The mean absolute deviation (mean deviation about the mean) is a measure of the degree of dispersion of data values.

It involves measuring the absolute difference between the data values from the mean and taking the average of these differences. Thus the mean absolute deviation measures the “average” distance of the data values from the mean.

**How to Find the Mean Absolute Deviation:**

- We first calculate the mean using the formula,
**X̄**= ∑*x*/n._{i} - Calculate the differences in the data values from the mean (
*x*–_{i}**X̄**). - Take the absolute value of the above differences. This simply means ignoring the sign of the difference if it is negative.
- Take the average of the above absolute values.

**Mean Absolute Deviation Formula (Raw and Ungrouped Data):**

The mean absolute deviation for raw data can be calculated using the formula,

Mean Absolute Deviation = ∑|*x _{i}* –

**X̄**|/n.

If the data is given in ungrouped tabular form with frequencies f_{i} then the mean absolute deviation can be found using the formula,

Mean Absolute Deviation = ∑f_{i}|*x _{i}* –

**X̄**|/n.

**Example:**

Consider the data values 2, 3, 7, 8, and 9.

*Step 1*: We first obtain the mean as follows, **X̄**=(2+3+7+8+9)/5=5.8.

*Step 2*: We find the absolute differences in the data values from the mean,

- |2 – 5.8| = |-3.8| = 3.8
- |3 – 5.8| = |-2.8| = 2.8
- |7 – 5.8| = |1.2| = 1.2
- |8 – 5.8| = |2.2| = 2.2
- |9 – 5.8| = |3.2| = 3.2

*Step 3*: We now calculate the average of the above absolute values,

Mean Absolute Deviation = (3.8+2.8+1.2+2.2+3.2)/5= 13.2/5 = 2.64.

**Mean Absolute Deviation for Grouped Data:**

Suppose that the data is given in tabular form with class intervals and corresponding frequencies.

- We first calculate the class mark x
_{i}by taking the mid-value for each of the class intervals. - We then calculate the mean using the formula,
**X̄**= ∑f_{i}*x*/∑f_{i}_{i}. - The mean deviation for grouped data can then be calculated using the formula, Mean Absolute Deviation = ∑f
_{i}|*x*– X̄|/∑f_{i}_{i}.

**Example:**

Consider the following frequency distribution,

Class Intervals | Frequency |

0-10 | 2 |

10-20 | 3 |

20-30 | 4 |

30-40 | 1 |

40-50 | 5 |

*Step 1*: We calculate the class mark (mid-value) and the mean of the data.

Class Intervals | Class Mark x_{i} | Frequency f_{i} | f_{i}x_{i} |

0-10 | 5 | 2 | 10 |

10-20 | 15 | 3 | 45 |

20-30 | 25 | 4 | 100 |

30-40 | 35 | 1 | 35 |

40-50 | 45 | 5 | 225 |

∑f_{i} = 15 | ∑f_{i}x = 415 _{i} |

Mean = ∑f_{i}*x _{i}*/∑f

_{i}= 415/15 = 27.67.

*Step 2*: We calculate the absolute differences and take their average.

Class Intervals | Class Mark x_{i} | |x – X̄|=_{i}|x_i-27.67| | Frequency f_{i} | f_{i}|x – X̄|_{i} |

0-10 | 5 | 22.67 | 2 | 45.34 |

10-20 | 15 | 12.67 | 3 | 38.01 |

20-30 | 25 | 2.67 | 4 | 10.68 |

30-40 | 35 | 7.33 | 1 | 7.33 |

40-50 | 45 | 17.33 | 5 | 86.65 |

∑f_{i} = 15 | ∑f_{i}|x – X̄| = 188.01_{i} |

Mean Absolute Deviation = ∑f_{i}|*x _{i}* – X̄|/∑f

_{i}= 188.01/15 = 12.534.