# Median

Median is the most middle value in the arrayed data. It means that when the data are arranged, the median is the middle value if the number of values is odd and the mean of the two middle values if the number of values is even. A value which divides the arrayed set of data into two equal parts is called the median, and the values greater than the median are equal to the values smaller than the median. It is also known as a positional average. It is denoted by $$\widetilde X$$ read as *X*– tilde.

__Median from Ungrouped Data__

Median = Value of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item

__Example__:

Find the median of the values **4, 1, 8, 13, 11**

__Solution__:

Arrange the data **1, 4, 8, 11, 13**

Median = Value of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item

Median = Value of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item = $${\left( {\frac{6}{2}} \right)^{th}} = {3^{th}}$$item

Median = **8**

__Example__:

Find the median of the values **5, 7, 10, 20, 16, 12**

__Solution__:

Arrange the data **5, 7, 10, 12, 16, 20**

Median = Value of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item

Median = Value of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item = $${\left( {\frac{7}{2}} \right)^{th}} = {3.5^{th}}$$item

Median = $$\frac{{10 + 12}}{2} = 11$$

__Median from Ungrouped Data__

For the median of grouped data, we find the cumulative frequencies and then calculate the median number $$\frac{n}{2}$$. The median lies in the group (class) which corresponds to the cumulative frequency in which $$\frac{n}{2}$$ lies. We use the following formula to find the median.

\[Median = l + \frac{h}{f}\left( {\frac{n}{2} – c} \right)\]

Here

$$l$$= Lower class boundary of the model class

$$f$$= Frequency of the median class

$$n = \sum f $$= Number of values or total frequency

$$c$$= Cumulative frequency of the class preceding the median class

$$h$$= Class interval size of the model class

__Example__**:**

Calculate the median from the following data:

Group |
60 – 64 |
65 – 69 |
70 – 74 |
75 – 79 |
80 – 84 |
85 – 89 |

Frequency |
1 |
5 |
9 |
12 |
7 |
2 |

__Solution__**:**

Group |
f |
Class Boundary |
Cumulative Frequency |

60 – 64 |
1 |
59.5 – 64.5 |
1 |

65 – 69 |
5 |
64.5 – 69.5 |
6 |

70 – 74 |
9 |
69.5 – 74.5 |
15 |

75 – 79 |
12 |
74.5 – 79.5 |
27 |

80 – 84 |
7 |
79.5 – 84.5 |
34 |

85 – 89 |
2 |
84.5 – 89.5 |
36 |

$$Median = l + \frac{h}{f}\left( {\frac{n}{2} – c} \right)$$ $$\because {\left( {\frac{n}{2}} \right)^{th}}$$item $$ = {\left( {\frac{{36}}{2}} \right)^{th}} = {18^{th}}$$item

\[Median = 74.5 + \frac{5}{{12}}\left( {18 – 15} \right) = 74.5 + \frac{5}{{12}}\left( 3 \right) = 75.75\]

__Median from Discrete Data__

When the data follows a discrete set of values grouped by size, we use the formula $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$ item for finding the median. First we form a cumulative frequency distribution, and the median is that value which corresponds to the cumulative frequency in which $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$ item lies.

__Example__:

The following frequency distribution is classified according to the number of leaves on different branches. Calculate the median number of leaves per branch.

No of Leaves |
1 |
2 |
3 |
4 |
5 |
6 |
7 |

No of Branches |
2 |
11 |
15 |
20 |
25 |
18 |
10 |

__Solution__**:**

No of LeavesX |
No of Branchesf |
Cumulative Frequency C.F |

1 |
2 |
2 |

2 |
11 |
13 |

3 |
15 |
28 |

4 |
20 |
48 |

5 |
25 |
73 |

6 |
18 |
91 |

7 |
10 |
101 |

Total |
101 |

Median = Size of $${\left( {\frac{{n + 1}}{2}} \right)^{th}}$$item $$ = \frac{{101 + 1}}{2} = \frac{{102}}{2} = 51$$item

Median = **5 **because $${51^{th}}$$ item corresponds to **5**