Median

Finding it

Sort first. The median of $9, 2, 5$ is $5$, and reading the middle of the unsorted list would give $2$.

With an odd count there is one middle value:

$$3, \; 5, \; \mathbf{8}, \; 9, \; 12 \;\Rightarrow\; 8$$

With an even count, take the mean of the two middle values:

$$3, \; 5, \; \mathbf{8}, \; \mathbf{9}, \; 12, \; 15 \;\Rightarrow\; \frac{8 + 9}{2} = 8.5$$

Why it resists outliers

The median cares only about position. A value ten times too large counts exactly as much as one slightly too large — both are simply “above the middle”.

Data	Mean	Median
$5, 6, 7, 8, 9$	$7$	$7$
$5, 6, 7, 8, 94$	$24$	$7$

Replacing $9$ with $94$ more than triples the mean and leaves the median where it was. That stability is why house prices, salaries and household incomes are reported as medians — a few very large values are normal in that data.

Where else it appears

The median is the middle line of a box plot, and it splits the data in half to produce the quartiles: $Q_1$ is the median of the lower half and $Q_3$ the median of the upper half.

Pair the median with the interquartile range when reporting spread. Both ignore the tails, so they describe the data consistently.