Median from a List of Data

To find the median, the first step is always to arrange the values in order from smallest to largest: 2, 3, 5, 7, 9. With 5 values, the median is the value in position (5 + 1) ÷ 2 = 3rd, which is 5. A student who says the median is 9 has likely picked the 3rd value from the list as it was written down (7, 2, 9, 3, 5) without ordering first.

This reveals the “reading off the unordered list” misconception — students treat the list order as meaningful when, for the median, only the sorted order matters. You could show the same five numbers written in a different order (e.g. 5, 9, 3, 2, 7) and ask whether the median has changed — it hasn’t, because the values are the same.

Ordering the values gives 3, 5, 8, 12. There are 4 values (an even count), so no single value sits exactly in the middle. The median is found by taking the mean of the 2nd and 3rd values: (5 + 8) ÷ 2 = 6.5. This is the correct median even though 6.5 is not one of the original data values.

Students often believe the “median must be a value in the data set” misconception. This is true when the data set has an odd number of values, but with an even count the median is the average of the two central values and can fall between them. This question also targets students who don’t know the procedure for even-count data sets — some pick just one of the two middle values (answering 5 or 8) rather than averaging them.

Consider the data set 1, 2, 6, 10, 81. The values in order are 1, 2, 6, 10, 81. The median (middle value of 5) is 6. The mean is (1 + 2 + 6 + 10 + 81) ÷ 5 = 100 ÷ 5 = 20. So the mean is 20 and the median is 6 — very different. The large outlier (81) pulls the mean upward but has no effect on which value sits in the middle position.

This targets the “mean and median are the same thing” misconception. Many students conflate the two averages or assume they must always be close. The key insight is that the mean is sensitive to extreme values while the median is not — one very large or very small value can drag the mean far from the median without affecting the median at all.

The values in order are: 3, 3, 3, 7, 10. With 5 values, the median is the 3rd value, which is 3. It doesn’t matter that 3 appears three times — every occurrence counts as a separate data point when determining position. A student who answers 7 may have collapsed the duplicates into the set {3, 7, 10} and found the middle of those three distinct values instead.

This reveals the “remove duplicates before finding the median” misconception. Students sometimes treat repeated values as a single entry, thinking “I’ve already counted 3.” But in statistics, every data point occupies its own position. The five values represent five observations, and all five must be included when finding the middle position. If a survey of 5 people finds that 3 of them scored 3, those are three separate results.

Example: {3, 6, 8, 11, 14} — ordered, the 3rd value is 8.

Another: {1, 4, 8, 8, 20} — the 3rd value is 8.

Creative: {8, 8, 8, 8, 8} — every value is 8, so the median is trivially 8. Or {−100, −5, 8, 999, 1000} — extreme surrounding values don’t affect the middle position.

Trap: {2, 4, 6, 10, 18} — a student might offer this thinking it works because the mean is (2 + 4 + 6 + 10 + 18) ÷ 5 = 40 ÷ 5 = 8. But the median is the 3rd value = 6, not 8. This exploits the “confusing mean with median” misconception.

Example: {3, 7, 9, 11, 14, 20} — the 3rd and 4th values are 9 and 11. Median = (9 + 11) ÷ 2 = 10.

Another: {1, 5, 10, 10, 15, 25} — the 3rd and 4th values are both 10. Median = (10 + 10) ÷ 2 = 10.

Creative: {−50, −1, 8, 12, 100, 200} — median = (8 + 12) ÷ 2 = 10. The wild surrounding values are irrelevant.

Trap: {2, 5, 10, 13, 17, 19} — a student might choose this because 10 appears in the data. But the 3rd and 4th values are 10 and 13, so the median = (10 + 13) ÷ 2 = 11.5, not 10. This exploits the “the median is just a value I can see in the list” misconception — for even-count data, you must average the two middle values, not just pick one that looks central.

Example: {1, 2, 3, 4, 20} — median = 3, mean = 30 ÷ 5 = 6. Different.

Another: {1, 1, 2, 10, 11} — median = 2, mean = 25 ÷ 5 = 5. Different.

Creative: {1, 1, 1, 1, 96} — median = 1, mean = 100 ÷ 5 = 20. A single outlier creates a huge gap between mean and median.

Trap: {2, 4, 6, 8, 10} — a student might offer any “nice” set of numbers, but for this symmetric set: median = 6, mean = 30 ÷ 5 = 6. They are the same! This exploits the “any set of different numbers will have different mean and median” misconception — symmetric or evenly spaced data sets always have mean = median.

Example: {4, 4, 4, 7, 10} — ordered: 4, 4, 4, 7, 10. Median = 3rd value = 4. Smallest = 4.

Another: {1, 1, 1, 5, 8} — median = 1 = smallest.

Creative: {0, 0, 0, 0, 1000} — median = 0 = smallest. The first three values being equal forces the median down to the minimum, no matter how large the remaining values are.

Trap: {3, 3, 5, 7, 9} — a student might offer this because 3 is the smallest and appears twice, so it “feels” dominant. But the median is the 3rd value = 5, not 3. Another excellent trap is {5, 5, 2, 7, 9}. A student might think “5 appears the most and is the median, so the lowest distinct number is 2… wait, 2 is the smallest.” This exploits the “confusing frequency (mode) with position (median)” misconception and forces them to reconcile the unordered list trap with the smallest value trap simultaneously.

The median depends only on the values in the data set, not on the order they happen to be written. Rearranging {4, 1, 7} into {7, 4, 1} or {1, 7, 4} does not change the sorted order (1, 4, 7) and therefore does not change the median (4). This is because finding the median always starts with sorting — the presentation order is discarded.

Students who believe this is sometimes or always true are revealing the “the list order matters” misconception — the same error that leads students to pick the middle item from an unordered list. Establishing that the median is a property of the set of values (not their arrangement) is foundational.

True case: {1, 3, 5, 7, 9} has 5 values. The median is the 3rd value = 5, which is in the data set. This is always the case for data sets with an odd number of values — the median is literally one of the values.

False case: {2, 3, 8, 11} has 4 values. The median = (3 + 8) ÷ 2 = 5.5, which does not appear in the data set. When the data has an even number of values and the two middle values differ, the median falls between them and is not in the data set. This targets the “median must be in the data” misconception.

With 7 values, the correct median position is found using the formula (n + 1) ÷ 2. Here, (7 + 1) ÷ 2 = 4th. This means 3 values sit below the median and 3 sit above it, confirming it is the true middle. This is always the case regardless of what the 7 values are.

Students who get this wrong often use the “wrong position formula” misconception, such as n ÷ 2 = 3.5 (leading them to arbitrarily pick the 3rd or 4th value with no clear rule), or simply counting to “roughly the middle” without a systematic method. Establishing the (n + 1) ÷ 2 formula and verifying it with small cases builds reliable technique.

True case: {1, 3, 5} — median = 3, a whole number.

False case: {1, 2, 5, 8} — median = (2 + 5) ÷ 2 = 3.5, not a whole number. This happens whenever the data set has an even number of values and the two middle values add to an odd total. Students who say “always” are revealing the “whole-number inputs always give whole-number outputs” misconception — they forget that the averaging step can produce a decimal.

True case: For the data set {1, 2, 3}, the median is 2. If we add the value 4, the set becomes {1, 2, 3, 4} and the median changes to 2.5.

False case: For the data set {1, 5, 5, 5, 9}, the median is 5. If we add another 5, the set becomes {1, 5, 5, 5, 5, 9} and the median remains 5. This tests a student’s structural understanding of position and how dynamic data affects central tendency.

The student has applied the “pick the middle from the unordered list” misconception. They correctly identified that the median of 5 values is the 3rd, but they counted from the list as it was written (12, 4, 8, 1, 6) without sorting first.

The correct approach is to order the values: 1, 4, 6, 8, 12. The 3rd value in the ordered list is 6, not 8. Finding the median always requires sorting the data first — the order the numbers were originally written in is irrelevant.

The student got the correct answer (4) but through the “pick the middle from the unordered list” misconception. They picked the 3rd value from the list as written (6, 1, 4, 9, 2) without ordering. The correct method is to sort first: 1, 2, 4, 6, 9 — the 3rd value happens to also be 4, so the answer is coincidentally correct.

This reasoning would fail on many other lists — for example, applied to the list 9, 1, 3, 7, 5, the student would answer 3, but the correct median is 5 (after ordering: 1, 3, 5, 7, 9). Getting the right answer by luck doesn’t mean the method is sound.

The student has used the “wrong position formula for even-count data” misconception. They divided the count by 2 and used the result as a position (2nd value = 5), rather than averaging the two central values.

The student’s formula (n ÷ 2) only identifies one of the two middle positions. The correct formula to find the median position is (n + 1) ÷ 2. For even-count data, this will give a decimal result (like 2.5), which reminds you to average the values at the 2nd and 3rd positions. Here that gives values 5 and 8, so the true median is 6.5.

The student has applied the “remove duplicates before finding the median” misconception. They reduced {5, 5, 5, 8, 12} to the set of distinct values {5, 8, 12} and then found the median of that reduced set.

But in statistics, every data point counts — the original data set has 5 values, not 3. Keeping all five values in order: 5, 5, 5, 8, 12. The median is the 3rd value = 5. Each occurrence of 5 represents a separate observation and must be included. If three students scored 5 on a test, those are three separate results, not one.

The student has applied the “zero is nothing” misconception. They treated 0 as an absence of data rather than a valid data point occupying a position in the list.

Zero must be included when finding the median. The correct ordered list of all 5 values is −4, −1, 0, 2, 5. The median is the 3rd value, which is 0.

The student got the correct answer (6) but through the “confusing mean with median” misconception. They calculated the mean by summing the values and dividing by the count.

Because this specific data set is perfectly symmetrical, the mean and median happen to be exactly the same. However, this method will fail on almost any other data set. This highlights the danger of arriving at the right answer using the wrong mathematical method.