In a continuous frequency distribution, the median of the data is 21.If each observation is increased by 5 find the new median

Given :

In a continuous frequency distribution, the median of the data is 21.

And

If each observation is increased by 5 .

Then the new median is also increased by 5 , so new median = 21 + 5 = 26 ( Ans )

We can understand it As :

Suppose there are 20 children and the number of toys with each child is given below :

3, 5, 6, 6, 8, 8, 9, 9, 10, 11, 11, 12, 12, 12, 14, 15, 15, 18, 18, 20

First we form a continuous frequency table , As :

Number of toys | 0 - 5 | 5 - 10 | 10 - 15 | 15 - 20 | 20-25 |

Frequency | 1 | 7 | 7 | 4 | 1 |

Here frequency is number of children that have toys in between ( 0 -5 , 5 - 10 , 10- 15 , 15 - 20 , 20-25)

CLASS INTERVALS | Frequency | Cumulative Frequency |

0-5 | 1 | 1 |

5-10 | 7 | 8 |

10-15 | 7 | 15 |

15-20 | 4 | 19 |

20-25 | 1 | 20 |

Now, median class is 10-15

lower limit of median class = 10

Frequency of median class, f = 7

Cumulative frequency of the class preceeding median class, cf = 8

Class size, h = 5

$\mathrm{Median}=\mathrm{l}+\frac{\mathrm{n}/2-\mathrm{cf}}{\mathrm{f}}\times \mathrm{h}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=10+\frac{10-8}{7}\times 5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=10+\frac{10}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{80}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

When each of the given observation is increased by 5, then we get

8,10,11,11,13,13,14,14,15,16,16,17,17,17,19,20,20,23,23,25

CLASS INTERVALS | FREQUENCY | CUMULATIVE FREQUENCY |

0-5 | 0 | 0 |

5-10 | 1 | 1 |

10-15 | 7 | 8 |

15-20 | 7 | 15 |

20-25 | 4 | 19 |

25-30 | 1 | 20 |

Now, median class is 15-20

lower limit of median class = 15

Frequency of median class, f = 7

Cumulative frequency of the class preceeding median class, cf = 8

Class size, h = 5

$\mathrm{Median}=\mathrm{l}+\frac{\mathrm{n}/2-\mathrm{cf}}{\mathrm{f}}\times \mathrm{h}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=15+\frac{10-8}{7}\times 5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{115}{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{increase}\mathrm{in}\mathrm{median}=\frac{115}{7}-\frac{80}{7}=\frac{35}{7}=5$

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