30,\ 27,\ 36,\ 32,\ 95

(a) Find the mean and the median. (b) The $95$ is found to belong to a swimmer who stopped mid-race, and is removed. Find the new mean and median. (c) Which average was more representative of the original data? Explain.

Solution

(a) $\sum x = 480$ , so the mean is $\tfrac{480}{13} = 36.9$ s (3 s.f.). Ordered: $26, 27, 28, 29, 30, 31, 32, 33, 35, 36, 38, 40, 95$ . The median is the $7$ th value, $32$ s.

(b) Without $95$ : $\sum x = 385$ for $12$ values, mean $= 32.1$ s (3 s.f.). Median: halfway between the $6$ th and $7$ th, $\tfrac{31 + 32}{2} = 31.5$ s.

(c) The median. In (a), $12$ of the $13$ times are $40$ s or less, yet the mean ( $36.9$ s) is larger than all but two of the times, because the single extreme value pulls it up. The median ( $32$ s) barely changes when the extreme value is removed.

An unknown frequency and the median (exam-hard)

The scores of some students are shown.

Score	$1$	$2$	$3$	$4$	$5$
Frequency	$4$	$7$	$k$	$6$	$3$

Given that the median score is $3$ , find the smallest possible value of $k$ .

Solution

$n = 20 + k$ . The first $11$ values are $1$ or $2$ ; the scores of $3$ occupy positions $12$ to $11 + k$ .

The median is $3$ when the middle position (or both middle positions, if $n$ is even) lies in positions $12$ to $11 + k$ .

Try small values:

$k = 1$ : $n = 21$ , median is the $11$ th value, which is $2$ . No.
$k = 2$ : $n = 22$ , median is the mean of the $11$ th and $12$ th values, $\tfrac{2 + 3}{2} = 2.5$ . No.
$k = 3$ : $n = 23$ , median is the $12$ th value, which is $3$ . Yes.

For larger $k$ the middle position stays within the block of $3$ s, so the smallest value is $k = 3$ .

Reconstructing a data set (exam-hard)

Five positive integers have median $6$ , mean $7$ , a single mode $4$ , and range $9$ . Find the five integers.

Solution

Write them in order $a \le b \le c \le d \le e$ . The median is $c = 6$ .

The mode is $4$ , and $4 < 6$ , so $4$ must appear at least twice among $a, b$ : $a = b = 4$ . Then $d$ and $e$ must be different from each other (otherwise there would be a second mode).

Range: $e - a = 9$ , so $e = 13$ . Mean: the total is $5 \times 7 = 35$ , so $d = 35 - 4 - 4 - 6 - 13 = 8$ .

The integers are $4, 4, 6, 8, 13$ . Check: the mode is $4$ alone, the median is $6$ , the mean is $7$ and the range is $9$ .

Watch out

Using $\tfrac{n+1}{2}$ for a cumulative frequency graph. For raw data and frequency tables use positions ( $\tfrac{n+1}{2}$ th value). For a cumulative frequency graph of grouped data read at $\tfrac{n}{2}$ .

Watch out

Not ordering the data first. The median of $12, 15, 9, 22, 18$ is not $9$ (the middle of the list as written); it is $15$ , the middle of $9, 12, 15, 18, 22$ .

Watch out

Reading the frequency, not the value. In a frequency table, the median is a value of the variable (number of children), never a frequency or a cumulative frequency.

Exam tip

State the positions: " $n = 50$ , so the median is the mean of the $25$ th and $26$ th values". It earns the method mark even if you misread the table.
For grouped data say "estimate". For raw data the median is exact.
When asked for "a measure of spread" to go with the median, give the IQR. When asked to compare, quote both groups' values.
Questions that say "explain why the median might be preferred to the mean" want: the data contain an extreme value (or are skewed), which affects the mean but not the median.

Summary

Median: the $\tfrac{n+1}{2}$ th ordered value; mean of the two middle values when $n$ is even.
$Q_1$ , $Q_3$ : medians of the lower and upper halves (exclude the median when $n$ is odd).
$\text{IQR} = Q_3 - Q_1$ , the spread of the middle $50\%$ ; range $=$ largest $-$ smallest.
Frequency tables: use cumulative frequencies to locate positions.
Grouped data: estimate from a cumulative frequency graph at $\tfrac{n}{2}$ , $\tfrac{n}{4}$ , $\tfrac{3n}{4}$ .
The median and IQR are not affected by extreme values; use them for skewed data or data with outliers.

Practice questions

Question

Find the median and interquartile range of $7, 3, 9, 12, 5, 8, 10, 4, 6$ .
Find the median and interquartile range of $2.4, 3.1, 1.8, 2.9, 3.5, 2.2, 4.0, 2.6$ .
The shoe sizes of $40$ people are: size $4$ ( $2$ people), $5$ ( $5$ ), $6$ ( $9$ ), $7$ ( $11$ ), $8$ ( $7$ ), $9$ ( $4$ ), $10$ ( $2$ ). Find the median and interquartile range.
For the shoe sizes in question 3, calculate the mean and say whether the mean, median or mode would be most useful to a shoe shop deciding which size to stock most of.
The masses of $60$ parcels are grouped as $0 \le m < 2$ ( $12$ ), $2 \le m < 4$ ( $20$ ), $4 \le m < 8$ ( $18$ ), $8 \le m < 12$ ( $10$ ) (kg). Estimate the median mass by linear interpolation.
The salaries of the $9$ employees of a small company are (in thousands of dollars): $18, 20, 21, 21, 23, 24, 26, 28, 150$ . Find the mean and the median, and explain which better represents a typical salary.
The numbers of pets owned by some families are shown: $0$ pets ( $5$ families), $1$ ( $9$ ), $2$ ( $8$ ), $3$ ( $k$ ), $4$ ( $4$ ). The median number of pets is $2$ . Find the set of possible values of $k$ .
Seven integers have mean $10$ , median $9$ , a single mode $7$ , range $12$ , smallest value $5$ and interquartile range $8$ . Find the seven integers.

Answers

Ordered: $3, 4, 5, 6, 7, 8, 9, 10, 12$ ; $n = 9$ . Median $=$ $5$ th $= 7$ . Lower half $3, 4, 5, 6$ : $Q_1 = 4.5$ . Upper half $8, 9, 10, 12$ : $Q_3 = 9.5$ . IQR $= 5$ .
Ordered: $1.8, 2.2, 2.4, 2.6, 2.9, 3.1, 3.5, 4.0$ ; $n = 8$ . Median $= \tfrac{2.6 + 2.9}{2} = 2.75$ . $Q_1 = \tfrac{2.2 + 2.4}{2} = 2.3$ . $Q_3 = \tfrac{3.1 + 3.5}{2} = 3.3$ . IQR $= 1.0$ .
Cumulative frequencies: $2, 7, 16, 27, 34, 38, 40$ . Median: mean of the $20$ th and $21$ st values, both size $7$ , so the median is $7$ . $Q_1$ : median of values $1$ to $20$ , mean of the $10$ th and $11$ th, both $6$ , so $Q_1 = 6$ . $Q_3$ : median of values $21$ to $40$ , mean of the $30$ th and $31$ st, both $8$ , so $Q_3 = 8$ . IQR $= 2$ .
$\sum fx = 8 + 25 + 54 + 77 + 56 + 36 + 20 = 276$ , so the mean is $\tfrac{276}{40} = 6.9$ . The mode (size $7$ ) is the most useful: the shop wants the size most people take, and $6.9$ is not a shoe size.
$n = 60$ , so read at $30$ . Cumulative frequencies $12, 32, 50, 60$ ; $30$ lies in $2 \le m < 4$ : median $\approx 2 + \tfrac{30 - 12}{20} \times 2 = 3.8\ \text{kg}$ .
$\sum x = 331$ , mean $= \tfrac{331}{9} = 36.8$ thousand dollars (3 s.f.). Median $=$ $5$ th value $= 23$ thousand dollars. The median: the one very large salary ( $150$ ) pulls the mean above eight of the nine salaries, while the median is unaffected.
$n = 26 + k$ . Positions: $0$ s are $1$ – $5$ , $1$ s are $6$ – $14$ , $2$ s are $15$ – $22$ , then $3$ s are $23$ to $22 + k$ . The median is $2$ when the middle position(s) lie in $15$ – $22$ . If $n$ is odd the median is the $\tfrac{27 + k}{2}$ th value: need $15 \le \tfrac{27 + k}{2} \le 22$ , i.e. $3 \le k \le 17$ . If $n$ is even both the $\tfrac{26 + k}{2}$ th and the next value must be in $15$ – $22$ : need $\tfrac{26 + k}{2} \ge 15$ and $\tfrac{26 + k}{2} + 1 \le 22$ , i.e. $4 \le k \le 16$ . Combining (and checking $k = 2$ gives median $1.5$ and $k = 18$ gives $2.5$ ): $3 \le k \le 17$ .
In order: $5, b, c, 9, e, f, g$ . Range $12$ gives $g = 17$ . The only values below the median are $5$ , $b$ and $c$ , and $7$ must appear at least twice, so $b = c = 7$ . With $n = 7$ , $Q_1$ is the $2$ nd value ( $7$ ) and $Q_3$ the $6$ th ( $f$ ), so $f - 7 = 8$ and $f = 15$ . The total is $7 \times 10 = 70$ , so $e = 70 - (5 + 7 + 7 + 9 + 15 + 17) = 10$ . The integers are $5, 7, 7, 9, 10, 15, 17$ . Check: ascending order, only $7$ repeats, median $9$ , mean $10$ , range $12$ , IQR $8$ .

The median

Quartiles

Why the IQR and not just the range

Percentiles

The median of a frequency table

The median of grouped data

Properties of the median

Worked examples

Practice questions

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