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Literacy rate (in %) |
45-55 |
55-65 |
65-75 |
75-85 |
85-98 |

Number of cities |
3 |
10 |
11 |
8 |
3 |

Answer :

In this case, the value of mid-point (x_{i}) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, \( u_i \ = \ \frac{x_i \ - \ A}{h} \)

\(=> \ u_i \ = \ \frac{x_i \ - \ 70}{10} \)

Class Interval | Frequency \( (f_i) \) | \( (x_i) \) | \( d_i \ = \ x_i \ - \ A \) | \(u_i \ = \ \frac{d_i}{h} \) | \( f_i u_i \) |

45-55 | 3 | 50 | -20 | -2 | -6 |

55-65 | 10 | 60 | -10 | -1 | -10 |

65-75 | 11 | 70 | 0 | 0 | 0 |

75-85 | 8 | 80 | 10 | 1 | 8 |

85-95 | 3 | 90 | 20 | 2 | 6 |

\( \sum f_i \ = \ 35 \) | \( \sum f_i u_i \ = \ -2 \) |

So,

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

\( = \ 70 \ + \ \frac{-2}{35} \ × \ 10 \)

\(= \ 69.42 \)

\(\therefore \) The mean literacy part is 69.42

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