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No of Students per teacher |
Number of states / U.T |

15-20 |
3 |

20-25 |
8 |

25-30 |
9 |

30-35 |
10 |

35-40 |
3 |

40-45 |
0 |

45-50 |
0 |

50-55 |
2 |

Answer :

The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30,

h = 5, f_{1} = 10,

f_{0} = 9 and

f_{2} = 3

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 30 \ + \ ( \frac{10 \ - \ 9}{2 \ × \ 10 \ - \ 9 \ - \ 3}) \ × \ 5 \)

\(= \ 30 \ + \ \frac{10 \ - \ 9}{20 \ - \ 9 \ - \ 3} \ × \ 5 \)

\(= \ 30 \ + \ \frac{1}{8} \ × \ 5 \ = \ 30 \ + \ 0.625 \)

\( = \ 30.625 \ = \ 30.6 \)

Calculation of mean :

Class Interval | Frequency \( (f_i) \) | Mid-point \( (x_i) \) | \( f_i x_i \) |

15-20 | 3 | 17.5 | 52.5 |

20-25 | 8 | 22.5 | 180.0 |

25-30 | 9 | 27.5 | 247.5 |

30-35 | 10 | 32.5 | 325.0 |

35-40 | 3 | 37.5 | 112.5 |

40-45 | 0 | 42.5 | 0 |

45-50 | 0 | 47.5 | 0 |

50-55 | 2 | 52.5 | 105.5 |

\( \sum f_i \ = \ 35 \) | \( \sum f_i x_i \ = \ 1022.5 \) |

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ \frac{1022.5}{35} \ = \ 29.2 \)

\(\therefore \) mean = 29.2

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