When to Use Mean, Median or Mode (With Worked Examples)

Quick answer: Use the mean when the data is roughly symmetric with no extreme values, the median when the data is skewed or contains an outlier, and the mode when the data is categorical or you need the most common value. In exams, if one value sits far away from the rest, the expected answer is almost always the median.

I once lost a mark on a one-mark question I was sure I had nailed. It asked why the median suited a set of house prices better than the mean, and I wrote "because the mean is affected by outliers." Nothing. The mark scheme wanted the outlier identified and its effect described using the numbers in the table. Since then I have treated "which average" questions as writing questions with a bit of arithmetic attached.

The three-question test that picks the right average

Ask three questions in order and stop at the first yes.

1. Is the data made of categories rather than numbers? Favourite subject, shoe size, blood group, eye colour. Use the mode. You cannot add up "blue" and divide by seven.
2. Is there an outlier, or is the data clearly skewed? One value miles from the rest, or a long tail in one direction. Use the median.
3. Neither? Use the mean. It is the only average that uses every single value, which makes it the most informative when nothing is distorting it.

That last point is why any question involving a total needs the mean. The mean mark of 12 students is 6.5, so the total is 12 x 6.5 = 78. A thirteenth student scores 9, so the new mean is 87/13 = 6.69 (2 dp). You cannot run that backwards with the median, which throws away the sizes of everything except the middle.

One data set, three completely different answers

Here are the minutes nine students spent on homework one evening: 20, 25, 25, 30, 35, 40, 45, 50, 210.

• Mean = 480/9 = 53.3 minutes (1 dp)
• Median = the 5th value once ordered = 35 minutes
• Mode = 25 minutes

The mean is larger than eight of the nine real values. Nobody worked anything like 53 minutes, so as a summary it is useless.

Now test how much each average moves when the outlier goes. Remove the 210 and the mean is 270/8 = 33.75 minutes, the median (30 + 35)/2 = 32.5 minutes. The mean shifted by 19.6 minutes. The median shifted by 2.5. That gap is the whole meaning of "the median is resistant to outliers", and quoting those two shifts earns the explain mark. If your arithmetic here is shaky, run the numbers through the math solver and check your ordering step, because ordering errors cost more median marks than anything else.

Estimated mean from a grouped frequency table

When data arrives grouped you cannot find the exact mean, so you estimate it using class midpoints. Here are marks out of 40 for 30 students.

• 0 < m ≤ 10, frequency 3, midpoint 5, product 15
• 10 < m ≤ 20, frequency 7, midpoint 15, product 105
• 20 < m ≤ 30, frequency 12, midpoint 25, product 300
• 30 < m ≤ 40, frequency 8, midpoint 35, product 280

Total frequency = 30. Total of the products = 700. Estimated mean = 700/30 = 23.3 marks (1 dp).

Three things get checked here every time:

• Write the word estimate. You assumed every student in a class scored exactly the midpoint, which almost none of them did.
• Divide by the total frequency (30), not the number of classes (4). Dividing by 4 gives 175, impossible on a paper out of 40. That sanity check has saved me twice.
• Midpoint = (lower bound + upper bound)/2, so for 20 < m ≤ 30 the midpoint is 25, not 20 and not 30.

Modal class, median class, and the unequal-width trap

For grouped data you name a class, not a value. The modal class above is 20 < m ≤ 30 because 12 is the highest frequency. For the median class, find the position (30 + 1)/2 = 15.5, then run cumulative frequencies: 3, 10, 22. The 15.5th value falls inside the third group, so the median class is also 20 < m ≤ 30.

The trap comes when class widths are unequal. Then the modal class is the one with the highest frequency density, not the highest frequency, and frequency density = frequency / class width. A class of width 20 with frequency 14 has density 0.7. A class of width 5 with frequency 9 has density 1.8. The narrow class wins even though its raw count is smaller. Papers set this deliberately, so check the widths before pointing at the biggest number.

How to write the "which average is most appropriate" answer

Give three things: name the average, name the feature of the data that forces that choice, and say what that feature does to the average you rejected, in context, with numbers.

Weak, and worth nothing: "The median, because the mean is affected by outliers."

Strong: "The median, because 210 minutes is an outlier. It pulls the mean up to 53.3 minutes, which is longer than eight of the nine students actually worked, so the mean is not typical of this group."

Standard scenarios worth memorising:

• A shop deciding which shoe size to stock uses the mode, because it wants the size most people buy, and an average size may not exist as a product.
• An employer advertising pay quotes the mean, because a few high salaries drag it above what most staff earn. A union would quote the median.
• To justify the mean, the reason to write is that it uses every value in the data set.

Comparing two data sets without dropping the comparison mark

A comparison answer needs one statement about average and one about spread, both written in context.

Say Class A has a median of 35 minutes and a range of 190 minutes, while Class B has a median of 28 minutes and a range of 22 minutes. The answer: "Class A's median is 7 minutes higher, so students in A typically spent longer on homework. Class A's range is much larger, so their times were far less consistent."

Two marks get lost here constantly. First, writing "A is higher" without saying higher what — always name the measure and the units. Second, comparing two averages and never mentioning spread. If there is an outlier, compare interquartile range instead of range, because the range is built from the two most extreme values and one of them is the problem. Build ten of these under time pressure with the quiz generator and the phrasing becomes automatic.

Test yourself

1. For the data 3, 4, 4, 5, 6, 40, which average best represents it, and what is your justification sentence?
2. The mean of 8 numbers is 12. The number 19 is removed. What is the new mean?
3. A grouped table has the class 0 to 10 with frequency 15, and the class 10 to 40 with frequency 24. Which is the modal class?

FAQ

Can a data set have more than one mode?

Yes. Two values tied for most frequent makes the data bimodal, and you quote both. If every value appears once there is no mode, and the correct answer is "no mode" — never write that the mode is 0, because 0 is a value and it is not in your list.

Is the median always the safer choice?

No, and picking it automatically will cost you. When the data is roughly symmetric with no extreme values, the mean is the better answer because it uses all the data, and a mark scheme asking you to justify the mean expects exactly that reason. The median only wins when something is distorting the mean.

Do I use n/2 or (n + 1)/2 for the median position?

Use (n + 1)/2 when working with a list of individual values, so for 9 values the median is the 5th. Use n/2 when reading a median off a cumulative frequency graph or working with a large grouped set, because there you are locating a position on a curve rather than picking out one item.

Why is my estimated mean different from the real mean?

Because grouping destroys the original values. The estimate assumes every item sits on its class midpoint, so it is slightly out whenever values inside a class are unevenly spread. That is expected, and the phrase "estimated mean" tells the examiner you understand why.

In short, categories mean mode, outliers mean median, and everything else means mean. Do the arithmetic carefully, but spend your real effort on the justification sentence, because that is where the marks quietly sit. More worked practice is on the maths hub.