Three averages describe where the data sit, and the range and interquartile range describe how spread out they are. Knowing which average to use, and why, earns as many marks as the arithmetic.
Mean \(= \dfrac{\text{sum of values}}{\text{number of values}}\); median = middle value once the data are IN ORDER, at position \(\dfrac{n+1}{2}\); mode = most common; range = largest \(-\) smallest. (Extended) Interquartile range = upper quartile \(-\) lower quartile.
Fig. 1: the eight values 7, 7, 8, 9, 10, 12, 15, 18 on a dot plot, drawn to scale. The large value 18 pulls the mean (10.75) to the right of the median (9.5).
Section 01
Mean, median, mode and range.
In the lab, drag one dot far to the right and watch the four measures. The mean chases the dot, the median barely moves, and the range grows. Then press 'Add an outlier' and compare how much each measure changes.
Mean: add up all the values and divide by how many there are.
Median: put the values in order; the median is the middle one, at position \(\dfrac{n+1}{2}\). With an even number of values it is halfway between the middle two.
Mode: the value that occurs most often. There can be more than one mode, or none.
Range: largest value \(-\) smallest value. It is a single number, not '7 to 18'.
Worked example All four measures
Find the mean, median, mode and range of 12, 7, 9, 15, 7, 10, 18, 8.
Order the data: 7, 7, 8, 9, 10, 12, 15, 18. There are \(n = 8\) values.
Median: position \(\dfrac{8+1}{2} = 4.5\), halfway between the 4th and 5th values, 9 and 10, so \(9.5\).
Mode: 7 appears twice, every other value once.
Range: \(18 - 7 = 11\).
Answer: Mean 10.75, median 9.5, mode 7, range 11 (Fig. 1)
Worked example A missing value from the mean
The mean of five numbers is 8. Four of the numbers are 6, 9, 7 and 11. Find the fifth number.
Total of all five numbers = mean \(\times\) how many \(= 8 \times 5 = 40\).
Total of the four known numbers: \(6 + 9 + 7 + 11 = 33\).
Fifth number: \(40 - 33 = 7\).
Answer: 7
Your turn
A sixth number is added to the five numbers above, and the mean rises to 9. What is the sixth number?
New total \(= 9 \times 6 = 54\). Old total \(= 40\). Sixth number \(= 54 - 40 = 14\).
Examiner trap
Order the data BEFORE finding the median. The middle of the unsorted list 12, 7, 9, 15, 7, 10, 18, 8 is 15 and 7, which gives a wrong 'median' of 11. And \(\dfrac{n+1}{2}\) is the POSITION of the median, not its value.
Section 02
Which average fits?
Each average has a job. Exam questions often ask which one is most suitable and why, so learn the reasons, not just the methods.
Average
Good because
Weakness
Mean
uses every value
dragged by extreme values (outliers)
Median
not affected by extreme values
ignores most of the actual values
Mode
the only average for qualitative data; shows the most popular
may not exist or may not be near the centre
Worked example An outlier
Five people at a small company earn $18,000, $20,000, $21,000, $22,000 and $95,000. Which average best represents a typical salary?
Median: the middle of the ordered list is $21,000.
Four of the five people earn far less than $35,200: the one large salary drags the mean up.
Answer: The median, $21,000, because the mean is distorted by the extreme value $95,000.
Examiner trap
'The mean, because it uses all the values' is not enough when there is an outlier. Look at the data: if one value is far from the rest, the median is the better choice, and you must say why.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
What is the mode of 4, 7, 7, 2, 9, 7, 4?
Quick check+10 XP
What is the range of 13, 5, 21, 8?
Quick check+10 XP
What is the median of 5, 9, 2, 8, 6?
Quick check+10 XP
Which average should be used for the favourite colours of a class?
Section 03
Quartiles and the interquartile range.
Extended only. The quartiles cut the ordered data into four equal parts. The interquartile range (IQR) measures the spread of the middle half of the data, so unlike the range it ignores extreme values. In the lab, tick 'quartiles and IQR' and add an outlier: the range jumps but the IQR hardly moves.
With the data in order and \(n\) values: lower quartile (LQ) at position \(\dfrac{n+1}{4}\), median at \(\dfrac{n+1}{2}\), upper quartile (UQ) at \(\dfrac{3(n+1)}{4}\).
Equivalently, the LQ is the median of the lower half and the UQ is the median of the upper half.
IQR = UQ \(-\) LQ. A smaller IQR means more consistent data.
Worked example Quartiles of 11 values (Extended)
Find the median, quartiles and interquartile range of 3, 5, 6, 8, 9, 11, 12, 14, 15, 18, 21.
The data are already in order, with \(n = 11\).
Median: position \(\dfrac{11+1}{2} = 6\), the 6th value is 11.
LQ: position \(\dfrac{11+1}{4} = 3\), the 3rd value is 6.
UQ: position \(\dfrac{3 \times 12}{4} = 9\), the 9th value is 15.
IQR \(= 15 - 6 = 9\).
Answer: Median 11, LQ 6, UQ 15, IQR 9
Worked example Compare with median and IQR (Extended)
Journey times (minutes). Route A: 12, 15, 18, 20, 22, 25, 30. Route B: 16, 17, 18, 19, 21, 22, 24. Compare the two routes.
Each list has \(n = 7\): LQ is the 2nd value, median the 4th, UQ the 6th.
Route A: median 20, LQ 15, UQ 25, so IQR \(= 10\).
Route B: median 19, LQ 17, UQ 22, so IQR \(= 5\).
Average: Route B is slightly quicker on average (\(19 < 20\)). Spread: Route B is more consistent (IQR \(5 < 10\)).
Answer: Route B is quicker on average and its times are more consistent.
Your turn
(Extended) Find the interquartile range of 14, 9, 22, 17, 11, 25, 19.