Thickness of computer chips | 5 | 10 | 15 | 20 | 25 | 30 | |

Number of chips | Company A | 10 | 15 | 24 | 20 | 18 | 13 |

company B | 12 | 18 | 20 | 22 | 24 | 4 |

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Login NowMeasure of dispression is a descriotive statistical measure used to measure the variation or spreed or scatter less in the data set.

It gives additional information that enables to judge the reliability of measure of central tendency.

The measure of dispersion is important because it determines the margin of error. You’ll have when measuring the central tendency.

Solution Part:

For Company A:

x |
f |
fx |
fx^{2} |

5 | 10 | 50 | 250 |

10 | 15 | 150 | 1500 |

15 | 24 | 360 | 5400 |

20 | 20 | 400 | 10000 |

25 | 18 | 450 | 11250 |

30 | 13 | 390 | 11700 |

N = 100 | Σfx = 1800 | Σfx^{2} = 40100 |

Now,

\(\bar{x} = \frac{\sum fx}{N}\) = \(\frac{1800}{100}\) = 18

Standard deviation(σ) = \(\sqrt{\frac{\sum fx^2}{N} – \left ( \frac{\sum fx}{N} \right )^2}\)

= \(\sqrt{\frac{40100 }{100} – \left ( \frac{1800}{100} \right )^2}\)

= 8.77

And, Co-varience (c.v) = \(\frac{σ}{\bar{x}} \times 100%\)

= \(\frac{8.77}{18} \times 100%\)

= 48.72%

For Company b:

x |
f |
fx |
fx^{2} |

5 | 12 | 60 | 300 |

10 | 18 | 180 | 1800 |

15 | 20 | 300 | 4500 |

20 | 22 | 440 | 8800 |

25 | 24 | 600 | 15000 |

30 | 4 | 120 | 3600 |

N = 100 | Σfx = 1700 | Σfx^{2} = 34000 |

Now,

\(\bar{x} = \frac{\sum fx}{N}\) = \(\frac{1700}{100}\) = 17

Standard deviation(σ) = \(\sqrt{\frac{\sum fx^2}{N} – \left ( \frac{\sum fx}{N} \right )^2}\)

= \(\sqrt{\frac{34000}{100} – \left ( \frac{1700}{100} \right )^2}\)

= 7.14

And, Co-varience (c.v) = \(\frac{σ}{\bar{x}} \times 100%\)

= \(\frac{7.14}{17} \times 100%\)

= 42%

Since, the standard deviation and covarience of company B is less than that of company A. so, company B is considered more consistent in terms of thickness of computer chip.

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## Discussion