## Finding the Mean From a Frequency Table

To find the mean of a list of numbers we add the numbers up and divide by how many numbers there are. For example the mean 1, 4, 6, 6, 7 isHowever sometimes we have a list in the form of a table. The table below is a fictitious summary of the number of goals in a series of football matches.

Number of Goals | Number of Games |

0 | 3 |

1 | 4 |

2 | 3 |

3 | 1 |

We could find the mean by writing out the list that the table represents:

0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3 and finding the mean of the list:

to 2 decimal places.

If the list is longer however this would be tedious. A more efficient method is to insert an extra column and row as below and find the total number of goals and total number of games using the extra row and column.

Number of Goals | Number of Games or Frequency | Total Number of Goals =Number of Goals*Frequency |

0 | 3 | 0 |

1 | 4 | 4 |

2 | 3 | 6 |

3 | 1 | 3 |

| Sum Total of Games=11 | Sum Total of Goals=13 |

As before, mean=(Sum Total of Goals)/(Sum Total of Games)= 13/11=1.18 to 2 d.p.

In the above example the number of goals in each game is known, but sometimes we have frequency tables with intervals, as in the table below, which shows the average length of some fictitious French snails.

Lengths (cm) | Frequency |

0-10 | 6 |

10-20 | 17 |

20-30 | 45 |

30-40 | 20 |

40-50 | 4 |

50 and above | 0 |

Now we really need to use the above method to avoid wring out a long list. We take the lengths as the midpoint of each interval:

Lengths (cm) | Frequency | Midpoint | Midpoint*Frequency |

0-10 | 6 | 5 | 30 |

10-20 | 17 | 15 | 255 |

20-30 | 45 | 25 | 1125 |

30-40 | 20 | 35 | 700 |

40-50 | 4 | 45 | 180 |

50 and above | 0 | NA | 0 |

| Number of Snails=92 | | Sum of Lengths=2290 |

Then the to 2 d.p.

This is only an estimate because we took for example all 6 lengths between 0 and 10 as 5. They could have been all 6 or 9 or any combination of 6 numbers between 0 and 10.