Mean, median, mode and range

Four numbers that describe a list

Data analysis questions on the GRE start from a handful of summary numbers. Each answers a different question about a list of values.

Measure	What it is	What it tells you
Mean	the sum divided by the number of values	the "balance point" of the data
Median	the middle value once the list is in order	the value that splits the data in half
Mode	the value that occurs most often	the most common value (a list can have several, or none)
Range	the greatest value minus the least	how far apart the extremes are

The mean, median and mode measure the centre of the data; the range measures its spread.

The mean is really a statement about the sum

$$\text{mean} = \frac{\text{sum}}{n} \quad\Longleftrightarrow\quad \text{sum} = \text{mean} \times n$$

The second form is the one that solves problems. If five numbers average 12, you know nothing about any single number, but you know for certain that they add up to 60. Missing-value questions, "a new value is added" questions and "a value is removed" questions all become simple once you switch from averages to sums, do the arithmetic on the sums, and only divide at the end.

The mean has a useful balancing property: the distances of the values above the mean exactly cancel the distances of the values below it. For 4, 7 and 10 the mean is 7, and the deviations $-3$, $0$ and $+3$ add up to zero. This is why adding a value equal to the current mean leaves the mean unchanged, a value above the mean pulls it up, and a value below pulls it down.

Averages of groups: weight by size

When two groups are combined, the combined mean is not the average of the two means unless the groups have the same size. Convert each group to a sum, add the sums, and divide by the total count:

$$\text{combined mean} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$$

The combined mean always lies between the two group means, closer to the mean of the larger group. That gives a quick check on any answer.

The median needs an ordered list

Put the values in order first. With an odd number of values $n$, the median is the value in position $\frac{n+1}{2}$. With an even number, there are two middle values, and the median is their mean, which may not be in the list at all.

For 3, 5, 8, 9, 12, 14, 20, 31 (eight values) the middle pair is 9 and 12, so the median is $10.5$. The mean is $\frac{102}{8} = 12.75$, larger than the median because the single large value 31 pulls the mean towards it.

That is the key difference between the two. The mean uses every value, so one extreme value (an outlier) can move it a lot. The median uses only the order, so it barely notices an outlier: replace 31 by 310 and the mean jumps to 47.625, while the median stays 10.5.

Frequency tables

Data often comes as a table of values and how many times each occurs. Treat it as a compressed list.

• Count $n$ by adding the frequencies.
• Mean: multiply each value by its frequency, add the products, divide by $n$.
• Median: run a cumulative count down the table until you reach the middle position.
• Mode: the value with the largest frequency (the value, not the frequency).

Never take the median or mean of the frequency column; the frequencies are counts, not data values.

How the measures react to changes

Change to every value	Mean, median, mode	Range
add a constant $c$	each increases by $c$	unchanged
multiply by $k > 0$	each is multiplied by $k$	multiplied by $k$

Shifting the whole list moves its centre but not the distance between its extremes. Doubling every value and then adding 5 turns a mean of 12 into $2 \times 12 + 5 = 29$, a median of 10 into 25, and a range of 8 into 16 (the "add 5" step does nothing to the range).

Comparing without computing

In a quantitative comparison you rarely need exact values. Ask which way the data leans. A list with a long tail of large values usually has a mean above its median; a perfectly symmetric list has mean equal to median. When an unknown value appears, find the range of values it can take and check whether the comparison holds at both ends.