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Worked example

Let us assume you have the following observations of bird weight (in grams), which have been divided into class intervals, but you not know their individual values, nor their mean.

Weight class |
100 - 120 | 120 - 130 | 130 - 140 | 140-170 |

Class mid-point |
110 | 125 | 135 | 155 |

Number of birds |
23 | 15 | 6 | 2 |

Lacking their mean, we can estimate it by multiplying the mid-point of each class by the number of observations it contains.

= |
__(110×23) + (125×15) + (135×6) + (155×2)__ |
= |
__5525__ |
= **120.1** |

23 + 15 + 6 + 2 |
46 |

Now we can work out the deviation of each mid-class from the sample mean. Then multiply the square of this deviation by the number of observations it refers to.

Class mid-point |
110 | 125 | 135 | 155 |

Deviation from mean ( y ) |
− 10.1 | 4.9 | 14.9 | 34.9 |

Squared deviation ( y^{2} ) |
102.0 | 24.01 | 222.0 | 1218 |

Number of birds ( f ) |
23 | 15 | 6 | 2 |

fy^{2} |
2346 | 360.2 | 1332 | 2436 |

We then estimate the variance of this sample as

s^{2} = 6474.2 */* 45, or **143.9**