Which centre to report for money-like data

Two centres that disagree

On the twelve amounts the mean is 511.67 and the median 445. They are not both wrong; they measure different things. The mean is the balance point — every dollar pulls on it. The median is the halfway rank — only the order of values matters, not how far out the extremes sit. When they disagree, the direction of the disagreement tells you the shape of the distribution.

Right-skewed (positively skewed) data has a long tail of large values on the right — a few big orders, many small ones. The tail pulls the mean up, so mean > median. Left-skewed data has the tail on the left and mean < median. In a symmetric distribution the two coincide. The word "skewed right" names where the tail is, not where the peak is; a right-skewed histogram has its peak on the left.

Worked example — remove the one big order

The amounts have one conspicuous value: order 1005 at $1,320, the Alberta online order. The next largest is $980. Take 1005 out and recompute on eleven orders:

• with 1320: mean 511.67, median 445
• without 1320: mean 438.18, median 420

The mean moved 73.48; the median moved 25, and only because the pair of middle values changed from (420, 470) to a single middle value of 420. One order out of twelve shifted the mean by 14 % of its value. That is the whole argument: the mean is sensitive to extreme values and the median is resistant to them. In R, mean(orders$amount[orders$order_id != 1005]) prints [1] 438.1818 and the median line prints [1] 420.

A compromise exists — the trimmed mean drops a share from each end before averaging; mean(orders$amount, trim = 0.1) drops the lowest and highest of the twelve and prints [1] 473 — but the report-writing rule is simpler.

Which to report

Money-like variables — household income, house prices, order amounts, claim sizes — are almost always right-skewed: there is a floor at zero and no ceiling. For them report the median, and pair it with the IQR (Lesson 4.4) as the spread, because the IQR is rank-based in the same way. Report the mean with the standard deviation when the data is roughly symmetric, or when the total matters — a budget is mean × count, and no median gets you there. On a quiz the phrase household income is the cue: median with IQR.

Read three pairs, then compute one

For each mean–median pair, name the skew: (a) mean 52, median 48 → right; (b) mean 3.1, median 3.1 → symmetric; (c) mean 71, median 80 → left. Then the numeric: the mean of the eleven amounts with order 1005 removed, to 2 dp.

CRISP-DM: choosing the centre to report is data understanding → explore data; choosing it for the memo is deployment → produce final report.