# Mean Absolute Deviation

The measure of spread represents the amount of dispersion in a data-set. i.e how spread-out are the values of data-set around the central value(example- **mean/mode/median**).It tells how far away the data points tend to fall from the central value.

- The lower value of the measure of spread reflects that the data points are close to the central value. In this case, the values in a data-set are more consistent.
- Further, the distance of the data points from the central-value, the greater is the spread. whereas here, the values are not much consistent.

Using the above diagram, we can infer that the narrow distribution represents a lower spread, and the broad distribution represents a higher spread.

## Range

The range is the simplest measure of variation. It is defined and calculated as the **difference between the largest and smallest values** of the data-set.

Range = largest value – smallest value |

- A small value of range means the data is quite consistent and most of the data-points lie near to the mean.
- Whereas a higher range means the data is quite inconsistent and the data has extreme values, and the data-points don’t lie near to the mean.
- The range doesn’t consider each value of the dataset. Thus, it gives a rough idea of the data-set and its variability.

### Examples

**Example 1: The data given are: 8, 10, 4, 1, 15. Calculate the range of the given data?**

**Solution**:

The data in ascending order is = 1 4 8 10 15.

Range = largest – smallest

= 15 – 1

Range = 14

**Example 2: What is the range of these integers?**

**14, -18, 7, 0, -5, -8, 15, -10, 20**

**Solution: **

The data in ascending order is = -18, -10, -8, -5, 0, 7, 14, 15, 20

Range = largest – smallest

= 20 – (-18)

Range = 38

**Example 3: Calculate the range of the given data:**

**8, 10, 5 , 14 , 42, 3566**

**Solution: **

The data in ascending order is = 5, 8, 10, 14, 42, 3566

Range = largest – smallest

= 3566 – 5

Range = 3561

## Mid-Range

The mid-range is the value midway between the largest and smallest value of a data-set. It is calculated as the mean of the largest value and smallest value of the data-set.

Mid-Range = (largest value + smallest value)/2 |

### Examples

**Example 1: The data given is 8, 10, 5, 9, 11. Calculate the mid-range of the given data?**

**Solution: **

The data in ascending order is = 5 8 9 10 11

Mid-Range = (largest value + smallest value)/2

= (5 + 11)/2

= 16/2

mid-range = 8

**Example 2: You take 7 statistics tests over the course of a semester. You score 94, 88, 74, 84, 91, 87 and 79. What is the mid-range of your scores?**

**Solution:**

The scores in ascending order is = 74 79 84 87 88 91 94

Mid-Range = (largest value + smallest value)/2

= (94 + 74)/2

= 168/2

mid-range= 84

**Example 3: The height of 8 students in centimeters is given as 120, 132, 117, 126, 110, 135, 150, and 143. Calculate the mid-range of the given data?**

**Solution:**

The scores in ascending order is = 110 117 120 126 132 135 143 150

Mid-Range = (largest value + smallest value)/2

= (150 + 110)/2

= 260/2

mid-range = 130

**Mean Absolute Deviation (MAD)**

The mean absolute deviation (MAD) of a data-set is the average distance between each data point of the data-set and the mean of data. i.e it represents the amount of variation that occurs around the mean value in the data-set. It is also a measure of variation. It is calculated as the average of the sum of the absolute difference between each value of the data-set and the mean.

MAD = (∑ |x_{i} – mean| ) ÷ n |

where 1 < i < n and n is the number of data-points in the data-set.

### Examples

**Example 1: The data-set is 11 , 15 , 18 , 17 , 12 , 17. Calculate the mean absolute deviation of the given data-set?**

**Solution:**

Step 1: Calculating the meanx̅ = (x

_{1}+ x_{2}+ x_{3}+ …… + x_{n}) / nx̅ = (11 + 15 + 18 + 17 + 12 + 17 ) / 6

x̅ = 15

The mean of the given data = 15

Step 2: Calculating the absolute difference between each data-point and mean.

Data-Point | Absolute Difference from mean |
---|---|

11 | |11 – 15| = 4 |

12 | |12 – 15| = 3 |

15 | |15 – 15| = 0 |

17 | |17 – 15| = 2 |

17 | |17 – 15| = 2 |

18 | |18 – 15| = 3 |

Step 3: Adding the Absolute Difference together(∑ |x

_{i}– mean| ) = 4 + 3 + 0 + 2 + 2 + 3

(∑ |x_{i }– mean| ) = 14

Step 4: Dividing the sum of absolute difference and the number of data-points.MAD = (∑ |x

_{i}– mean|) ÷ nMAD = 14/6

MAD = 2.33

Hence, we can conclude that, on average, each data-point is 2 distance away from the mean.

**Example 2: The following table shows the number of oranges that grew on Nancy’s orange tree each season**

Season | Number of Oranges |
---|---|

Winter | 5 |

Summer | 17 |

Spring | 24 |

Fall | 10 |

**Find the mean absolute deviation (MAD) of the data set?**

**Solution:**

Step 1: Calculating the meanx̅ = (x1 + x2 + x3 + …… + xn) / n

x̅ = (5 + 17 + 24 + 10) / 4

x̅ = 56/4

The mean of the given data = 14

Step 2: Calculating the absolute difference between each data-point and mean

Data-Point | Absolute Difference from mean |
---|---|

5 | |5 – 14| = 9 |

17 | |17 – 14| = 3 |

24 | |24 – 14| = 10 |

10 | |10 – 14| = 4 |

Step 3:Adding the Absolute Difference together(∑ |xi – mean| ) = 9 + 3 + 10 + 4

(∑ |xi – mean| ) = 26

Step 4: Dividing the sum of absolute difference and the number of data-pointsMAD = (∑ |x

_{i}– mean| ) ÷ nMAD = 26 / 4

MAD = 6.5

**Example 3: Consider the following data-set**

Name of the student | Marks in Maths |
---|---|

Chetan | 90 |

Shubham | 74 |

Riya | 80 |

Manu | 92 |

**Calculate the mean absolute deviation of the given data?**

**Solution:**

Step 1: Calculating the meanx̅ = (x1 + x2 + x3 + …… + xn) / n

x̅ = (90 + 74 + 80 + 92) / 4

x̅ = 336/4

The mean of the given data = 84

Step 2: Calculating the absolute difference between each data-point and mean

Data-Point | Absolute Difference from mean |
---|---|

90 | |90 – 84| = 6 |

74 | |74 – 84| = 10 |

80 | |80 – 84| = 4 |

92 | |92 – 84| = 8 |

Step 3: Adding the Absolute Difference together(∑ |xi – mean| ) = 6 + 10 + 4 + 8

(∑ |xi – mean| ) = 28

Step 4: Dividing the sum of absolute difference and the number of data-pointsMAD = (∑ |xi – mean|) ÷ n

MAD = 28 / 4

MAD = 7

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