Mean, median, mode, and skew

Three centres, how each is found, and what a single outlier does to the mean but not to the median.

Three answers to "what is typical?"

Mean $\overset{x}{ˉ} = \frac{\sum x _{i}}{n}$ — the balance point; every value pulls on it in proportion to its distance.
Median — the middle value once the data are sorted. Locate it at position $\frac{n + 1}{2}$ : a whole number when $n$ is odd, and a half-position when $n$ is even, which means "average the two values either side".
Mode — the most frequent value. Useful for categorical data and for spotting a bimodal shape; rarely the answer for a measured variable.

Worked example — Maple Ledger's overdue invoices

Maple Ledger Bookkeeping has ten invoices past due. Days overdue: 3, 7, 2, 12, 5, 9, 4, 30, 6, 7.

Mean. $\sum x = 85$ , so $\overset{x}{ˉ} = 85/10 = 8.5$ days.

Median. Sort: 2, 3, 4, 5, 6, 7, 7, 9, 12, 30. With $n = 10$ the position is $(10 + 1) /2 = 5.5$ , so average the 5th and 6th sorted values: $(6 + 7) /2 = 6.5$ days.

Mode. 7 appears twice, every other value once: mode = 7.

Three different numbers for one "typical" delay. Which one to report depends on what the 30 is doing.

What the 30 does

Suppose the 30-day invoice is collected and drops out of the list. Nine values remain: $\sum x = 55$ , $\overset{x}{ˉ} = 55/9 = 6.11$ . The mean has fallen by 2.4 days. The median is now the 5th of nine sorted values (2, 3, 4, 5, 6, 7, 7, 9, 12) — 6 days, down by half a day.

One value moved the mean from 8.5 to 6.11 and the median from 6.5 to 6. The mean is sensitive to an extreme value because the extreme value's full distance enters the sum; the median is resistant because only the rank of the extreme value matters, and 30 has the same rank as 13 would. For a manager reporting "typical days overdue" while one account is in dispute, the median is the honest centre; the mean is the right one for anything that scales with the total — total interest cost, for example, which the 30 really does drive.

Reading skew from the centres

When a distribution is right-skewed, the tail of large values pulls the mean up past the median: mean > median is the signature of right skew, mean < median of left skew, and mean ≈ median of symmetry. Maple Ledger's 8.5 > 6.5 says right-skewed before any histogram is drawn — and the histogram would confirm it: a hump at 2–9 and one value far out to the right.

The rule is a diagnostic, not a theorem — there are odd shapes where it fails — but on the paper "the mean exceeds the median, so the distribution is right-skewed" is the expected sentence.

The trap in the position rule

$(n + 1) /2$ gives a position, not a value. For $n = 10$ it is 5.5, and the median is the average of the 5th and 6th sorted values, not "5.5" and not the average of the 5th and 6th values in the order they were given. Sort first, every time.