Once you’ve collected your data you need to be able to draw conclusions from it and there are a number of methods you can use to achieve this.
To explain each method the following data will be used:
Farmer Briggs decides to count the number of spots on his cows. He comes up with the following data:
8  5  6  7  8  8  9  7  8  6 
To make the information clearer he decides to put the information into a table:
Number of spots 
Number of cows 
5 6 7 8 9 
1 2 2 4 1 
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The Mean
Farmer Briggs wants to compare the number of spots on his cows to the cows in other neighbouring farms.
To do this he needs to figure out an average and the spread of his results.
The mean
The mean is the average of a set of results.
To calculate this you need to add together all the results then divide by the total number of results.
For Farmer Briggs’ results:
8 + 5 + 6 + 7 + 8 + 8 + 9 + 7 + 8 + 6 = 72
Total number of results = 10
Mean = 72/10 = 7.2
You can keep your answer as a decimal number but it can make more sense (or you might be asked in a question) to round the answer.
So, in this case, the average number of spots would be 7.
In this case, there weren’t too many results to work with so it was quite simple to work out the mean in this way. However, if you have more than 100 results, for example, it can be easier to use a frequency table to figure out the average:
Number of spots  Number of cows  Frequency of spots 
5 6 7 8 9 
1 2 2 4 1 
5 x 1 = 5 6 x 2 = 12 7 x 2 = 14 8 x 4 = 32 9 x 1 = 9 
Total 
10 
72 
Mean = 72/10 = 7.2
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The median
The median is the middle number once all the results have been put in numerical order.
If you have an odd number of results this is simply the middle number. For example:
5 7 7 8 8 9 10 10 10 11 12
Median = 9
However, if you have an even number of results, like Farmer Briggs and his cows, then you need to calculate the mean of the two central numbers. So, for Farmer Briggs:
5 6 6 7 7 8 8 8 8 9
The mean of 7 and 8:
(7 + 8) 2 = 7.5
A frequency table can also be used to calculate the median:
Number of spots  Number of cows 
5 6 7 8 9 
1 2 2 4 1 
Farmer Briggs knows that the total number of cows is 10
He knows then that the median result will be between the 5^{th} and 6^{th} results.
Using the table he can see that it must be between 7 and 8 spots so the median is 7.5.
You can use the following equation to figure out the median:
The median will always be the n + 1 ^{th} result where ‘n’ is the number of results.
So, Farmer Briggs could also have figured out the median like this:
(10 + 1) 2 = 5.5
The median is the 5.5^{th} result.
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The mode
The mode is the number which appears most in the data collected. If you arrange Farmer Briggs’ data in numerical order it will be clear to see which number appears the most frequently:
5 6 6 7 7 8 8 8 8 9
It’s obvious that 8 appears the most often so the mode is 8.
A frequency table makes it even clearer:
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The range
The range is the difference between the highest and lowest number.
If you look at Farmer Brigg’s results:
8 5 6 7 8 8 9 7 8 6
You can see that the largest number is 9 and the smallest is 6
To find the range subtract the smaller number from the larger number:
9 – 6 =3
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Comparing distributions
Farmer Briggs spoke to one of his neighbours, Farmer Brown, so that he could compare the mean number of spots on his cows.
Farmer Brown found he had the same mean number of spots as Farmer Briggs: 7.2.
They assumed that they must, therefore, have similar result in terms of spots per cow. However, this didn’t turn out to be the case as you can see below:
Farmer  Results  
Briggs 
8 
5 
6 
7 
8 
8 
9 
7 
8 
6 
Brown 
12 
2 
3 
8 
15 
6 
7 
10 
7 
2 
As you can see, although the mean was the same the results from both farmers are very different.
Farmer Brown has cows with only a few spots and cows with a lot more spots. This means that his range is much larger then Farmer Briggs':
15 – 2 = 13
Therefore by giving both the average and the spread you can get a better picture of the results.
Statistical measures
In your exam you’ll be expected to know how to:
 – use data to find patterns and exceptions
 – compare distribution and make inferences.
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The advantages and disadvantages of the mean, median and mode
You may find that using all three averages for a set of results isn’t necessary.
The table below sets out the main advantages and disadvantages for each one.
Average 
Advantages 
Disadvantages 

Mean  All data collected is used for the final answer  If there are results which are very large or very small then the end answer may not be representative to the whole  
Median  The answer isn’t affected by very large or very small results  If there a lot of results then this can be time consuming to calculate  
Mode  If the results don’t involve numbers then this can be used  There are three main disadvantages:

For example, the table below shows how much age ranges of people who came to a circus. Can you see what the disadvantage would be to use the modal class or the mean?
Age range  Number of people 
10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 
39 37 38 10 7 
The modal class is 10 19 years old. However, the ranges of 20 – 29 and 30 – 39 years old are really close.
In terms of the mean, due to the fact that most people who came were aged under 40 the answer would be distorted.
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