Sum

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

No. of bulbs
produced (Thousands) |
30 - 40 | 40 - 50 | 50 - 60 | 60 - 70 | 70 - 80 | 80 - 90 | 90 - 100 |

No. of factories | 12 | 35 | 20 | 15 | 8 | 7 | 8 |

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#### Solution

Class
(Number of bulbs produced in thousands) |
Frequency (Number of factories) f _{i} |
Cumulaive frequency less than the upper limit |

30 - 40 | 12 | 12 |

40 - 50 | 35 | 47 |

50 - 60 (Median Class) |
20 | 67 |

60 - 70 | 15 | 82 |

70 - 80 | 8 | 90 |

80 - 90 | 7 | 97 |

90 - 100 | 8 | 105 |

N = 105 |

From the above table, we get

L (Lower class limit of the median class) = 50

N (Sum of frequencies) = 105

h (Class interval of the median class) = 10

f (Frequency of the median class) = 20

cf (Cumulative frequency of the class preceding the median class) = 47

Now, Median = \[L + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\]

\[= 50 + \left( \frac{\frac{105}{2} - 47}{20} \right) \times 10\]

= 50 + 2.75

= 52.75 thousand lamps

= 52750 lamps

Hence, the median of the productions is 52750 lamps.

Concept: Tabulation of Data

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