Mean or Median: Which to Use

What outliers do to the mean but not the median - worked through seven house prices, and why salaries and house prices are usually reported as medians.

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Mean or Median: Which to Use — Moosa Academy

Both the mean and the median claim to describe the centre of a set of data, and on ordinary data they agree closely. The difference only shows itself when a few extreme values appear — and then choosing the wrong one can give a number that describes nobody at all.

Concept Two ways to find the centre
The mean — add all the values and divide by how many there are.
The median — sort the values and take the one in the middle.

The key difference is how each reacts to an outlier — a value lying far away from the rest of the data. The mean uses every value, so an outlier pulls it. The median only cares about position, so an outlier barely touches it.

Example Ordinary data: the mean works well

Five house prices in a neighbourhood, in riyals:

700,000 — 850,000 — 900,000 — 1,100,000 — 1,200,000
Step 1 — the total:  4{,}750{,}000
Step 2 — divide by five:  4{,}750{,}000 \div 5 = 950{,}000
⟹ mean = 950,000 riyals

The values sit close together, so the mean lands in a sensible place. Anyone told "about 950,000" would form an accurate picture of the neighbourhood.

Example Add two outliers: the mean distorts

Now add one luxury house at 2,500,000 and one old house at 300,000. Sorted, the seven prices are:

300K the five 2.5M median
The two hollow points are the outliers. They sit far from the cluster, and it is the mean — not the median — that gets dragged towards them.
300,000 — 700,000 — 850,000 — 900,000 — 1,100,000 — 1,200,000 — 2,500,000
Step 1 — the total:  7{,}550{,}000
Step 2 — divide by seven:  7{,}550{,}000 \div 7 \approx 1{,}078{,}571
⟹ mean ≈ 1,078,571 riyals

This figure is higher than six of the seven houses. It is arithmetically correct and practically misleading — the single luxury house has pulled it upwards.

Example The median holds steady

Take the same seven prices and find the median instead.

Step 1 — sort them (already done above)
Step 2 — count them: seven values, an odd number
Step 3 — the middle one is the fourth:  900{,}000
⟹ median = 900,000 riyals

The outliers changed the mean by more than 128,000, but they did not move the median at all. Making the luxury house twice as expensive would still leave the median at 900,000 — only its position in the order matters, not its size.

Note A third option: remove the outliers

If the extreme values genuinely do not belong to the group being described, they can be removed before taking the mean:

Drop 300,000 and 2,500,000, leaving the original five
 4{,}750{,}000 \div 5 = 950{,}000
⟹ mean = 950,000 riyals, sensible again

This must be done honestly. Removing values simply because they are inconvenient is not analysis — the justification has to be that they belong to a different group.

Note Why this matters in practice

Consider salaries at a company. If one director earns far more than everyone else, the mean salary rises for all the wrong reasons, and an applicant reading it would be badly misled. The median salary answers the question they actually asked: what does a typical employee earn? This is why income and house-price figures are usually reported as medians.

Summary
  1. The mean adds every value and divides; the median takes the middle of the sorted list.
  2. An outlier is a value lying far from the rest of the data.
  3. On closely grouped data both agree: the five houses give a mean of 950,000.
  4. Two outliers pushed the mean to about 1,078,571 — above six of the seven houses.
  5. The median stayed at 900,000, because only position matters, not size.
  6. Use the mean for even data, the median when outliers are present, or remove outliers first if they honestly do not belong.