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Mean absolute deviation (MAD) is a variable that indicates the average distance between their observation and the mean. Mean absolute deviation is necessary to calculate the mean data. Mean measures the average of the observation, while deviation refers to the variance of the previous data. Thus, mean absolute deviation refers to the average distance of each observation from the mean of given data information. When we combine mean and deviation, we can define deviation as the average distance for each observation from the mean of data.
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Keyterms: Mean, Mean absolute deviation, Standard deviation, Calculations, Data point, Statistics
Mean Absolute Deviation
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Mean absolute deviation (MAD) is a variable that indicates the average distance between their observation and the mean. Mean absolute deviation uses initial data units to make interpretations easier. Large numbers indicate that data points are spread out from the average of the data.
In contrast, lower values are associated with closely related data points. The rate of total deviation is also known as the average deviation. This definition of deviation sounds similar to standard deviation (SD). Although they both measure variability, they have different calculations.
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Mean Absolute Deviation Formula
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Mean Absolute Deviation formula is given by,
\(MAD = \frac{\sum |X - \mu|}{N}\)
Where,
|X – µ| = Absolute deviation
X = Value of the data point
µ = Mean
N = Sample size
Statistics include total deviation. The deviation is the difference between a data point and the mean. The total deviation rate automatically removes any subtraction that occurs.
Steps to Calculate Mean Absolute Deviation
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We can obtain the mean deviation from the mean by following the below methods.
- Find the mean of the given observations.
- Calculate the difference between each observation and the calculated mean.
- Evaluate the mean of the differences obtained in the second step.
\(\frac{1}{n} \sum^n_{i=1} |x_i -m| = \frac{|x_1-m| + |x_2 -m|+...+|x_n -m|}{n}\)
Calculation of Mean Absolute Deviation
Suppose a deviation from the average value a is given as (x - a), where x is the perception of any data set. To find the average deviation, we need to find the sum of all deviations in a data set. Since the average inclination is between high values and minimum set data, we can see that some deviations will be positive and some will be negative. The sum of such deviations will give zero.
Also Read: Types of Probability
Example of Mean Absolute Deviation
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Consider this example where students of a class scored the following marks in an exam - 55, 65.70, 70, 72, 85, 90, 93, 100. Here, we should find the mean absolute deviation.
Let us first find the mean of the given data,
Mean = (55+65+70+70+72+85+90+93+100)/ 10
Mean = 790/ 10
Mean = 79
Now, let us calculate the deviation,
| Marks | Deviation |
|---|---|
| 55 | 55 - 79 = - 24 |
| 65 | 65 - 79 = - 14 |
| 70 | 70 - 79 = - 9 |
| 70 | 70 - 79 = - 9 |
| 72 | 72 - 79 = - 7 |
| 85 | 85 - 79 = 6 |
| 90 | 90 - 79 = 11 |
| 90 | 90 - 79 = 11 |
| 93 | 93 - 79 = 14 |
| 100 | 100 - 79 = 21 |
Now calculate the mean absolute deviation.
Mean Absolute Deviation = |- 24 - 14 - 9 - 9 - 7 + 6 + 11 + 11 + 14 + 21 = 0|/ 10
Mean Absolute Deviation = (24 + 14 + 9 + 9 + 7 + 6 + 11 + 11 + 14 + 21)/ 10
Mean Absolute Deviation = 126/ 10
Mean Absolute Deviation = 12.6
This gives us a clear picture of the deviation of observation. Therefore, we can say that each point in the data set measures a distance of 12.6 from the mean of 79.
Things to Remember
- Mean is often used to measure central tendency. It represents the amount of data collection provided and also, it is suitable for accurate and continuous data.
- The mean deviation indicates the scattering of various series objects from their average value. The values of the series extremes are not significantly affected.
- In mathematics, mean absolute deviation (MAD) is a strong measure of the variability of a consistent sample of quantitative data.
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Sample Questions
Ques. Calculate the mean absolute of the following table. (5 marks)

Ans. X = 85+75+80/3 = 80
| Name of the students | Max. Marks (percentage) | Deviation |
|---|---|---|
| Anmol | 85 | 85 - 80 = 5 |
| Kushagra | 75 | 75 - 80 = -5 |
| Garima | 80 | 80 - 80 = 0 |
Mean Deviation = Sum of all the deviation from mean/Total number of observations
Mean Deviation = 5 + (-5) + 0 = 0
Mean Absolute Deviation = 5 + ?-5 ?+ 0 / 3 = 10/3 = 3.333
Ques. Calculate the mean absolute deviation of the given observations mentioned below. (5 marks)

Ans. Mean = 46 + 40 + 46 + 44/4 = 44
| Name of subjects | Maximum marks | Deviation |
|---|---|---|
| Maths | 46 | 46 - 44= 2 |
| Science | 40 | 40 - 44= -4 |
| English | 46 | 46 - 44 = 2 |
| Computer | 44 | 44 - 44 = 0 |
Mean Absolute Deviation = 2+ |-4|+ 2 / 4 = 2
This gives us a clear idea about the deviation of the observations from the measure of the central tendency.
Ques. What is the use of absolute mean deviation in daily lives? (2 marks)
Ans. Many practitioners use means in their daily lives, for example, teachers give tests to students and then measure the results to see if the score is high, average, or very low. Each scale tells different variations. A complete deviation can further help determine the distance between each score and the average initial score.
Ques. Will there be a negative value to the mean absolute deviation? (2 marks)
Ans. If the mean absolute deviation is zero, the total deviation is all zero. The deviation from the mean definition is determined by MAD and the distance is never negative. If all the absolute deviations are zero, the only way for MAD can average is to zero.
Ques. Find the median of the data using an empirical formula, when it is given that mode = 35.3 and mean = 30.5. (CBSE 2014) (5 marks)
Ans. Mode = 3(Median) – 2(Mean)
35.3 = 3(Median) – 2(30.5)
35.3 = 3(Median) – 61
96.3 = 3 Median
Median = 96.3/ 3
Median = 32.1
Ques. Show that the mode of the series obtained by combining the two series S1 and S2 given below is different from that of S1 and S2 taken separately: (CBSE 2015) (5 marks)
S1: 3, 5, 8, 8, 9, 12, 13, 9, 9
S2: 7, 4, 7, 8, 7, 8, 13
Ans. In S1 : Number 9 occurs 3 times (maximum)
∴ Mode of S1 Series = 9
In S2 : Number 7 occurs 3 times (maximum)
∴ Mode of S, Series = 7
After combination:
In S1 & S2 : No. 8 occurs 4 times (maximum)
∴ Mode of S1 & S2 taken combined = 8
So, the mode of S1 & S2 combined is different from that of S1 & S2 taken separately.
Ques. From the following frequency distribution, find the median class: (2015) (3 marks)

Ans.
| Cost of living index | No.of weeks | c.f |
|---|---|---|
| 1400-1550 | 8 | 8 |
| 1550-1700 | 15 | 23 |
| 1700-1850 | 21 | 44 |
| 1850-2000 | 8 | 52 |
| 52 |
Here, n = 52
n/2 = 52/2 =26
∴ Median class 1700 – 1850.
Ques. Following table shows the sale of shoes in a store for one month: (2014) (3 marks)

Find the model size of the shoes sold.
Ans. Maximum no. of pairs sold = 25 (size 5)
∴ Modal size of shoes = 5
Ques. Weekly household expenditure of families living in a housing society are shown below: (2014) (2 marks)

Find the upper limit of the modal class.
Ans. Maximum frequency = 48
∴ Modal class=9,000 – 12,000
Upper limit of the modal class=12,000
Ques. Convert the following frequency distribution to a ‘more than’ type cumulative frequency distribution. (2012) (2 marks)

Ans.
| Marks Obtained | No.of Students | c.f |
|---|---|---|
| 0-20 | 5 | 5 |
| 20-40 | 9 | 14 |
| 40-60 | 12 | 26 |
| 60-80 | 8 | 34 |
| 80-100 | 6 | 40 |
| 40 |
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