__7.1c Median__**1.**The

**median of a group of data**refers to the value which is at the

middle of the data after the data has been arranged according to

grouped data and ungrouped data.

**(A) Ungrouped Data**

**Example 1:**

Find the median for each of the sets of data given below.

**(a)**15, 18, 21, 25, 20, 18

**(b)**13, 6, 9, 17, 11

*Solution:***(a) Arrange the data in the ascending order**

15, 18, 18, 20, 21, 25

$$\text{Median}={T}_{\frac{n+1}{2}}={T}_{\frac{6+1}{2}}={T}_{3\frac{1}{2}}=\frac{18+20}{2}=19$$

**(b)**6, 9, 11, 13, 17

$$\text{Median}={T}_{\frac{n+1}{2}}={T}_{\frac{5+1}{2}}={T}_{3}=11$$

**(B) Grouped Data (without Class Interval)**

**Example 2:**

The frequency table shows the marks obtained by 40 students in a

biology test.

Marks |
50 |
55 |
60 |
65 |
70 |

Number of students |
6 |
8 |
15 |
10 |
1 |

**Solution:**$$\text{Median}={T}_{\frac{n+1}{2}}={T}_{\frac{40+1}{2}}={T}_{20\frac{1}{2}}=60\leftarrow (20\frac{1}{2}\text{thtermis60)}$$

**(C) Grouped Data (with Class Interval)**

*m*= median

*L*= Lower boundary of median class

*N*= Number of data

*F*= Total frequency before median class

*f*= Total frequency in median class

_{m}*c*= Class size = (Upper boundary – lower boundary)

**1.**The median can be determined from an accumulative frequency

table and the ogive.

**2.**The ogiveis an accumulative graph; the median, quartiles and the

range between quartilescan be determined from it.

**Example 3:**

The grouped frequency distribution was obtained from 100 students regarding the scores in their test shown as below.

Scores |
Frequency |

5 – 9 |
4 |

10 – 14 |
10 |

15 – 19 |
19 |

20 – 24 |
26 |

25 – 29 |
21 |

30 – 34 |
12 |

35 – 39 |
8 |

*Solution:***Method 1: using formula**

Scores |
Frequency |
Cumulative Frequency |

5 – 9 |
4 |
4 |

10 – 14 |
10 |
14 |

15 – 19 |
19 |
33 |

20 – 24 |
26 |
59 |

25 – 29 |
21 |
80 |

30 – 34 |
12 |
92 |

35 – 39 |
8 |
100 |

$$\begin{array}{l}Step\text{}1\\ \text{Medianclassisgivenby}{T}_{\frac{n}{2}}=\text{}{T}_{\frac{100}{2}}={T}_{50}\\ \therefore \text{Medianclassistheclass}20-24\\ \\ Step\text{}2\\ m=L+\left(\frac{\frac{N}{2}-F}{{f}_{m}}\right)c\\ m=19.5+\left(\frac{\frac{100}{2}-33}{26}\right)5\\ m=19.5+3.269=22.77\end{array}$$