The median of a distribution is the value of the variable which divides it into two equal parts. It is the value such that the number of observations above it is equal to the number of observations below it The median is thus a positional average.

In the case of ungrouped data, if the number of observations is odd then the median is the middle value after the values have been arranged in ascending or descending order of magnitude. In the case of an even number of observations, there are two middle terms, and the median is obtained by taking the arithmetic mean of the middle terms.

**Steps to Calculate the Median For Grouped Data:**

- The grouped frequency distribution consists of class intervals and corresponding frequencies.
- Sum up the values all all the frequencies and denote it by the letter N. We have that \sum f_i = N.
- Calculate the (less than) cumulative frequencies for the given distribution. The cumulative freuquencies can be obtained by summing up the current and preceeding frequencies for each class interval.
- Calculate the value N/2 and note down the value of the cumulative frequency just greater than N/2.
- The class corresponding to the cumulative frequency just greater than N/2 is called the median class and the value of median is obtained by the formula given below.

**Formula for Median of Grouped Data:**

The median for data given in the form of a grouped frequency distribution table can be calculated using the formula, \text{Median }= l + \frac{h}{f}(N/2 - c)

where, l \text{ is the lower limit of the median class }\\ h \text{ is the size of the median class }\\ f \text{ is the frequency of the median class }\\ c \text{ is the cumulative frequency of the class interval preceding the median class }

**Example:**

Suppose we are given a grouped frequency distribution table as follows,

Class Interval | Frequency |

100-110 | 9 |

110-120 | 11 |

120-130 | 13 |

130-140 | 18 |

140-150 | 19 |

150-160 | 14 |

160-170 | 6 |

** Solution**: We first sum up all the frequencies to find N. We also calculate the cumulative frequencies as shown below.

Class Interval | Frequency | Cumulative Frequency |

100-110 | 9 | 9 |

110-120 | 11 | 9+11 = 20 |

120-130 | 13 | 9+11+13 = 33 |

130-140 | 18 | 51 |

140-150 | 19 | 70 |

150-160 | 14 | 84 |

160-170 | 6 | 90 |

N = 90 |

The value of N/2 is equal to N/2 = 90/2 = 45. The cumulative frequency just greater than N/2 is 51. Hence the corresponding class interval 130-140 is the median class. We have that l =130, h = 10, f = 18 and c = 33. On substituting these values in the above formula we have that, \text{Median }= l + \frac{h}{f}(N/2 - c) \text{Median }= 130 + \frac{10}{18}(45 - 33) \text{Median }= 130 + \frac{10\times 12}{18} = 130 + 6.6667 = 136.6667 We have thus found the required value od the median. Out of the 90 values in the abive data set around half of them lie below 136.6667 and around half of these values lie above the median.