Mean, Median, Mode and Range from a List of Data
Questions designed to stretch thinking, reveal misconceptions, and spark mathematical reasoning.
Convince Me That…
Students must construct a mathematical argument for why each statement is true.
Consider the dataset {3, 3, 3, 5, 11}. The values in order are 3, 3, 3, 5, 11, so the median (middle value) is 3. The mean is (3 + 3 + 3 + 5 + 11) ÷ 5 = 25 ÷ 5 = 5. So the mean is 5 but the median is only 3. This challenges the “mean and median are interchangeable” misconception.
The mean is pulled towards extreme values while the median is not. The single large value of 11 drags the mean upwards, but the median stays anchored at the middle position. Any dataset where the values are bunched towards one end with a “tail” stretching the other way (a skewed distribution) will have the mean and median differ — sometimes dramatically.
Take {1, 1, 9} and {1, 9, 9}. Both have a range of 9 − 1 = 8. But the mean of {1, 1, 9} is 11 ÷ 3 ≈ 3.67, while the mean of {1, 9, 9} is 19 ÷ 3 ≈ 6.33. Same range, very different means. This disproves the “range determines the mean” misconception.
The range only uses two values — the largest and smallest — so it tells you nothing about how the remaining data is distributed between those extremes. You can fill in the “gap” with values close to the minimum, close to the maximum, or anywhere in between, giving completely different means each time.
Start with {2, 5, 5, 5, 8}. The mean is 25 ÷ 5 = 5, the median is 5, and the mode is 5. Now remove the 8 to get {2, 5, 5, 5}. The mean is now 17 ÷ 4 = 4.25, but the median is still (5 + 5) ÷ 2 = 5 and the mode is still 5. This targets the “changing the data affects all measures equally” misconception.
The mean uses every single value in its calculation, so removing any value (other than one exactly equal to the current mean) will shift it. The median depends only on the position of the middle value(s), and the mode depends only on frequencies — neither was affected by removing the 8 from the end.
Consider {1, 2, 3, 4, 100}. The mean is 110 ÷ 5 = 22. The range is 100 − 1 = 99. So the range (99) is much larger than the mean (22). This challenges the “range is always smaller than the mean” misconception.
A single extreme outlier can make the range enormous without dramatically changing the mean. The range is a measure of spread and the mean is a measure of centre — there is no fixed relationship between their sizes.
Give an Example Of…
Think carefully — the fourth box is a trap! Give a non-example that looks right but isn’t.
Example: {4, 4, 5, 7, 10} — mean = 30 ÷ 5 = 6, mode = 4
Another: {4, 4, 6, 7, 9} — mean = 30 ÷ 5 = 6, mode = 4
Creative: {4, 4, 4, 8, 10} — mean = 30 ÷ 5 = 6, mode = 4. Three 4s makes the mode even clearer, with the other values pushed higher to compensate.
Trap: {4, 6, 6, 6, 8} — mean = 30 ÷ 5 = 6, but mode = 6, not 4. A student might think “I included a 4, so the mode is 4” but 6 appears three times while 4 appears only once. The mode is the most frequent value, not just any value you include.
Example: {3, 5, 7, 9, 13} — median = 7, range = 13 − 3 = 10
Another: {1, 6, 7, 8, 11} — median = 7, range = 11 − 1 = 10
Creative: {2, 2, 7, 7, 12} — median = 7, range = 12 − 2 = 10. Repeated values still give the right median and range.
Trap: {2, 4, 7, 9, 10} — median = 7, but range = 10 − 2 = 8, not 10. A student might think the range is the largest value (10) rather than the largest minus the smallest (10 − 2 = 8). This is the “range = largest value” misconception.
Example: {5, 5, 8} — mode = 5, mean = 18 ÷ 3 = 6 > 5
Another: {5, 5, 5, 9} — mode = 5, mean = 24 ÷ 4 = 6 > 5
Creative: {1, 5, 5, 5, 100} — mode = 5, mean = 116 ÷ 5 = 23.2 — the outlier makes the mean massively exceed the mode.
Trap: {3, 5, 5, 7} — mode = 5, but mean = 20 ÷ 4 = 5. The mean equals 5, not greater than 5. A student might assume adding any numbers above 5 guarantees the mean is above 5, but the below-5 values pull it back down. For the mean to exceed the mode, the total of values above 5 must outweigh the total of values below 5.
Example: {1, 1, 5, 6, 7} — mode = 1, median = 5, mean = 20 ÷ 5 = 4. All three are different: 1, 5, 4.
Another: {1, 1, 2, 6, 10} — mode = 1, median = 2, mean = 20 ÷ 5 = 4. All different: 1, 2, 4.
Creative: {3, 3, 5, 8, 11} — mode = 3, median = 5, mean = 30 ÷ 5 = 6. All different (3, 5, 6) — notice the mean is the largest of the three averages here, which surprises students who expect the mode or median to be biggest.
Trap: {3, 5, 5, 5, 7} — mode = 5, median = 5, mean = 25 ÷ 5 = 5. All three measures are 5 — they’re all the same, not all different! A student might think “I used different numbers, so the averages must be different” but symmetric data clustered around a central value often produces identical averages.
Always, Sometimes, Never
Is the statement always true, sometimes true, or never true? Students should justify their decision with examples.
This targets the “mean is always the biggest average” misconception. When data is skewed with high outliers, the mean can exceed the mode — for example, {1, 1, 5, 8} has mode = 1 and mean = 3.75, so the mean is greater. But when data is skewed the other way, the mode can exceed the mean — for example, {3, 7, 7, 7} has mode = 7 and mean = 6, so the mean is smaller. They can also be equal: {5, 5, 5} has mean = mode = 5.
Since the relationship can go either way, this is sometimes true.
This targets the “adding data always changes the spread” misconception. The range is determined by the largest and smallest values: range = max − min. The median always lies between the minimum and maximum (or equals one of them). So adding the median as a new data point can never create a new maximum or a new minimum — it just adds a value within the existing spread.
For example, {2, 5, 11} has median 5 and range 9. Add 5 to get {2, 5, 5, 11}: range is still 11 − 2 = 9. This works for every dataset — because min ≤ median ≤ max, inserting the median can never extend the range.
This targets the “adding the mean preserves all measures” misconception. Adding the mean as a new value always preserves the mean itself, but the median may or may not change because it depends on position, not the total.
True case: {1, 3, 5} has mean = 3 and median = 3. Add 3: the new dataset {1, 3, 3, 5} has median = (3 + 3) ÷ 2 = 3. Median unchanged. False case: {1, 2, 9} has mean = 4 and median = 2. Add 4: the new dataset {1, 2, 4, 9} has median = (2 + 4) ÷ 2 = 3. Median changed from 2 to 3.
This targets the misconception that “identical data values can still produce different measures.” If the range is 0, then max − min = 0, so every value must be identical. Call that value k. The mean is k (sum = nk, divided by n). The mode is k (only value present). Therefore mean = mode = k.
There is no counterexample — a range of 0 forces all values to be the same, which forces the mean, median, and mode all to be equal.
Odd One Out
Which is the odd one out? Can you make a case for each one? There’s no single right answer!
Explain the Mistake
Each example contains a deliberate error targeting a common misconception. Can you find where and why the reasoning goes wrong?
Answer: Median = 9, Range = 8
Reasoning: “The median is the middle number. There are 5 numbers so I want the 3rd one. Counting along: 7, 2, 9 — so the median is 9. The range is 9 minus 1 which is 8.”
The student has fallen for the “median without ordering” misconception — they picked the 3rd value from the data as written, rather than first arranging the values in order. The correct first step is to sort the data: 1, 2, 4, 7, 9. Now the 3rd (middle) value is 4, not 9. The median is 4.
Interestingly, the range calculation of 9 − 1 = 8 is correct, since 9 is the largest value and 1 is the smallest regardless of order. But the median requires the data to be ordered first — this is the step students most commonly skip.
Answer: Mean = 7, Median = 7 ✓
Reasoning: “The mean and the median are basically the same thing — they’re both the average. I put the numbers in order and the middle one is 7, so the mean and median are both 7.”
The student gets the correct answer — the mean is (3 + 5 + 7 + 9 + 11) ÷ 5 = 35 ÷ 5 = 7, and the median is indeed 7 — but their reasoning reveals the “mean = median” misconception. They found only the median (the middle value) and assumed the mean must be the same.
This happens to work here because the data is evenly spaced (an arithmetic sequence), making it symmetric. But for a dataset like {1, 2, 3, 4, 100}, the median is 3 while the mean is 22 — wildly different. The student needs to understand that the mean requires summing all values and dividing by the count, which is a completely different process from finding the middle value.
Answer: Mean = 6, Mode = 6
Reasoning: “I added them up — 4 + 4 + 6 + 8 + 8 = 30 — and divided by 5 to get 6. The mode is the one in the middle, which is also 6.”
The student calculated the mean correctly (30 ÷ 5 = 6), but has confused mode with median — the “mode means middle” misconception. The mode is the most frequently occurring value, not the middle value. In this dataset, 4 appears twice and 8 appears twice, while 6 appears only once.
The dataset is actually bimodal, with modes of 4 and 8. The value 6 is the median (the middle value when the data is ordered), not the mode. Mode is about frequency; median is about position.
Answer: Mode = 8, Range = 8
Reasoning: “The mode is 8 because it appears the most — it comes up twice. The range is the biggest number in the list, which is also 8.”
The student correctly identified the mode as 8 (it appears twice, more than any other value). However, they applied the “range = largest value” misconception. The range is not the largest value — it is the largest value minus the smallest value: 8 − 1 = 7, not 8.
Students often confuse range with the maximum because they remember “range” as meaning “the biggest” rather than “the difference between the biggest and smallest.” A helpful check: if the range were simply the biggest number, then {8, 8, 8} would have a range of 8, but those values have no spread at all — the range should be 0.