1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
---|---|---|---|---|---|

Number of workers | 12 | 14 | 8 | 6 | 10 |

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

**Solution:**

Daily income | Cumulative frequency |
---|---|

Less than 120 | 12 |

Less than 140 | 26 |

Less than 160 | 34 |

Less than 180 | 40 |

Less than 200 | 50 |

2. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Weight (in kg) | Number of students |
---|---|

Less than 38 | 0 |

Less than 40 | 3 |

Less than 42 | 5 |

Less than 44 | 9 |

Less than 46 | 14 |

Less than 48 | 28 |

Less than 50 | 32 |

Less than 52 | 35 |

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

**Solution:**

Weight (in kg) | Frequency | Cumulative Frequency |
---|---|---|

36-38 | 0 | 0 |

38-40 | 3 | 3 |

40-42 | 2 | 5 |

42-44 | 4 | 9 |

44-46 | 5 | 14 |

46-48 | 14 | 28 |

48-50 | 4 | 32 |

50-52 | 3 | 35 |

Since `N = 35` and `n/2 = 17.5` hence median class = Less than 46-48

Here; `l = 46`, `cf = 14`, `f = 14` and `h = 2`

Median can be calculated as follows:

`text(Median)=l+(n/2-cf)/(f)xxh`

`=46+(17.5-14)/(14)xx2`

`=46+1/2=46.5`

This value of median verifies the median shown in ogive.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |
---|---|---|---|---|---|---|

Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |

Change this distribution to a more than type distribution, and draw its ogive.

**Solution:**

Production yield | Cumulative frequency |
---|---|

More than 50 | 100 |

More than 55 | 98 |

More than 60 | 90 |

More than 65 | 78 |

More than 70 | 54 |

More than 75 | 16 |

Copyright © excellup 2014