# Mean, Median, Mode and Range

## Mean, Median, Mode and Range Revision

**Mean, Median, Mode, and Range**

The **mean, median, **and **mode** are different types of average and the** range** tells us how spread out our data is.

**Mean**

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:

\text{Mean} = \dfrac{\text{Sum of items}}{\text{Total number of items}}

**Advantages and Disadvantages**

**Advantage**– every bit of data is used in calculating the**mean**, so it represents all the data.**Disadvantage**– it is highly affected by outliers. An outlier is a piece of data that doesn’t quite fit with the rest of them.

**Note:** A better way to calculate the **mean** is to remove outliers before calculating it. The question will specifically tell you to do this if it is required.

**Median**

The **median** is often referred to as “the middle”, which is precisely what it is.

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

**Method 1:** Put the numbers **in order** from smallest to largest and find the middle value/middle two values. 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 there is one left, then that is the **median**; if there are two values, left then the **median** is the halfway point between the two.

**Method 2:** If n is the number of values in the list, then work out the value of \frac{n+1}{2}. The **median** is that number of values along in the list.

**Advantages and Disadvantages**

**Advantage**– it is not affected by outliers.**Disadvantage**– it does not consider all the data.

**Mode**

The **mode** is the most common value. To find it, look for which **value appears most often**. 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 be no repeats at all, in which case there is simply **no mode**.

**Advantages and Disadvantages**

**Advantage**– it is not affected by outliers.**Disadvantage(s)**– Firstly, there may not actually be a mode. 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?

**Range**

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.

\text{Range} = \text{Biggest value} - \text{Smallest Value}

**Note:** The **range** is highly affected by outliers. So a better way to calculate the **range** is to remove outliers before calculating it. The question will specifically tell you if this is required.

**Comparing Two Data Sets**

You may need to use what you have learnt about averages to compare two given sets of data.

The things to look out for when comparing data are:

- Averages (
**mean**,**median**,**mode**) – which set of data has the higher or lower averages. - Spread (interquartile
**range**, quartiles,**range**, minimum/maximum values) – which set of data is**more varied**, or**less consistent**.

**Example 1: Finding the Mean, Median and Mode**

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.

**[3 marks]**

**Mean:** There are 9 data points. First add the numbers together and then divide the result by 9.

56 + 79 + 77 + 48 + 90 + 68 + 79 + 92 + 71 = 660

\text{Mean} = \dfrac{660}{9}=73.3 (1 dp)

**Median:** Firstly, put the numbers in ascending order.

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.

\xcancel{48}, \xcancel{56}, \xcancel{68}, \xcancel{71}, 77, 79, 79, 90, 92

Counting along the list, we get that the median is 77.

**Mode:** We can see very clearly from the ordered list that there is only one repeat, 79, so the mode is 79.

**Example 2: Calculating the Range**

Find the **range** of 12, 8, 4, 16, 15, 15, 5, 15, 10, 8

**[1 mark]**

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

Largest - Smallest = 16-4=12, so the **range is 12**

**Example 3: Finding the Mean – Applied Questions**

There were 5 members of a basketball team who had a **mean points** score of 12 points each per game.

One of the team members left, causing the **mean point** score to reduce to 10 points each per game.

What was the **mean score** of the player that left?

**[2 marks]**

**Step 1:** Find the total for the original number of players: 5\times12=60

**Step 2:** Find the total after once the mean has changed, so 4\times10=40

**Step 3:** Calculate the difference between these two totals as that difference has been caused by the person who left: 60-40=20

Therefore the **mean** score of the person who left was 20 points per game. The same method applies if a new person/amount is added, you find the old and new totals and the difference is always due to the thing which caused the change.

**Example 4: Comparing Two Data Sets**

Two Maths teachers are comparing how their Year 9 classes performed in the end of year exams.

Their results are as follows:

Class A: 76, 35, 47, 64, 95, 66, 89, 36, 84

Class B: 51, 56, 84, 60, 59, 70, 63, 66, 50

Give 3 comparisons between the results of these 2 maths classes

**[3 marks]**

Let’s firstly think about averages. We can start with the mean,

Mean of Class A: \dfrac{76+35+47+64+95+66+89+36+84}{9}=65.\dot{7}

Mean of Class B: \dfrac{51+56+84+60+59+70+63+66+50}{9}=62.\dot{1}

So, Class A has a higher mean.

As neither of the classes have any repeated scores, we can’t use the mode. Therefore, let’s look at median,

Order Class A: 35, 36, 47, 64, 66, 76, 84, 89, 95

Middle score: 66

Order Class B: 50, 51, 56, 59, 60, 63, 66, 70, 84

Middle score: 60

So, Class A has a higher median.

Now, let’s look at range to consider the spread of data:

Class A range: 95-35=60

Class B range: 84-50=34

Class A has a bigger range.

In conclusion, Class A generally has higher scores, with both the mean and median being higher than class B. Class B’s scores are more concentrated and less spread out than Class A.

## Mean, Median, Mode and Range Example Questions

**Question 1:** Find the mode and range of the list of numbers below:

\begin{aligned}&280, 350, 320, 400, 350,\\&490, 590, 470, 280, 410, 350\end{aligned}

**[2 marks]**

It is not necessary to order the numbers, but it may help, especially in working out the range. In ascending order, these values are:

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

Since the number 350 occurs 3 times, it is the most common value, so:

\text{mode} = 350

The range is the difference between the lowest and the highest value. The lowest value is 280 and the highest is 590, so:

\text{range} = 590-280=310.

**Question 2:** Dani recorded the heights of members of her extended family, to the nearest cm. Find the median of their heights (listed below):

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

**[1 mark]**

First of all, since we have been asked to work out the median, we need to order the set of values:

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

There are 8 values in total, so we need to know which value, or values, we need in order to find the median.

Since there is an even number of values, there is not one single middle value, so you will need to find the two middle values. To find the median value, we can use the following formula:

\dfrac{n + 1}{2} where n represents the total number of values.

In this question, we have 8 values, so:

\frac{8 + 1}{2}=4.5

The answer 4.5 tells us that the median is half-way between the 4th value and the 5th value. The 4th value is 168 and the 5th value is 170, so the median is 169.

**NOTE: **if you struggle to work out the half-way value, add up the two numbers and divide by 2 (in other words, work out the mean of these two values).

**Question 3:** Below is a list of recorded reaction times in seconds.

\begin{aligned}&0.25, 0.34, 0.39, 0.38, 0.39,\\&1.67, 0.28, 0.30, 0.42, 0.46\end{aligned}

**a)** Calculate the mean of these values.

**b)** Identify any outliers and explain how the value of the mean would change if any outliers were to be removed from the calculation of the mean.

**[3 marks]**

a) In order to calculate the mean, we need to add up all the values and divide by 10 (since there are 10 values in total). The sum of the reaction times is

0.25+0.34+0.39+0.38+0.39 +\, 1.67+0.28+0.3+0.42+0.46 = 4.88

Then

\text{Mean}=\dfrac{4.88}{10}=0.488

b) 1.67 is the outlier as it is vastly higher than all the other values.

If this outlier were removed, then the mean would be lower.

**Question 4:** The mean length of 7 planks of wood is 1.35m. When an extra plank of wood is added, the mean length of a plank of wood increased to 1.4m.

What is the length of the extra plank of wood that was added?

**[2 marks]**

In most questions involving the mean, we are given the total and need to work out the mean from the total. In this question, we have been given the mean, so we are going to have to calculate the total from the mean.

If the mean length of 7 planks of wood is 1.35m, then the total length of all these planks of wood combined can be calculated as follows:

7 \times 1.35m = 9.45m

When the extra plank of wood is added, the mean length of a plank of wood increases to 1.4m. This means there are now 8 planks of wood, with a combined length of:

8 \times 1.40m = 11.2m

Therefore, by adding this additional plank of wood, the combined length has increased from 9.45m to 11.2m, so the length of this extra plank of wood is therefore:

11.2m - \, 9.45m = 1.75m

**Question 5:** In a rowing team, the weight of 8 women, in kilograms, is:

63, \, 60, \, 57, \, 66, \, 62, \, 65, \, 69, \, 58

In order to be a more competitive team, the coach has said that each team member should try to increase their overall muscle mass which will result in a 2\% gain in overall body weight.

What will be the mean weight of the team if all 8 are successful in precisely meeting this 2\% weight gain?

**[2 marks]**

In this question, we do not need to work out a 2\% increase in weight for each individual team member (it would not be wrong to do so, just unnecessarily time-consuming).

The combined weight of all 8 members is:

63 + 60+57+66+62+65+69+58 = 500kg

If each team member increases their weight by 2\%, then this is the same as the team increasing their combined weight by 2\%. Therefore, if the team is successful in achieving this 2\% weight gain, then the combined weight of the team can be calculated as follows:

1.02\times500 = 510kg

Since there are 8 team members in total, then mean weight following this weight gain is:

510kg \div \, 8 =63.75kg

**Question 6**: A farmer is growing potatoes. He grows some in a greenhouse and some outside and wants to compare which environment produces bigger potatoes. Here are the results of measuring his potatoes, in \text{cm}:

Outside potatoes: 9, 8, 11, 14, 16, 13, 12

Greenhouse potatoes: 5, 6, 17, 16, 14, 4, 8

The farmer looks at the data and says “the greenhouse has the largest potato so this must be the best growing conditions”.

Decide whether you agree with the farmer, giving 2 reasons. Describe how his experiment could be improved.

**[3 marks]**

As the farmer wants to know the best conditions, it’s better to look at averages rather than range.

We cannot use mode as there are no repeated values.

Let’s use mean first,

Outside potatoes: \dfrac{9+8+11+14+16+13+12}{7}=11.8...

Greenhouse potatoes: \dfrac{5+6+17+16+14+4+8}{7}=10

Therefore, the outside potatoes had a higher mean size.

Now, let’s look at median:

Outside potatoes ordered: 8, 9, 11, 12, 13, 14, 16

Median: 12

Greenhouse potatoes ordered: 4, 5, 6, 8, 14, 16, 17

Median: 8

Therefore, the farmer is wrong as both the mean and median are higher for the potatoes outside so it is likely this is the best growing environment. However, to be more sure on his conclusion, it would be helpful to use more potatoes to get a more accurate average.