1^{st}Judge |
2 | 1 | 4 | 6 | 5 | 8 | 9 | 10 | 7 | 3 |

2^{nd} Judge |
4 | 3 | 2 | 5 | 1 | 6 | 8 | 9 | 10 | 7 |

3^{rd} Judge |
5 | 8 | 4 | 7 | 10 | 2 | 1 | 6 | 9 | 3 |

Indicate which pair of judges has the nearest approaches to music

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We know,

\(r_{12} = 1 – \frac{6 \sum d_1^2}{n(n^2 – 1)}\)

\(r_{13} = 1 – \frac{6 \sum d_2^2}{n(n^2 – 1)}\)

\(r_{23} = 1 – \frac{6 \sum d_3^2}{n(n^2 – 1)}\)

Now,

x | y | z | d_{1} = x – y |
d_{2} = y – z |
d_{3} = z – x |
d_{1}^{2} |
d_{2}^{2} |
d_{3}^{2} |

2 | 4 | 5 | -2 | -1 | 3 | 4 | 1 | 9 |

1 | 3 | 8 | -2 | -5 | 7 | 4 | 25 | 49 |

4 | 2 | 4 | 2 | -2 | 0 | 4 | 4 | 0 |

6 | 5 | 7 | 1 | -2 | 1 | 1 | 4 | 1 |

5 | 1 | 10 | 4 | -9 | 5 | 16 | 81 | 25 |

8 | 6 | 2 | 2 | 4 | -6 | 4 | 16 | 36 |

9 | 8 | 1 | 1 | 7 | -8 | 1 | 49 | 64 |

10 | 9 | 6 | 1 | 3 | -4 | 1 | 9 | 16 |

7 | 10 | 9 | -3 | 1 | 2 | 9 | 1 | 4 |

3 | 7 | 3 | -4 | 4 | 0 | 16 | 16 | 0 |

Σd_{1} = 0 |
Σd_{2} = 0 |
Σd_{3} = 0 |
Σd_{1}^{2} = 60 |
Σd_{2}^{2} = 206 |
Σd_{3}^{2} = 204 |

Now,

r_{12} = \(1 – \frac{6 \times 60}{10 \times 99}\) = 0.636

r_{13} = \(1 – \frac{6 \times 206}{10 \times 99}\) = -0.248

r_{23} = \(1 – \frac{6 \times 204}{10 \times 99}\) = -0.236

Since, r_{12} > r_{13} and r_{23}. so, 1st and 2nd judge has nearest approaches to music.

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