Averages and spread: mean, median, mode and range
Calculating and comparing measures of central tendency and spread, and understanding the effect of an outlier
About three lessons of 45 to 60 minutes
1 number can't tell the whole story
A news report says 'the average wage in this town is £80,000'. Sounds comfortable, until you learn that 1 billionaire lives there, and everyone else earns around £35,000. The MEAN is genuinely £80,000, but it is a terrible description of what a 'typical' person actually earns, because 1 extreme value dragged it far away from everyone else.
This is exactly why statisticians rarely rely on a single number. The mean, median and mode each answer a slightly different question, and the range tells you how spread out the data is. Choosing the right one, and knowing when an outlier is quietly distorting your answer, is what turns raw numbers into an honest description of reality.
- Average wage skewed by 1 billionairethe mean is pulled far above what almost everyone actually earns; the median would tell a truer story
- A shoe shop's most commonly sold sizethe mode, the single most frequent value, is what a shop actually needs to restock
- A heatwave week's daily temperaturesthe range shows how volatile the week was, not just how hot it was on average
- Median house price in a street with 1 mansionthe median resists the mansion's influence far better than the mean does
What students will be able to do
Students will calculate the mean, median, mode and range of a data set, compare 2 data sets using their mean and range, explain how an outlier affects the mean more than the median, and find a missing value in a data set from its mean.
- I can calculate the mean of a data set.
- I can find the median of a data set, for both an odd and an even number of values.
- I can find the mode of a data set, or explain that there is no mode.
- I can calculate the range of a data set.
- I can compare 2 data sets using their mean and their range.
- I can explain why an outlier affects the mean more than the median, and find a missing value from the mean.
Standards this unit teaches
- KS3 Maths: StatisticsUK National Curriculum (England)Statistics
Statutory requirement (Department for Education, "National curriculum in England: mathematics programmes of study", updated 28 September 2021, Key stage 3, "Statistics" strand, https://www.gov.uk/government/publications/national-curriculum-in-england-mathematics-programmes-of-study/national-curriculum-in-england-mathematics-programmes-of-study): pupils should be taught to "describe, interpret and compare observed distributions of a single variable through: appropriate graphical representation involving discrete, continuous and grouped data; and appropriate measures of central tendency (mean, mode, median) and spread (range, consideration of outliers)".
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
Words to teach and display
- Mean
- the total of all the values divided by how many values there are, the 'share out equally' average
- Median
- the middle value when the data is ordered from smallest to largest (or the mean of the 2 middle values, if there is an even number of values)
- Mode
- the value that appears most often in a data set; a data set can have no mode, 1 mode, or more than 1
- Range
- the largest value minus the smallest value, a simple measure of how spread out a data set is
- Outlier
- a value that is much larger or much smaller than the rest of the data set
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. Mean, median, mode and range
ConcreteThe mean, median and mode are 3 different 'averages', 3 different ways of describing a typical value, and the range describes how spread out the data is. Each answers a different question, so it matters which 1 (or more) a situation actually calls for.
The mean shares the total out equally: add every value, then divide by how many values there are. The median is the middle value once the data is ordered. The mode is whichever value appears most often (a data set can have no mode at all, if every value is different). The range is simply the largest value minus the smallest.
Find the mean, median, mode and range of: 4, 6, 6, 8, 10, 14.
- Mean: sum = 4 + 6 + 6 + 8 + 10 + 14 = 48. Mean = 48 / 6 = 8.
- Median: the data is already ordered. With 6 values, average the 2 middle ones: (6 + 8) / 2 = 7.
- Mode: 6 appears twice, every other value once, so the mode is 6.
- Range: largest - smallest = 14 - 4 = 10.
Answer: Mean = 8, median = 7, mode = 6, range = 10.
- Why might the median differ from the mean for the same data set?
- What does it mean if a data set has no mode?
2. Comparing data sets and outliers
PictorialTo compare 2 data sets, use the mean (or median) to compare a typical value, and the range to compare consistency, a SMALLER range means the data is more tightly clustered, more consistent. An outlier is a value far from the rest of the data, and it can distort the mean while barely moving the median.
Every value in a data set counts equally toward the TOTAL, so 1 extreme value pulls the mean noticeably toward it. The median only cares about POSITION in the ordered list, so a single extreme value barely shifts it, at most it changes which value ends up in the middle.
A data set is 12, 14, 15, 16, 18, 52. The value 52 is an outlier. Find the mean and median with and without the outlier.
- Mean WITH the outlier: (12 + 14 + 15 + 16 + 18 + 52) / 6 = 127 / 6 = 21.2 (1 d.p.).
- Mean WITHOUT the outlier: (12 + 14 + 15 + 16 + 18) / 5 = 75 / 5 = 15.
- Median WITH the outlier (sorted: 12, 14, 15, 16, 18, 52): middle 2 values are 15 and 16, so (15 + 16) / 2 = 15.5.
- Median WITHOUT the outlier (12, 14, 15, 16, 18): the middle value is 15.
Answer: Mean drops sharply, from 21.2 to 15, but the median barely moves, from 15.5 to 15. The mean is pulled toward the outlier; the median is not.
- Why is the median less affected by an outlier than the mean?
- If you wanted to fairly describe 'a typical value' for a data set with an outlier, which average would you choose, and why?
3. Finding a missing value from the mean
AbstractThe mean formula rearranges just like any other formula: total = mean x count. So if you know the mean, the count, and all but 1 of the values, you can find the missing value: missing = total - sum of the known values.
This is a genuinely algebraic use of an average, exactly the kind of 'rearrange formulae to change the subject' skill KS3 algebra also develops, just applied to statistics instead of a geometric formula.
The mean of 5 numbers is 20. Four of the numbers are 18, 19, 21 and 22. Find the missing number.
- Total = mean x count = 20 x 5 = 100.
- Sum of the known values = 18 + 19 + 21 + 22 = 80.
- Missing value = total - known sum = 100 - 80 = 20.
Answer: The missing number is 20.
- Why does 'mean x count' give you the total of all the values?
- If the mean of 4 numbers is 15, what is their total?
Common misconceptions and how to address them
MisconceptionFinding the median by picking the middle number in the list AS GIVEN, without sorting first.
Why it happens: Students treat the data in the order it was written or collected, rather than realising 'median' specifically requires the values to be in numerical order first.
How to address it: ALWAYS sort the values from smallest to largest before finding the middle. For 9, 2, 7, 4, 5, sorted is 2, 4, 5, 7, 9, and the median is 5 (the middle of the SORTED list), not 7 (the middle of the original, unsorted list).
MisconceptionWith an even number of values, picking 1 of the 2 middle values as the median instead of averaging them.
Why it happens: The 'find the middle value' rule learned for an odd number of values does not obviously extend to a data set with 2 middle values.
How to address it: With an EVEN number of values, the median is the MEAN of the 2 middle values, not either one alone. For 4, 6, 6, 8: the middle values are 6 and 6, so the median is (6 + 6) / 2 = 6. For 4, 6, 8, 10: the middle values are 6 and 8, so the median is (6 + 8) / 2 = 7.
MisconceptionIf every value in a data set is different, picking whichever value looks 'closest to the middle' as the mode, rather than saying there is no mode.
Why it happens: Wanting to give a definite answer, some students select a plausible-looking value rather than accepting that a mode only exists when a value repeats.
How to address it: The mode is specifically the MOST FREQUENT value. If every value appears exactly once, there genuinely is no mode. 'No mode' is a valid, correct answer, not a gap that needs filling.
MisconceptionAutomatically deleting an outlier from a data set before calculating anything, without being asked to.
Why it happens: Students recognise an outlier as unusual and instinctively want to ignore it, but an outlier is still a real, valid data point unless a question specifically asks for it to be excluded.
How to address it: Only remove an outlier from a calculation when explicitly asked (e.g. 'find the mean without the outlier'). Otherwise, include every value; an outlier is genuine data, even though it changes the mean a lot.
Guided practice (with answers)
1. Find the mean of: 5, 8, 8, 11, 13.
Answer: 9, because (5 + 8 + 8 + 11 + 13) / 5 = 45 / 5 = 9.
2. Find the median of: 3, 7, 2, 9, 5.
Answer: 5, because sorted the data is 2, 3, 5, 7, 9, and the middle value is 5.
3. Find the mode of: 6, 6, 9, 12, 12, 12.
Answer: 12, because it appears 3 times, more than any other value.
4. Find the range of: 14, 8, 22, 3, 17.
Answer: 19, because 22 - 3 = 19.
5. A data set has mean 10 with 4 values; 3 of them are 7, 9 and 12. Find the missing value.
Answer: 12, because total = 10 x 4 = 40, known sum = 7 + 9 + 12 = 28, so missing = 40 - 28 = 12.
6. A data set is 5, 6, 7, 8, 50. Does removing the outlier 50 change the mean or the median more?
Answer: The mean changes far more (from 15.2 to 6.5, a change of 8.7) than the median (from 7 to 6.5, a change of 0.5).
Independent practice worksheets
Practise mean, median, mode and range, comparing data sets and outliers, and finding a missing value from the mean, with computed, never-wrong answer keys.
Differentiation
- Always write the data set sorted, smallest to largest, before finding the median, every single time, even when it looks unnecessary.
- Use a simple written formula card: mean = total / count, median = middle (sorted), mode = most frequent, range = largest - smallest.
- Practise 'no mode' and 'more than 1 mode' data sets explicitly, so both are seen as normal, correct outcomes, not something to fear.
- For the missing-value problems, always write out 'total = mean x count' as the very first line before doing anything else.
- Investigate how adding 1 new value to a data set changes the mean, median, mode and range differently, by testing several different new values.
- Explore data sets with more than 1 outlier, and discuss whether the median remains a fair 'typical value' description.
- Find a missing value given the MEDIAN (not the mean) of a small data set, a genuinely harder reverse problem since the median depends on ORDER, not a total.
- Research why some official statistics (e.g. median house price, median income) are reported using the median rather than the mean, and write a short justification.
Assessment: exit ticket
A three-question exit ticket sampling the mean, the median, and finding a missing value.
1. Find the mean of: 6, 9, 9, 12, 14.
Answer: 10, because (6 + 9 + 9 + 12 + 14) / 5 = 50 / 5 = 10.
2. Find the median of: 8, 3, 15, 10.
Answer: 9, because sorted the data is 3, 8, 10, 15, and the median is (8 + 10) / 2 = 9.
3. The mean of 4 numbers is 15. Three of them are 11, 14 and 17. Find the missing number.
Answer: 18, because total = 15 x 4 = 60, known sum = 11 + 14 + 17 = 42, so missing = 60 - 42 = 18.
Teacher notes and timings
- Rough timing across 3 lessons: Lesson 1 mean/median/mode/range (section 1), Lesson 2 comparing data sets and outliers (section 2), Lesson 3 finding a missing value from the mean plus the exit ticket (section 3).
- Mode is deliberately computed from the ACTUAL data every time, including the 'no mode' and tied-mode cases, rather than defaulting to a value, so the answer key is honest about when no mode genuinely exists.
- The outlier worked example in section 2 uses real computed differences (8.7 for the mean, 0.5 for the median), not an assumed 'the mean always changes more' rule, since with a small enough data set the median can occasionally move more; every worksheet item computes this rather than asserting it.
- Language to keep repeating: sort before finding the median; the mode is the most frequent value, or there may be none; the mean is pulled toward an outlier, the median is not; total = mean x count.
- Use Student view to project this lesson. Print saves the full teacher unit, including answers and teacher notes; use the linked independent-practice worksheets for student handouts.