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Mean, median and mode answer different questions about the centre or most common values of a dataset. They are not interchangeable names for the same calculation. Choosing among them requires attention to the variable, the distribution and the question.

The mean uses every numerical value, the median uses their ordered positions, and the mode identifies the most frequent value or category. Statistics Canada’s mean guidance shows how raw observations and frequency tables support the same underlying calculation.

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Calculate the arithmetic mean

For nn numerical observations x1,…,xnx_1,\ldots,x_n, the arithmetic mean is

x‾=x1+x2+⋯+xnn=∑i=1nxin.\bar{x}=\frac{x_1+x_2+\cdots+x_n}{n}=\frac{\sum_{i=1}^{n}x_i}{n}.

Consider the ten synthetic journey times used in these statistics examples, measured in minutes:

8,10,12,12,14,16,18,20,22,48.8,\ 10,\ 12,\ 12,\ 14,\ 16,\ 18,\ 20,\ 22,\ 48.

Their sum is 180, so x‾=180/10=18\bar{x}=180/10=18 minutes. If the same total journey time were spread equally across the ten records, each would receive 18 minutes. This interpretation explains why the mean need not be one of the observed values in a different dataset.

An equivalent property is that deviations from the mean add to zero. Values below 18 contribute negative deviations; values above it contribute positive deviations. Their total balances. Squared deviations do not cancel, which is why they are useful in measures of spread.

Find the median by ordering the data

The median separates an ordered numerical dataset into lower and upper portions. With an odd number of values, use the middle observation. With an even number, the usual numerical convention averages the two middle observations.

For ten values, the central positions are fifth and sixth. Here those values are 14 and 16:

median=14+162=15 minutes.\operatorname{median}=\frac{14+16}{2}=15\text{ minutes}.

Order the observations first. Averaging the middle entries of an unsorted list is not a median calculation. Ties can also mean that more than half the observations equal or fall below the median; the phrase “exactly half are smaller” is not generally correct.

The median is less sensitive than the mean to how far an extreme value lies from the rest. It is not completely immune to changing or removing observations, because those changes can alter the central positions.

Identify the mode and report ties

The mode is the most frequent observed value. In the journey-time list, 12 appears twice and every other value once, so the mode is 12 minutes. The mean, median and mode are therefore 18, 15 and 12 minutes respectively.

A dataset can have several tied most frequent values. For 1,1,2,3,31,1,2,3,3, both 1 and 3 are modes. If every value appears once, school resources often say there is “no mode,” meaning there is no uniquely repeated most frequent value. More formally, every observed value ties for the maximum frequency. State the convention rather than pretending one number is specially typical.

Modes can also describe nominal categories. If bus is the most frequent travel mode, it is the modal category. It makes sense to count that category even though calculating an average of its arbitrary numerical code does not.

Mean, median and mode answer different questions

Different centres in the same listTen synthetic journey times form a dot plot. Mean 18, median 15 and mode 12 are marked separately.01020304050Each dot is one observationJourney time (minutes)Mode 12Median 15Mean 18
Synthetic journey times, n = 10. Repeated observations at 12 minutes are stacked.
Diagram explanation and exact data

The mode is the most frequent observed value, 12. The median lies halfway between 14 and 16, giving 15. The mean is 180 divided by 10, giving 18; the high value at 48 pulls it above the median.

Mean, median and mode answer different questions: exact reference data
Observation Minutes
1 8
2 10
3 12
4 12
5 14
6 16
7 18
8 20
9 22
10 48

See what an extreme value changes

Replace 48 in the synthetic list with 18. The new total is 180−48+18=150180-48+18=150, so the mean becomes 15 minutes. The fifth and sixth values remain 14 and 16, so the median remains 15 minutes.

This example shows how a large value pulls the mean upward. It does not imply that the original 48 should be removed. A genuinely long journey may be exactly what the study needs to understand. Use Outliers to distinguish investigation from selective deletion.

For right-skewed data, the mean is often above the median, but that relationship is not a theorem that identifies every possible distribution. Inspect the observations or plot rather than inferring the entire shape from two numbers.

Use frequencies and weights correctly

When distinct values xix_i occur with frequencies fif_i, calculate

x‾=∑fixi∑fi.\bar{x}=\frac{\sum f_i x_i}{\sum f_i}.

Suppose a synthetic borrowing table records 0 books for 2 learners, 1 book for 3 learners and 2 books for 5 learners. The total is

0(2)+1(3)+2(5)=13 books,0(2)+1(3)+2(5)=13\text{ books},

across 2+3+5=102+3+5=10 learners. The mean is 1.3 books per learner, not (0+1+2)/3=1(0+1+2)/3=1. The latter ignores how many observations each value represents.

Combining group means

If one group of 10 has mean 70 and another group of 30 has mean 80, the combined mean is

10(70)+30(80)10+30=310040=77.5.\frac{10(70)+30(80)}{10+30}=\frac{3100}{40}=77.5.

The unweighted average of the two means is 75, which would give the smaller group the same influence as the larger one. Group means may be combined using their sample sizes only when they describe compatible measurements and non-overlapping observations. Survey weights can require a different calculation.

Deliberately assigned weights

Weights can also represent a chosen assessment rule rather than a frequency. If two components have values 80 and 90 and weights 40% and 60%, the weighted result is

0.40(80)+0.60(90)=86.0.40(80)+0.60(90)=86.

This is the result of that declared rule. It is not the arithmetic mean of two equally weighted observations. Identify what the weights mean before using them.

Choose a summary for the purpose

A mean is useful when totals and equal-share reasoning matter. A median can describe a central experience when a few large or small values strongly influence the mean. A mode identifies the most common category or recorded value. None of these tells you the full spread or whether there are several clusters.

For a numerical report, state which measure you used and keep the unit. “Median journey time: 15 minutes” is clearer than “average: 15.” Include sample size and a spread measure when those affect interpretation.

Avoid comparing means from incompatible reference groups, periods or measurement rules. A carefully calculated average cannot make incompatible data comparable.

Practice questions

  1. Find the mean, median and mode of 2,3,3,4,82,3,3,4,8.
  2. Combine a group of 10 with mean 70 and a group of 30 with mean 80.
  3. In the journey-time example, replacing 48 by 18 changes the mean to what value, and does it change the median?
Check your answers
  1. The mean is (2+3+3+4+8)/5=4(2+3+3+4+8)/5=4. The middle value is 3, so the median is 3. The mode is also 3 because it occurs twice.
  2. The combined mean is (10×70+30×80)/40=77.5(10\times70+30\times80)/40=77.5. Use the group sizes as weights, not an equal average of the two means.
  3. The total becomes 150, giving mean 150/10=15150/10=15 minutes. The median remains (14+16)/2=15(14+16)/2=15 minutes because the middle positions are unchanged.

Related topics

Range, Variance and Standard Deviation explains spread around a centre. Quartiles and Interquartile Range develops an ordered-data alternative, and Histograms shows why a distribution needs more than one summary.

Sources and further reading