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# The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Profit (in lakhs in Rs) | Number of shops (frequency) |
---|---|

More than or equal to 5 | 30 |

More than or equal to 10 | 28 |

More than or equal to 15 | 16 |

More than or equal to 20 | 14 |

More than or equal to 25 | 10 |

More than or equal to 30 | 7 |

More than or equal to 35 | 3 |

Given:

We have to draw both ogives for the above data and hence obtain the median.

Solution:

We first prepare the cumulative frequency distribution table by more than method as given below:

Profit (in lakhs in Rs) | Number of shops (Cumulative frequency) |

More than or equal to 5 | 30 |

More than or equal to 10 | 28 |

More than or equal to 15 | 16 |

More than or equal to 20 | 14 |

More than or equal to 25 | 10 |

More than or equal to 30 | 7 |

More than or equal to 35 | 3 |

Represent profits(in lakhs) along X-axis and cumulative frequency along Y-axis.

Plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) on the graph and join them in free hand to get a more than ogive.

We first prepare the cumulative frequency distribution table by less than method as given below:

Represent profits(in lakhs) along X-axis and cumulative frequency along Y-axis.

Plot the points (10, 3), (15, 7), (20, 10), (25, 14), (30, 16), (35, 28) and (40, 30) on the graph and join them in a free hand to get a less than ogive.

Total number of days $N = 30$

This implies,

$\frac{N}{2} = \frac{30}{2}=15$

Draw a line parallel to X-axis at the point of intersection of both ogives, which further intersect at (0, 12) on Y-axis.

Draw a line perpendicular to X-axis at point of intersection of both ogives, which further intersect at (17.50, 0) on X-axis.

Therefore, 17.50 is the required median using ogives.

Hence, the median is 17.50 lakhs.