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Mean, Median, Mode and Range
Rung 3 of 4 · The traps

The Four Traps That Cost Marks

The methods are easy. The marks get lost in four sneaky details — and every one of them shows up in exams on purpose.


ExplorePick a trap, guess the answer, then “reveal the catch”. Hit “new example” to drill it.
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Knowing the four recipes isn't enough — examiners write questions that punish you for rushing. Here are the four banana skins, in the order people slip on them.

Trap 1 — the Median Needs Ordering First

This is the big one. If the list is 8, 2, 6, 2, 9 and you grab the middle as written, you'll say 6 — and you might get lucky, or you might not. The median is the middle of the ordered list, 2, 2, 6, 8, 9. Always sort before you point at the middle. In the toy, watch the raw strip and the ordered strip disagree.

Trap 2 — an Even Number of Values

With an even count there's no single middle — there are two middle values, and the median is the number halfway between them. For 3, 5, 8, 12 the middle two are 5 and 8, so the median is (5 + 8) ÷ 2 = 6.5. The answer can be a value that isn't even in the list, and that's fine.

Say it plainly: always order first; with an even count, average the middle two; the mode can be none or several; and one outlier hijacks the mean.

Trap 3 — the Mode Can Be Missing, or Doubled Up

If every value appears exactly once, there is no mode — and "no mode" is the correct answer, not zero and not "all of them". And if two values tie for most common, there are two modes (we call that bimodal). Don't assume there's always exactly one.

Trap 4 — One Outlier Drags the Mean

Hit the outlier trap in the toy. A tidy cluster around 6 or 7 plus one value of 50 sends the mean shooting up, even though almost nobody is near it. The median doesn't budge, because it only cares about position, not size. That's why for lopsided data the median is usually the fairer summary — the headline of the final rung.

Us, Thinking Out Loud

Why can the median be a number that isn't in the list?

Give me a real situation where one outlier would make the mean misleading.