What you need to know

Mean, Median, Mode, and Range

The mean, median, and mode are different types of average. We use them when have a bunch of numbers, often data that we’ve collected, and we want to get a feel for how big/small that group of numbers in. Are they very high? Are they only a little bigger then zero? Are they closer to 10 or 20 in general? These are the questions we’re answer when finding an average.


The mean is the most popular kind of average. To find the mean we must add up all the numbers we’re finding the average of, and then divide by how many numbers there are in that list.


The median is often referred to as “the middle”, which is precisely what it is. To find the median of a list of numbers, we put the numbers in order from smallest to largest and find the middle value/middle two values. If there is a middle value, then that is the median; if there are two middle values, then the median is the halfway point between the two.

There are two common ways of finding the middle value(s):

  • Cross out the smallest number and the largest number, then cross out the next smallest and largest, keeping going crossing out pairs of number like this until you have one or two left.
  • If n is the number of values in the list, then work out the value of \frac{n+1}{2}. If the answer is a whole number such as 5, then the median is the 5th point along the ordered list. If the answer is a decimal (ending in .5) such as 12.5, then the median is the halfway point between the 12th and 13th value along.


The mode is the most common value. To find it, look for which value appears most often. Note: there might be two values which are tied for the most appearances, in which case we say the data is bimodal, or alternatively there might no repeats at all, in which case there is simply no mode.

Example: 9 people take a test. Their scores out of 100 are:

56, 79, 77, 48, 90, 68, 79, 92, 71

Work out the mean, median, and mode of their scores.

First up, the mean. The question tells that there are 9 data points, so we must add the numbers together and divide the result by 9.

\text{Mean } = \dfrac{56 + 79 + 77 + 48 + 90 + 68 + 79 + 92 + 71}{9}=73.3\text{ (1dp)}

So, the mean is 73.3. Next up, the median. Firstly, we have to put the numbers in ascending order. This looks like

48, 56, 68, 71, 77, 79, 79, 90, 92

There are 9 numbers, and \frac{9+1}{2}=5, so the median must be the 5th term along. Counting along the list, we get that the median is 77. Finally, the mode. We can see very clearly from the ordered list that there is only one repeat: 79, so we must have that the mode is 79.

It’s important to understand that the mean, median, and mode all come with their advantages and disadvantages. To understand these, you need to know another definition – outliers. An outlier is a piece of data that doesn’t quite fit with the rest of them. For example, in the list

500, 520, 493, 480, 530, 7,

7 is an outlier. Outliers can occur for all kinds of reasons and often it’s just human error. Now we know what an outlier, let’s see what’s good and what’s not about the mean, median, and mode.


  • Advantage – every bit of data is used in calculating the mean, so it represents all the data. The mean is generally considered the better of the 3 averages, but it certainly isn’t always.
  • Disadvantage – it is highly affected by outliers. The fact that every bit of data is used to calculate the mean can be a weakness. The list we just saw: 500, 520, 493, 480, 530, 7 – what is its mean? Is it representative of where the data is? How about if you take the 7 out and then find the mean, is that a better average?*


  • Advantage – it is not affected by outliers.
  • Disadvantage – it does not consider all the data. Consider the values 1, 1, 2, 3, 12, 14, 15 – what is the median? Does it actually represent the middle of these numbers?


  • Advantage – it is not affected by outliers.
  • Disadvantage(s) – firstly, it sometimes is impossible to find. Secondly, it does not consider all of the data. Consider the values 32, 35, 35, 128, 201, 176, 295 – what is the mode? Does it represent the “average” of the data?


The range is not another average – it is a measure of spread. This means the range is a way of telling us how spread out the data is. To calculate it, we subtract the smallest value from the biggest value.

Example: Find the range of 12, 8, 4, 16, 15, 15, 5, 15, 10, 8.

A good way to make sure you haven’t missed any numbers in determining the biggest and smallest value is to order them. Doing this, we get

4, 5, 8, 8, 10, 12, 15, 15, 15, 16.

Therefore, the smallest value is 4 and the largest is 16 and 16-4=12, so the range is 12.

Sadly, the range also has its disadvantages – it is highly affected by outliers, just like the mean. If I included the value 100 in the list above, suddenly the range would be 100-4=96, which isn’t really a fair description of how spread out the data are since the rest are quite close together.

*A better way to calculate both mean and range is to remove outliers before calculating them. A question may ask you to redo calculations of the mean/range with outliers removed, or it may ask you to identify how these values are affected by outliers. Get to know your outliers.

All that said, if you’re asked to find the mean/range of a bunch of numbers, then don’t go removing any numbers you think might be outliers unless the question leads you to it.

Mean Median Mode and Range Questions

It’s not necessary to order the numbers to find the mode and range but it may help. In ascending order, these values are


280, 280, 320, 350, 350, 350, 400, 410, 470, 490, 590.


The most common number is 350, so \text{mode } = 350.


The lowest value is 280 and the highest is 590, so \text{range } = 590-280=310.

Firstly, we must order them. This looks like


154, 163, 164, 168, 170, 179, 185, 188


There are 8 terms, and \frac{8+1}{2}=4.5, so the median is halfway between the 4th term and the 5th term. The 4th term is 168 and then 5th term is 170, so the halfway point is 169.


NOTE: if it isn’t obvious what the halfway value is, add up the two numbers and divide by 2.

a) We need to add up all the values and divide by how many there are, which in this case is 10.


\text{Mean }=\dfrac{0.25+0.34+0.39+0.38+0.39+1.67+0.28+0.3+0.42+0.46}{10}=0.488


b) Clearly, 1.67 is much higher than the rest of the values so it is an outlier. If this outlier were removed, then the mean would be lower.

Whether you are looking for Mean revision notes or resources for Mode and Median, Maths Made Easy has brought together a comprehensive list of revision resources for teachers, students and tutors to utilise. Take a look at our Mean, Median, Mode and Range worksheets and see which ones you want to add in to your collection. Fro more exceptional GCSE Maths revision materials, go to our homepage.

Need some extra help? Find a Maths tutor now

Or, call 020 3633 5145