The median of a set of data is the value that satisfies when half the values are above and half below. So, to find this out it is enough to sort the elements from smallest to largest and choose the central value.

- Case 1: The number of values is odd.

We have the following $$5$$ values: $$10, 5, 12, 3,14$$.

We sort them: $$3, 5, 12, 14, 19$$ and we take the central value: $$12$$ satisfies that there are two values above and two below.

- Case 2: The number of values is even.

The $$6$$ values are $$4, 6, 5, 7, 3, 9$$. We sort them: $$3, 4, 5, 6, 7, 9$$ and we take both central values, $$5$$ and $$6$$.

The median will be the average of both: $$$\displaystyle \frac{5+6}{2}=5,5$$$

Find the median of the following list of numbers of a Russian roulette: $$$0,3,14,16,19,20,21,22,30,36$$$ We calculate the average of both central numbers, $$19$$ and $$20$$.

The median is $$19,5$$

## Calculation of the median using stages

When we have a lot of data it is much more comfortable to use stages for the calculation of the median.

The grades from the biolgy exami of $$30$$ students are the following:

$$$3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7,$$$ $$$ 8, 8, 8, 10, 10$$$

Calculate the table of absolute and cumulative frequencies:

Mark | Absolute frequency | Cumulative frequency |

$$3$$ | $$2$$ | $$2$$ |

$$4$$ | $$5$$ | $$7$$ |

$$5$$ | $$8$$ | $$15$$ |

$$6$$ | $$6$$ | $$21$$ |

$$7$$ | $$4$$ | $$25$$ |

$$8$$ | $$3$$ | $$28$$ |

$$10$$ | $$2$$ | $$30$$ |

Having an even number of elements, there will be two central values, $$15$$ and $$16$$. Now, we have to find which mark these positions correspond to.

The $$15$$th position is the last mark with $$5$$. The $$16$$th is the first one with $$6$$. Then the median is: $$$\displaystyle \frac{6+5}{2}=5,5$$$