1. The three averages
Mode, median and mean
- An average is a single value standing for a whole set of data, and three of them are used here
- The mode is the value occurring most often
- A set can have more than one mode, or none if every value appears equally often
- It is the only average usable with categorical data such as colours
- Put the data in order and the value sitting in the centre is the median
- With an even number of values there is no single middle, so take the midpoint of the two central ones
- Forgetting to order the data first usually gives the wrong median
- Add every value together and divide by how many there were, and the result is the mean
- It need not come out whole, and rounding it to an integer for the sake of tidiness is wrong
- The range is not an average at all: it measures spread, and equals the largest value minus the smallest
Which one to use
- The mean uses every value, which makes it the most representative, but also the most distorted by an extreme value
- The median ignores how far away the extremes are, so it stands up better when the data contains one
- The mode is the only choice for non-numerical data, and is useful where the most common value is what matters
- Where a question asks which average is most appropriate, look for an extreme value and say so
Worked example
All three averages and the range
Find the mode, median, mean and range of 7, 3, 9, 3, 8, 5.
Solution:
- Put the values in order first: 3, 3, 5, 7, 8, 9
- The mode is the value appearing most often, which is 3
- There are 6 values, so the median is midway between the 3rd and 4th
- Median = (5 + 7) ÷ 2 = 6
- The total is 7 + 3 + 9 + 3 + 8 + 5 = 35
- Mean = 35 ÷ 6 = 5.8333…, which is 5.83 to 3 significant figures
- Range = 9 − 3 = 6
2. Averages from a frequency table
Using the frequencies
- A frequency table lists each value alongside how many times it occurs, so the frequencies must be used as counts
- Read the mode as the value against the biggest frequency, not that frequency figure itself
- The mean needs the total of all the data, found by multiplying each value by its frequency and adding
- Mean = Σ(frequency × value) ÷ total frequency
- The median is found by counting through the frequencies until the middle position is reached
- With n values the median sits at position (n + 1) ÷ 2
- Adding an extra column for frequency × value keeps the working visible and is what the method marks reward
Worked example
Averaging when values repeat
The table records the number of goals scored in each of 20 matches. Find the mean number of goals.
| Number of goals | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Number of matches | 5 | 8 | 4 | 3 |
Solution:
- Multiply each number of goals by its frequency
- 0 × 5 = 0, 1 × 8 = 8, 2 × 4 = 8, 3 × 3 = 9
- Total goals = 0 + 8 + 8 + 9 = 25
- Total matches = 5 + 8 + 4 + 3 = 20
- Mean = 25 ÷ 20 = 1.25 goals
- The answer is not a whole number, which is perfectly acceptable for a mean
3. Grouped data
Why it is only an estimate
- Grouped data records how many values fall in each class interval but not the values themselves
- Without the original values the exact total cannot be found, so the mean can only be estimated
- The estimate assumes every value in a class sits at the midpoint of that class
- Find a midpoint by adding the two endpoints and halving
- The phrase "estimate of the mean" in a question is the signal that the data is grouped
The method
- Add a midpoint column, then a frequency × midpoint column
- Total the frequency column and the frequency × midpoint column
- Divide the second total by the first
- Whichever class holds the largest frequency is the modal class
Worked example
An estimate of the mean from grouped data
The table shows the time t minutes taken by 50 people. Calculate an estimate of the mean, and write down the modal class.
| Time (t minutes) | 0 < t ⩽ 10 | 10 < t ⩽ 20 | 20 < t ⩽ 30 | 30 < t ⩽ 40 |
|---|---|---|---|---|
| Frequency | 8 | 14 | 18 | 10 |
Solution:
- Midpoints are (0 + 10) ÷ 2 = 5, then 15, 25 and 35
- Multiply each frequency by its midpoint
- 8 × 5 = 40, 14 × 15 = 210, 18 × 25 = 450, 10 × 35 = 350
- Total of frequency × midpoint = 40 + 210 + 450 + 350 = 1050
- Total frequency = 8 + 14 + 18 + 10 = 50
- Estimate of the mean = 1050 ÷ 50 = 21 minutes
- A frequency of 18 is the largest in the table, so the modal class is 20 < t ⩽ 30
Exam tip
Work out each midpoint from its own class
An estimate of the mean is set in 9 of the 21 papers, and in 8 of them the classes differ in width and the midpoints rise in uneven steps. Work each one out from the two ends of its own class: 30 < t ⩽ 35 has midpoint 32.5, but 35 < t ⩽ 50 has 42.5. The midpoints earn a method mark of their own, and one scheme gives nothing for class widths used instead.
4. Quartiles and spread
Splitting the data into quarters
- The median divides ordered data in half; the quartiles divide it into quarters
- Split the ordered data in half at the median: the middle of the bottom half is the lower quartile, and the middle of the top half is the upper quartile
- Where the number of values is odd, leave the median itself out of both halves
- The interquartile range is the upper quartile minus the lower quartile
- The IQR measures the spread of the middle half of the data, so it is unaffected by extreme values, which is exactly why it is preferred to the range when one is present
Worked example
Quartiles and interquartile range
Find the median, the quartiles, the range and the interquartile range of 4, 6, 7, 9, 11, 12, 14, 15, 18, 20, 23, 28.
Solution:
- The data is already in order and there are 12 values
- The median lies between the 6th and 7th: (12 + 14) ÷ 2 = 13
- The lower half is 4, 6, 7, 9, 11, 12, whose median is (7 + 9) ÷ 2 = 8, the lower quartile
- The upper half is 14, 15, 18, 20, 23, 28, whose median is (18 + 20) ÷ 2 = 19, the upper quartile
- Range = 28 − 4 = 24
- Interquartile range = 19 − 8 = 11
- The IQR is much smaller than the range, since it ignores the extremes at both ends
5. Comparing two sets of data
One average and one spread
- A proper comparison quotes an average and a measure of spread, and interprets both
- A higher mean or median means the values are typically larger
- A smaller range or interquartile range means the values are more consistent or less spread out
- Say what this means in context rather than only stating the numbers
- Where one set contains an extreme value, compare medians and interquartile ranges rather than means and ranges
Working backwards from a mean
- Knowing a mean and how many values there are gives the total, since total = mean × number of values
- That total is what makes a missing value findable
- If five numbers have a mean of 12, their total is 60, so a missing fifth value is 60 minus the sum of the other four
- The same idea handles an added or removed value, by adjusting both the total and the count
Worked example
Working backwards from a mean
Five numbers have a mean of 12. Four of them are 8, 15, 11 and 9. Find the fifth.
Solution:
- The mean is the total divided by how many values there are, so the total can be recovered from it
- Total = mean × number of values = 12 × 5 = 60
- Add the four known values: 8 + 15 + 11 + 9 = 43
- The fifth value is the part of the total not yet accounted for
- 60 − 43 = 17
- Check by finding the mean of all five: (8 + 15 + 11 + 9 + 17) ÷ 5 = 60 ÷ 5 = 12
- Reconstructing the total is the whole method, and it works for any "find the missing value" version of this question
Worked example
Two missing values from the mean and the range
Seven whole numbers have a mean of 11 and a range of 16. Five of them are 3, 8, 9, 12 and 15. The other two are p and q, where p is greater than q. Find p and q.
Solution:
- Total of all seven = mean × number of values = 11 × 7 = 77
- The five known numbers add to 3 + 8 + 9 + 12 + 15 = 47, so p + q = 77 − 47 = 30
- The known numbers run only from 3 to 15, a spread of 12, which is less than the range of 16, so one of the missing numbers must be the largest or the smallest
- Range = largest − smallest, so largest = smallest + range
- Try the bigger missing number as the largest: p = 3 + 16 = 19
- Then q = 30 − 19 = 11
- Check: 11 lies between 3 and 19, so 3 is still the smallest and 19 the largest, and the range is 19 − 3 = 16
- Check the mean: (3 + 8 + 9 + 11 + 12 + 15 + 19) ÷ 7 = 77 ÷ 7 = 11