# 7.1 Advantages and disadvantages of using the arithmetical average

What are the advantages and disadvantages of using the arithmetical average?

Ever heard of families with 2.4 children? This is the national average but it means nothing – because you can’t have 0.4 of a child! This highlights one of the problems with averages: the value you get may not be a real value in terms of what you are talking about.

Another problem is that the average value will be affected by the highest and lowest values. For example, your football team could be having a really bad season, scoring nothing in nine games. The average number of goals scored per game in these nine games would be zero. Then, suddenly, they start to play very well and in the next match score ten goals. This would increase the average goals scored to one goal per match, which would make it look as though they’d scored a goal in every match when they hadn’t.

Averages are good, however, because they aren’t too complicated to work out (compared to some other statistical calculations) and they use all the available data.

Now try the following activity. Remember to refer to the example if you get stuck and to check your answers once you have completed the questions.

## Activity 11: Finding out averages

- The ages of four children in a family are 4, 6, 8 and 10 years. What is the average age?
- Find the average of the following data sets:
- a.4, 6, 11
- b.3, 7, 8, 4, 8
- c.8, 9, 10, 9, 4, 2
- d.11, 12, 13, 14, 15, 16

- The number of goals scored by a football team in recent matches were as follows:

2 | 3 | 0 | 1 | 3 |

2 | 3 | 2 | 1 | 3 |

- Work out the average number of goals per match.

### Answer

Check your answers with the answers below.

- First, add all of the ages together:
- 4 + 6 + 8 + 10 = 28

Then divide this total by the amount of data given:

- 28 ÷ 4 = 7

The average age is 7.

- You will find the following answers using the same calculation you used for question 1:
- a.Add all the numbers (4 + 6 + 11 = 21) and then divide this answer by the amount of data given (21 ÷ 3 = 7). The answer is 7.
- b.Add all the numbers (3 + 7 + 8 + 4 + 8 = 30) and then divide this answer by the amount of data given (30 ÷ 5 = 6). The answer is 6.
- c.Add all the numbers (8 + 9 + 10 + 9 + 4 + 2 = 42) and then divide this answer by the amount of data given (42 ÷ 6 = 7). The answer is 7.
- d.Add all the numbers (11 + 12 + 13 + 14 + 15 + 16 = 81) and then divide this answer by the amount of data given (81 ÷ 6 = 13.5). The answer is 13.5.

- The average number of goals per match is 2:
2 + 3 + 1 + 3 + 2 + 3 + 2 + 1 + 3 = 20

20 ÷ 10 = 2

Now have a go at another activity to check your knowledge.

## Activity 12: The maths test

- In a maths class the scores for a test (out of 10) were as follows:

5 | 6 | 6 | 4 | 4 |

7 | 3 | 5 | 6 | 7 |

8 | 6 | 2 | 8 | 5 |

4 | 5 | 6 | 5 | 6 |

- What is the average score?

- Some of the students felt that the teacher had been too harsh with their marks. The tests were remarked and the new results were as follows:

4 | 6 | 6 | 4 | 4 |

6 | 1 | 5 | 6 | 6 |

7 | 6 | 1 | 9 | 5 |

3 | 5 | 6 | 5 | 5 |

- Work out the average score for these new results. Which set of results gave the best marks? Was the teacher harsh with the first marking?

### Answer

First, add up the total number of marks:

- 5 + 6 + 6 + 4 + 4 + 7 + 3 + 5 + 6 + 7 + 8 + 6 + 2 + 8 + 5 + 4 + 5 + 6 + 5 + 6 = 108

Then divide this by the number of scores (or the number of students), which is 20:

- 108 ÷ 20 = 5.4

So the average score is 5.4 out of 10.

Again, first add up the total number of marks:

- 4 + 6 + 6 + 4 + 4 + 6 + 1 + 5 + 6 + 6 + 7 + 6 + 1 + 9 + 5 + 3 + 5 + 6 + 5 + 5 = 100

Then divide this total by 20:

- 100 ÷ 20 = 5

The best set of results was the first set. The teacher had not been marking it harshly.

## Summary

In this section you have:

- learned that the mean is one sort of average
- learned that the mean is worked out by adding up the items and dividing by the number of items
- understood that the mean can give a ‘distorted average’ if one or two values are much higher or lower than the other values.