**Mean Deviation about the Median and Mean Deviation from the Mean**

**Mean Deviation about the Median: **To find the mean deviation about the median we have to find the median of the given discrete frequency distribution. After arranging the observations in either ascending order, we shall find the sum of the frequency ∑fᵢ = N and compute the cumulative frequency then we shall identify the observations whose cumulative frequency is equal to or just greater then N/2. This is the median of the data.

Mean deviation(MD) = \(\frac{1}{N}\sum\limits_{i=1}^{n}{{{f}_{i}}|{{x}_{i}}-median|}\)

**Finding the Mean Deviation from the Mean for a Continuous Frequency Distribution: **Continuous frequency distribution is a series in which the data is classified into different class-intervals (without gap) along with their respect frequency (f_{i}).

**Example: **Mean deviation from the mean. Following table from the given tabular data

Sales (in Rs. Thousand) |
40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 | 80 – 90 |
90 – 100 |

Number of companies | 5 | 15 | 25 | 30 | 20 |
5 |

**Solution:**

Sales (in Rs. Thousand) |
Number of companies fᵢ | Mid of class Interval xᵢ | fᵢ xᵢ | |xᵢ – x̅| |
fᵢ |xᵢ – x̅| |

40 – 50 | 5 | 45 | 255 | 26 |
130 |

50 – 60 |
15 | 55 | 825 | 16 | 240 |

60 – 70 | 25 | 65 | 1625 | 6 |
150 |

70 – 80 |
30 | 75 | 2250 | 4 | 120 |

80 – 90 | 20 | 85 | 1700 | 14 |
280 |

90 – 100 |
5 | 95 | 475 | 24 | 120 |

∑ fᵢ = N = 100 | ∑ fᵢ xᵢ = 7100 |
∑fᵢ |xᵢ – x̅|= 1040 |

Here ∑fᵢ = N = 100

And \(\overline{x}=\frac{\sum{{{f}_{i}}{{x}_{i}}}}{N}\)

= 7100/100

x̅ = 71

Hence Mean deviation from the mean = 1/N [∑fᵢ |xᵢ – x̅|]

= [1/100] (1040)

= 10.4