The measurements (in mm) of the diameters of the head of the screws are given below:

Diameter (in mm) | No. of Screws |

33 — 35 | 10 |

36 — 38 | 19 |

39 — 41 | 23 |

42 — 44 | 21 |

45 — 47 | 27 |

Calculate mean diameter of head of a screw by ‘Assumed Mean Method’.

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#### Solution

Let A be the assumed mean.

A is taken as the class mark of the middle class.

Hence, let us take 40 as the assumed mean.

Then A = 40 and deviation d_{i} = x_{i} – A = x_{i} – 40

Diameter |
Class |
Deviations |
Number of |
f_{i}d_{i} |

(in mm) |
marks |
d_{i}=x_{i}-A |
Screws |
- |

- | x_{i} |
d_{i}=x_{i}-40 |
f_{i} |
- |

33-35 | 34 | -6 | 10 | -60 |

36-38 | 37 | -3 | 19 | -57 |

39-41 | 40=A | 0 | 23 | 0 |

42-44 | 43 | 3 | 21 | 63 |

45-47 | 46 | 6 | 27 | 162 |

Total |
- | - | ∑f_{i}=100 |
∑f_{i}d_{i}=108 |

Here,∑f_{i}d_{i}=108, ∑f_{i}=100

`bard=(Sigmaf_id_i)/(Sigmaf_i)=108/100=1.08`

`barx=A+bard`

=40+1.08

=41.08

Thus, the mean diameter of the screw heads is 41.08 mm.

Concept: Mean of Grouped Data

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