Averages and Spread
An average describes a typical value; a measure of spread describes how varied the values are. Calculate them accurately, then explain what they say about the data in its context.
Foundation & Higher. Quartiles and interquartile range are taught in the later Higher lessons. Use frequency tables and arithmetic with integers and decimals.
Mean, median, mode and range
Mean = sum of values ÷ number of values. The median is the middle of the ordered list; with an even count, average the two middle values. The mode is the most frequent value. The range = largest − smallest.
Worked example — summarise a small dataset
Times in minutes are 4, 7, 7, 8, 10, 12. They are already ordered. Sum = 48, so mean = 48/6 = 8 min. The middle values are 7 and 8, so median = (7 + 8)/2 = 7.5 min. Mode = 7 min; range = 12 − 4 = 8 min.
For an odd count, such as 3, 5, 8, 9, 12, the median is the third value, 8. A dataset can have more than one mode, or no unique mode. The mean need not be one of the observed values.
Choose an average and interpret outliers
An outlier is a value unusually far from the main pattern. Check whether it is a recording error or a genuine unusual case before deciding what to do with it.
Worked example — change one extreme value
Replace 12 by 42 in the six times above. The total rises from 48 to 78, so the mean rises from 8 to 13 min. The median remains 7.5 min; the range becomes 42 − 4 = 38 min.
The median better describes the centre of the main cluster here because the extreme time pulls the mean upwards. The mean is useful when every value should contribute, such as finding total time from mean time × number.
Use the mode for a most popular category, such as travel method; a mean of category labels is meaningless. Range is easy to calculate but uses only the two extreme values.
Use frequencies as weights
Each frequency tells you how often a value contributes. Multiply value x by frequency f, add the products, then divide by the total frequency.
| x | f | xf |
|---|---|---|
| 0 | 2 | 0 |
| 1 | 3 | 3 |
| 2 | 5 | 10 |
| 3 | 1 | 3 |
| 4 | 1 | 4 |
| All | 12 | 20 |
Worked example — find the mean and median from a frequency table
The total number of books is 20, shared across 12 students: mean = 20/12 = 5/3 ≈ 1.67 books per student. Do not divide by the five rows.
The median uses the 6th and 7th observations. The first two are 0; the next three are 1, reaching position 5; positions 6–10 are 2. Therefore median = 2. Mode = 2; range = 4 − 0 = 4.
You can use running frequencies to locate positions without rewriting the whole list. For an even total n, use positions n/2 and n/2 + 1; for an odd total use (n + 1)/2.
Work backwards and combine groups correctly
Worked example — find a missing value
Five scores have mean 12, so their total is 5 × 12 = 60. Four scores are 9, 10, 13 and 14, totalling 46. The missing score is 60 − 46 = 14.
Worked example — combine groups of different sizes
Eight students have mean score 14, and twelve have mean score 19. Their totals are 8 × 14 = 112 and 12 × 19 = 228. Combined mean = (112 + 228)/(8 + 12) = 17.
The average of 14 and 19 is 16.5, which gives equal weight to unequal groups. Rebuild the totals first.
Adding a new value equal to the old mean leaves the mean unchanged. Adding a value above the old mean increases it; use the new total and new count to calculate the exact change.
Estimate from grouped data
Grouping hides individual values. To estimate a mean, represent each class by its midpoint: (lower endpoint + upper endpoint) ÷ 2. Multiply midpoint m by frequency f.
| Class | f | m | fm |
|---|---|---|---|
| 0–10 | 4 | 5 | 20 |
| 10–20 | 12 | 15 | 180 |
| 20–40 | 16 | 30 | 480 |
| 40–60 | 8 | 50 | 400 |
| All | 40 | — | 1080 |
Worked example — estimate the mean journey time
Estimated total = 20 + 180 + 480 + 400 = 1080 min. Estimated mean = 1080/40 = 27 min. This treats each time as its class midpoint. The exact mean cannot be recovered from these frequencies.
The modal class is 20–40 minutes, with the largest frequency, 16. This is a class, not the value 30. The 20th and 21st observations lie in that same class because the running totals are 4, 16, 32, 40: it is also the median class. Its midpoint is not an exact median.
The actual smallest and largest values are unknown. The endpoints suggest an estimated overall range of 60 − 0 = 60 min, but do not call it an exact range. The Higher cumulative-frequency lesson estimates a median more carefully within a class.
Compare centre and spread in context
Worked example — support both parts of a comparison
Bus A has median journey time 22 min and range 10 min. Bus B has median 26 min and range 6 min. A is typically quicker because its median is 4 min lower. B is more consistent by the range because its range is 4 min smaller.
Do not say A is always quicker or B is better in every way. The summaries do not describe every individual journey. Compare the same measure, units and kind of journey; sample size and selection also affect how far you can generalise.
If every value increases by 3, the mean and median increase by 3 while the range stays unchanged. If every value is doubled, the mean, median and range double. A translation changes location; a positive scaling changes both location and spread.
Think Like a Mathematician
Two datasets can have the same mean and very different spreads. A useful comparison usually needs one measure of centre and one of spread, each tied to what the values actually represent.
Common Mistakes
Unsorted median: order the values before finding the middle.
Frequency means: divide by the number of observations, not rows.
Grouped means: label the answer as an estimate.
Unsupported comparison: quote a statistic and explain its contextual meaning.
Practice
1. Find mean, median, mode and range for 2, 3, 3, 6, 11.
Sum 25, count 5: mean 5. Ordered middle and mode are both 3. Range = 11 − 2 = 9.
2. Find the median of 9, 2, 5, 12, 4, 8.
Order: 2, 4, 5, 8, 9, 12. Median = (5 + 8)/2 = 6.5.
3. Values 1, 2 and 4 have frequencies 3, 2 and 5. Find the mean and median.
Total frequency 10; weighted total 3 + 4 + 20 = 27. Mean 2.7. Positions 5 and 6 are 2 and 4, so median 3.
4. Four scores have mean 15. Three are 11, 17 and 14. Find the fourth.
Required total 4 × 15 = 60; known total 42; missing score 18.
5. A group of 5 has mean 10; a group of 15 has mean 14. Find the combined mean.
(5 × 10 + 15 × 14)/20 = 260/20 = 13.
6. Classes 0–10, 10–20, 20–30 have frequencies 2, 5, 3. Estimate the mean.
Midpoints 5, 15, 25. Estimated mean = (10 + 75 + 75)/10 = 16.
7. For that grouped table, identify the modal and median classes.
Both are 10–20. Its frequency 5 is greatest; positions 5 and 6 lie there because running totals are 2, 7, 10.
8. Why might the median be more useful than the mean for a street with one extremely expensive house?
The expensive outlier pulls the mean upwards. The median is less affected and can better represent a typical house in the main group.
9. A has mean 31 s and range 8 s; B has mean 28 s and range 12 s. Compare.
B is faster on average by 3 s; A is more consistent by the range, which is 4 s smaller. These summaries do not say every B time is faster.
10. Every value increases by 5. A mean was 12 and a range was 9. State the new mean and range.
New mean 17; range remains 9 because both extremes increase by 5.
Frequently Asked Questions
Can a mean number of people or items be a decimal?
Yes. An average may be a decimal even when each actual count is a whole number.
Is the midpoint always the median of a class?
No. It is a convenient representative for an estimated mean. Individual values may be unevenly distributed within the class.
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