The mean, the median and the mode all try to answer the same question: what is a typical value in this data? They can give very different answers, and choosing the wrong one can mislead. This guide shows how each one is calculated, what each one reacts to, and how to pick the right summary for the data in front of you.
How each one is calculated
Mean
The mean (the usual average) adds all the values and divides by how many there are:
$$ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} $$
Median
Sort the values from smallest to largest. The median is the middle one. With an even number of values, it is the mean of the two middle ones. Half the data lies at or below the median and half at or above.
Mode
The mode is the value that appears most often. A data set can have one mode, several modes, or none at all when no value repeats.
Example 1. Seven commute times in minutes: 12, 15, 11, 15, 18, 15, 14. Sorted: 11, 12, 14, 15, 15, 15, 18.
$$ \text{mean} = \frac{100}{7} \approx 14.29 $$
$$ \text{median} = 15 $$
$$ \text{mode} = 15 $$
The median is the fourth of seven sorted values, and 15 appears three times, so it is the mode.
Example 2. Six values: 4, 8, 6, 5, 3, 7. Sorted: 3, 4, 5, 6, 7, 8. With six values the median is the mean of the third and fourth: $latex \frac{5 + 6}{2} = 5.5$. The mean is $latex \frac{33}{6} = 5.5$. No value repeats, so there is no mode.
Move the data and watch the summaries
In the plot below each dot is a data value. The dashed line is the median, the orange triangle is the mean (the point where the dots would balance on a ruler), and orange dots mark the mode. Drag a dot, then press the outlier button a few times.
Drag a dot sideways, click the plot to add one, or focus a dot and use the arrow keys. Orange dots are the mode. Try the outlier button.
With the starting data the mean is 4.3 and the median is 4. Adding one value of 20 moves the mean to 5.73 while the median stays at 4. The mean reacts to every value, including extreme ones. The median only cares about order.
Which one should you use?
| Situation | Best summary | Why |
|---|---|---|
| Data is roughly symmetric, no extreme values | Mean | It uses every value and matches the median anyway |
| Skewed data: incomes, house prices, waiting times | Median | A few huge values do not drag it away from the typical case |
| Categories or counts: shoe size, favorite color, most sold item | Mode | You cannot average colors, but you can find the most common one |
| You need a total: cost for n items, total hours | Mean | Total = mean × n, and only the mean keeps that link |
In a right-skewed distribution, a long tail of high values pulls the mean past the median, and the mode sits at the peak. The order is mode, then median, then mean. Left-skewed data reverses it.
Example 3: salaries. A small company pays six people $2,800, $3,000, $3,000, $3,200, $3,400 and $12,000 a month, the last being the owner.
$$ \text{mean} = \frac{27{,}400}{6} \approx 4{,}566.67 $$
$$ \text{median} = \frac{3{,}000 + 3{,}200}{2} = 3{,}100 $$
$$ \text{mode} = 3{,}000 $$
The mean of $4,566.67 is higher than what 5 of the 6 people earn, so it describes nobody. The median of $3,100 is a fair picture of the typical salary. Remove the owner and the mean drops to $3,080, almost the median ($3,000): a single value caused the gap.
Common mistakes
- Not sorting before finding the median. The median of 7, 3, 9 is 7, the middle after sorting (3, 7, 9), not the middle of the list as written.
- Writing 0 when there is no mode. No value repeats means there is no mode. Zero is a number that may or may not be in the data.
- Thinking the mode must be unique. Two values tied for most frequent give two modes (bimodal data).
- Averaging averages. The mean of two group means equals the overall mean only when the groups have the same size. Otherwise use a weighted average.
- Using the mean for skewed data. Reporting the mean salary of a company with one very high earner overstates what most people make.
Practice problems
Problem 1
Find the mean, median and mode of 6, 9, 3, 9, 8, 5, 9.
Answer
Sorted: 3, 5, 6, 8, 9, 9, 9. Mean: $latex \frac{49}{7} = 7$. The middle value is the median: 8. The value 9 appears three times.
Answer: mean 7, median 8, mode 9.
Problem 2
Find the median of 12, 7, 15, 9.
Answer
Sorted: 7, 9, 12, 15. Four values, so average the two in the middle: $latex \frac{9 + 12}{2} = 10.5$.
Answer: 10.5.
Problem 3
Five houses in a neighborhood sold for (in thousands of dollars) 210, 225, 230, 240, 1250. Which is the better summary of a typical price, the mean or the median?
Answer
Mean: $latex \frac{2{,}155}{5} = 431$. Median: 230.
Answer: the median (230). One expensive house pulls the mean up to 431, higher than four of the five sales.
Problem 4
The mean of five numbers is 12. Four of them are 10, 11, 13 and 15. What is the fifth?
Answer
The five numbers add up to $latex 5 \times 12 = 60$. The four known ones add up to $latex 10 + 11 + 13 + 15 = 49$.
Answer: $latex 60 - 49 = 11$, so the fifth number is 11.
Problem 5
Find the mode(s), mean and median of 1, 2, 2, 3, 3, 4.
Answer
Both 2 and 3 appear twice, so the data has two modes. Mean: $latex \frac{15}{6} = 2.5$. Median: $latex \frac{2 + 3}{2} = 2.5$.
Answer: modes 2, 3, mean 2.5, median 2.5.
Problem 6
Compare the mean and median of 4, 5, 6, 7, 8 before and after adding the value 100.
Answer
Before: mean 6, median 6. After (4, 5, 6, 7, 8, 100): mean $latex \frac{130}{6} \approx 21.67$, median $latex \frac{6 + 7}{2} = 6.5$.
Answer: the mean jumps from 6 to 21.67, while the median only moves from 6 to 6.5.
Frequently asked questions
Can the mean, median and mode be the same number?
Yes. In a perfectly symmetric data set with a single peak, such as 2, 3, 4, 4, 4, 5, 6, all three equal 4.
Which one is best for income?
The median. A few very high incomes pull the mean up, so the median shows what a typical person earns.
Can the mean be a value that is not in the data?
Yes, and it often is. The mean of 1 and 2 is 1.5. The median of an even-sized set can also fall between two data values. The mode, by contrast, is always one of the data values.
Do I need all three?
Not always, but looking at two or three gives a quick check. If the mean and the median differ a lot, the data is skewed or has outliers.
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