# Mean, Median, Mode and Range

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## 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.

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## 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:

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

• 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.

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## 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.

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

• 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?
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## 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.

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## 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:

1. Averages (mean, median, mode) – which set of data has the higher or lower averages.
2. Spread (interquartile range, quartiles, range, minimum/maximum values) – which set of data is more varied, or less consistent.
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## 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 $5$th 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 $79$.

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## 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 $12$

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## 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.

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## 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]

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.

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## Mean, Median, Mode and Range Example Questions

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:

$\text{range} = 590-280=310$.  Gold Standard Education

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 $4$th value and the $5$th value. The $4$th value is $168$ and the $5$th 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).  Gold Standard Education

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.  Gold Standard Education

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 $1.35$m, then the total length of all these planks of wood combined can be calculated as follows:

$7 \times 1.35$m $= 9.45$m

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

$8 \times 1.40$m $= 11.2$m

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

$11.2$m $- \, 9.45$m $= 1.75$m  Gold Standard Education

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:

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

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:

$1.02\times500 = 510$kg

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

$510$kg $\div \, 8 =63.75$kg  Gold Standard Education

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.  Gold Standard Education

## Mean, Median, Mode and Range Worksheet and Example Questions

### (NEW) Mean, Median, Mode and Range 1 Exam Style Questions - MME

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### (NEW) Mean, Median, Mode and Range 2 Exam Style Questions - MME

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## Mean, Median, Mode and Range Drill Questions

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### Averages From Frequency Tables - Drill Questions

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