A data set is summarised by one number for its centre and one for its spread. Choosing the right pair, calculating them from frequency tables and grouped data, and comparing two sets in context are all standard Paper 5 marks.
The mean \(\bar{x} = \dfrac{\sum fx}{\sum f}\) uses every value, so one extreme value can drag it. The median and interquartile range use only positions, so they resist outliers and skew. Quote a centre AND a spread, and compare them in context.
Fig. 1: one extreme salary drags the mean to 32.1, above six of the seven values. The median, 22, still sits in the middle of the typical salaries
Section 01
Three averages.
In the frequency-table tab of the lab, change one frequency and watch which averages move. Then try the examples: the mean of a frequency table is \(\dfrac{\sum fx}{\sum f}\), the median is found by counting along the cumulative frequencies, and the mode is the value with the highest frequency.
Worked example From a frequency table
The numbers of goals \(x\) scored in 25 matches are: \(x = 0\): 5 matches, 1: 8, 2: 6, 3: 4, 4: 2. Find the mean, median and mode.
\(\sum fx = 0 + 8 + 12 + 12 + 8 = 40\), so \(\bar{x} = \dfrac{40}{25} = 1.6\).
The median is the 13th value. Cumulative frequencies: 5, 13, 19, 23, 25. The 13th value is 1.
The highest frequency is 8, at \(x = 1\).
Answer: Mean 1.6, median 1, mode 1
Worked example A missing value
The mean of 8 numbers is 12. A ninth number is added and the mean becomes 13. Find the ninth number.
Total of the 8 numbers: \(8 \times 12 = 96\).
Total of the 9 numbers: \(9 \times 13 = 117\).
Ninth number \(= 117 - 96\).
Answer: 21
Worked example Combining two groups
In a test, 12 students in group A have mean mark 64 and 18 students in group B have mean mark 59. Find the mean mark of all 30 students.
Combined mean \(= \dfrac{768 + 1062}{30} = \dfrac{1830}{30}\).
Answer: 61
Examiner trap
Averaging the two means. \(\tfrac{1}{2}(64 + 59) = 61.5\) is wrong because the groups are different sizes; always go back to the totals \(\sum x\).
Section 02
Estimates from grouped data.
When data are grouped the raw values are lost, so every answer is an estimate. Use each class midpoint as if every value in the class sat there, and estimate the median by assuming the values are spread evenly through its class.
Time \(t\) (min)
\(0 \le t < 10\)
\(10 \le t < 20\)
\(20 \le t < 30\)
\(30 \le t < 50\)
Frequency
6
13
13
8
Midpoint
5
15
25
40
Worked example Estimated mean and median
Use the table to estimate the mean and the median time, and state the modal class.
Median: the 20th value. Cumulative frequencies 6, 19, 32, 40, so it is the 1st of the 13 values in \(20 \le t < 30\): \(20 + \dfrac{1}{13} \times 10 = 20.8\) minutes.
Modal class: compare frequency DENSITIES because widths differ. \(10 \le t < 20\) and \(20 \le t < 30\) both have density 1.3, higher than 0.6 and 0.4.
Answer: Mean about 21.8 min, median about 20.8 min; two modal classes, \(10 \le t < 20\) and \(20 \le t < 30\)
Examiner trap
Using the upper class boundary (or the class width) instead of the midpoint in \(\sum fx\). For \(30 \le t < 50\) the midpoint is 40, not 50 and not 20.
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
Find the mean of the frequency table \(x = 1\): 4, \(x = 2\): 5, \(x = 3\): 1.
Quick check+10 XP
A data set has one very large outlier. Which average best represents a typical value?
Quick check+10 XP
What value is used for the class \(20 \le x < 30\) when estimating a mean?
Quick check+10 XP
Which measure of spread is NOT affected by a single extreme value?
Section 03
Measuring spread.
The range uses only the two extreme values, so it is very sensitive to outliers. The interquartile range, \(Q_3 - Q_1\), measures the spread of the middle half of the data. The standard deviation (next lesson) uses every value and goes with the mean.
Situation
Centre
Spread
Roughly symmetrical, no outliers
Mean
Standard deviation
Skewed, or with outliers
Median
Interquartile range
Qualitative data, or the most common value is wanted
Mode
None
Worked example Choosing a measure
Seven salaries, in thousands of dollars, are 18, 20, 21, 22, 24, 25, 95 (Fig. 1). Find the mean and the median, and say which better represents a typical salary.
Mean \(= \dfrac{225}{7} = 32.1\) (3 s.f.).
Median: the 4th of 7 values, 22.
Six of the seven salaries are below the mean, because the outlier 95 pulls it up.
Answer: Mean 32.1, median 22; the median is more representative
Your turn
Find the interquartile range of 3, 7, 8, 11, 14, 15, 19, 22, 26, 30, 41.
11 values: \(Q_1\) is the 3rd value, 8, and \(Q_3\) is the 9th value, 26. IQR \(= 26 - 8 = 18\).
Examiner trap
Quoting a quartile POSITION instead of a value: for 11 values \(Q_3\) is the 9th value (26), not 9.
Section 04
Comparing data sets.
A comparison question wants two statements: one about average and one about spread (consistency), each written in the context of the question, and using matching measures for both sets.
Worked example Two runners
Ana's 100 m times in 9 races have median 12.6 s and IQR 0.55 s. Ben's have median 12.7 s and IQR 0.70 s. Compare their performances.
Average: Ana's median is lower, so she was generally faster.
Spread: Ana's IQR is smaller, so her times were more consistent.
Answer: Ana was faster on average and more consistent than Ben
Examiner trap
Saying 'Ana's data are higher' without context. In this question a lower time is better, so say 'faster', and refer to times, not just numbers.
Stage 2 · Exam
Exam-style questions
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Finish the four checks above to unlock the exam questions
Exam style+20 XP
Group A has 12 students with mean mark 64. Group B has 18 students with mean mark 59. Find the mean mark of all 30 students.
Exam style+20 XP
Estimate the mean of the grouped data \(0 \le x < 4\): 3, \(4 \le x < 6\): 7, \(6 \le x < 10\): 10.
Exam style+20 XP
For the frequency table \(x = 0, 1, 2, 3, 4\) with frequencies 3, 7, 5, 6, 9, what is the median?
Exam style+20 XP
Which pair of measures is most suitable for summarising 3, 5, 6, 8, 9, 40?
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Centre and spread, sorted.
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