IB Applications and Interpretation SL · Descriptive statistics · Section 4 of 5

Quartiles and box-and-whisker diagrams

3 worked examples · By Álvaro López

Key idea

Quartiles divide the ordered data into four groups with the same number of data, and the box-and-whisker diagram draws them together with the extreme values. They summarise a distribution without one odd value distorting it, something the mean cannot manage. In Paper 1 you calculate them by hand using the positions; in Paper 2 the GDC gives them to you, but you must know how to read them from a cumulative frequency graph.

Range

It is the simplest measure of spread, but it uses only two data values. A single very high or very low value makes it jump. That is why quartiles are also used: they are not pulled so much by the extremes.

Quartiles and percentiles

With the data ordered from smallest to largest, the quartiles split them into four parts with roughly the same number of data in each.

Percentiles are the same idea with any percentage. The percentile of order $k$ , $P_{k}$ , is the value that leaves $k %$ of the data below it. Therefore, $Q_{1} = P_{25}$ , $Q_{2} = P_{50}$ and $Q_{3} = P_{75}$ .

Interquartile range

Five-number summary and box-and-whisker diagram

The five-number summary of a distribution is the minimum, $Q_{1}$ , the median, $Q_{3}$ and the maximum. With these five values the box-and-whisker diagram is drawn.

Box-and-whisker diagram of 11 scores with minimum 2, Q1 equal to 5, median 8, Q3 equal to 11 and maximum 13, with no outliers

The box contains the central $50 %$ of the data, and each whisker covers, roughly, a quarter of the data. If the median is not in the centre of the box, or one whisker is much longer than the other, the data are skewed towards one side.

Outliers

A datum far away from the rest should not stretch the whisker out to it. For that purpose, outliers are defined.

The limits are calculated from the quartiles, not from the median or from the extremes.

Quartiles from the cumulative frequency graph

With data grouped into intervals the individual data are not known, so the quartiles are estimated with the cumulative frequency graph. The upper limit of each interval is marked on the horizontal axis, and the cumulative frequency on the vertical axis. The curve starts at the point $(lower limit of the first class, 0)$ and joins the points with straight segments.

Cumulative frequency graph of 40 data with estimated quartiles Q1 = 14, Q2 = 22 and Q3 = 27 minutes, drawn with dashed lines over the polyline (0, 0), (10, 6), (20, 16), (30, 36) and (40, 40)

Worked examples

Example 1 Basic

Waiting times at a counter

At a counter, the number of minutes that each of the 9 customers waited in the morning has been recorded. The data, unordered, are:

$7, 21, 3, 12, 8, 18, 10, 15, 13$

a) Find the range.

b) Calculate $Q_{1}$ , $Q_{2}$ and $Q_{3}$ by hand, using the positions.

c) Calculate the interquartile range.

Solution

First we order the data: $3, 7, 8, 10, 12, 13, 15, 18, 21$ . There are $n = 9$ data.

a) $range = 21 - 3 = 18$ minutes.

b) The positions are $\frac{n + 1}{2} = 5$ , $\frac{n + 1}{4} = 2.5$ and $\frac{3 ( n + 1 )}{4} = 7.5$ .

$Q_{2}$ occupies the fifth place: $Q_{2} = 12$ .
$Q_{1}$ is halfway between the second datum (7) and the third (8): $Q_{1} = \frac{7 + 8}{2} = 7.5$ .
$Q_{3}$ is halfway between the seventh datum (15) and the eighth (18): $Q_{3} = \frac{15 + 18}{2} = 16.5$ .

c) $IQR = Q_{3} - Q_{1} = 16.5 - 7.5 = 9$ minutes.

Example 2 Medium

Test scores and an unusual result

A teacher has marked a test out of 50 points for 11 students. The scores are:

$41, 15, 22, 19, 14, 16, 23, 15, 21, 18, 20$

a) Find the five numbers that summarise the data.

b) Calculate the interquartile range.

c) Find the upper limit for outliers and state which values are outliers.

d) State how far each whisker reaches.

Solution

We order the data: $14, 15, 15, 16, 18, 19, 20, 21, 22, 23, 41$ . There are $n = 11$ data.

a) The positions are $\frac{n + 1}{4} = 3$ , $\frac{n + 1}{2} = 6$ and $\frac{3 ( n + 1 )}{4} = 9$ . So $Q_{1}$ is the third datum (15), $Q_{2}$ the sixth (19) and $Q_{3}$ the ninth (22). The minimum is $14$ and the maximum is $41$ .

The five numbers are $14, 15, 19, 22, 41$ .

b) $IQR = 22 - 15 = 7$ .

c) The limits are $Q_{1} - 1.5 \cdot IQR = 15 - 10.5 = 4.5 Q_{3} + 1.5 \cdot IQR = 22 + 10.5 = 32.5$ The datum $41$ exceeds $32.5$ , so it is an outlier. The $14$ lies within the limits, so it is not an outlier.

d) The lower whisker reaches the minimum, $14$ . The upper whisker does not reach $41$ , because that value is an outlier: it reaches $23$ , the largest datum that is not an outlier.

Example 3 Hard

Calls to the customer service

At a customer service centre, the time taken to answer each of the $n = 40$ calls received in one afternoon has been recorded. The times, in minutes, are grouped as follows:

Time (min)	$[0, 10 [$	$[10, 20 [$	$[20, 30 [$	$[30, 40 [$
Calls	6	10	20	4

The cumulative frequency graph is drawn and the quartiles are estimated by joining the points with straight segments.

a) Estimate $Q_{1}$ , $Q_{2}$ and $Q_{3}$ .

b) Calculate the interquartile range.

c) Estimate how many calls took more than 25 minutes to be answered.

Solution

First the cumulative frequencies are completed: $6$ , $6 + 10 = 16$ , $16 + 20 = 36$ and $36 + 4 = 40$ . The points of the graph are $(0, 0)$ , $(10, 6)$ , $(20, 16)$ , $(30, 36)$ and $(40, 40)$ .

a) The positions are $\frac{n}{4} = 10$ , $\frac{n}{2} = 20$ and $\frac{3 n}{4} = 30$ .

$Q_{1}$ : the horizontal line at $10$ cuts the segment from $(10, 6)$ to $(20, 16)$ . On that segment the curve rises $10$ calls in $10$ minutes, that is, one call per minute. From $6$ to $10$ there are $4$ calls missing, or $4$ minutes: $Q_{1} = 10 + 4 = 14$ minutes.
$Q_{2}$ : the horizontal line at $20$ cuts the segment from $(20, 16)$ to $(30, 36)$ . It rises $20$ calls in $10$ minutes, two per minute. From $16$ to $20$ there are $4$ calls missing, or $2$ minutes: $Q_{2} = 20 + 2 = 22$ minutes.
$Q_{3}$ : the horizontal line at $30$ cuts the same segment. From $16$ to $30$ there are $14$ calls missing; at two per minute that is $7$ minutes: $Q_{3} = 20 + 7 = 27$ minutes.

b) $IQR = 27 - 14 = 13$ minutes.

c) On the segment from $(20, 16)$ to $(30, 36)$ the curve rises two calls per minute. At $25$ minutes the cumulative frequency is $16 + 2 \cdot 5 = 26$ . Therefore, $40 - 26 = 14$ calls took more than $25$ minutes.

In the exam: write the positions $\frac{n}{4}$ , $\frac{n}{2}$ and $\frac{3 n}{4}$ that you use, and the segment on which you read. Without that mark, the value read cannot be checked.

Common mistakes

Summary

Range $= x_{m a x} - x_{m i n}$ . The quartiles $Q_{1}$ , $Q_{2}$ (median) and $Q_{3}$ split the ordered data into four parts: $Q_{1} = P_{25}$ , $Q_{2} = P_{50}$ and $Q_{3} = P_{75}$ .
$IQR = Q_{3} - Q_{1}$ is the width of the box, which contains the central $50 %$ of the data.
Five-number summary: minimum, $Q_{1}$ , median, $Q_{3}$ and maximum. With these values the box-and-whisker diagram is drawn.
Outliers are the data outside $[Q_{1} - 1.5 \cdot IQR, Q_{3} + 1.5 \cdot IQR]$ . Each whisker reaches the most extreme datum that is not an outlier.
With grouped data, on the cumulative graph find $\frac{n}{4}$ , $\frac{n}{2}$ and $\frac{3 n}{4}$ on the vertical axis and read the quartile on the horizontal axis. These are estimated values.

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