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.

Distribution of Data
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 = (∑ |xi – 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 mean
x̅ = (x1 + x2 + x3 + …… + xn) / n
x̅ = (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
(∑ |xi – mean| ) = 4 + 3 + 0 + 2 + 2 + 3
(∑ |xi – mean| ) = 14
Step 4: Dividing the sum of absolute difference and the number of data-points.
MAD = (∑ |xi – mean|) ÷ n
MAD = 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 mean
x̅ = (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-points
MAD = (∑ |xi – mean| ) ÷ n
MAD = 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 mean
x̅ = (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-points
MAD = (∑ |xi – mean|) ÷ n
MAD = 28 / 4
MAD = 7
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