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Question

What is the mean deviation from the mean of the numbers 10, 9, 21, 16, 24 ?

This question was previously asked in
NDA I 2016 GAT Previous Year Paper (17-Apr-2016)
The correct answer is

5.2

Calculate Mean Deviation from the Mean

The mean deviation from the mean is a measure of dispersion that tells us how spread out the numbers in a dataset are, on average, from their mean. It is calculated as the average of the absolute differences between each number in the dataset and the mean of the dataset.

Understanding Mean Deviation

Mean deviation provides a straightforward way to understand the typical distance of data points from the central value (the mean). It uses absolute values of deviations to avoid positive and negative differences cancelling each other out.

Steps to Calculate Mean Deviation from the Mean

  1. Calculate the mean (\(\bar{x}\)) of the dataset.
  2. Find the absolute difference between each number (\(x_i\)) in the dataset and the mean (\(|\text{x}_i - \bar{x}|\)).
  3. Calculate the mean of these absolute differences. This is the mean deviation from the mean.

Detailed Calculation for the Given Numbers

The given numbers are 10, 9, 21, 16, 24.

Let's follow the steps to find the mean deviation from the mean for these numbers.

Step 1: Calculate the Mean (\(\bar{x}\))

The mean is the sum of all numbers divided by the count of numbers.

Number of observations (\(n\)) = 5

Sum of numbers (\(\Sigma x_i\)) = 10 + 9 + 21 + 16 + 24 = 80

Mean (\(\bar{x}\)) = \(\frac{\Sigma x_i}{n} = \frac{80}{5} = 16\)

The mean of the numbers is 16.

Step 2: Find the Absolute Deviations from the Mean

Now, we find the absolute difference between each number and the mean (16).

Number (\(x_i\)) Mean (\(\bar{x}\)) Difference (\(x_i - \bar{x}\)) Absolute Difference (\(|x_i - \bar{x}|\))
10 16 10 - 16 = -6 |-6| = 6
9 16 9 - 16 = -7 |-7| = 7
21 16 21 - 16 = 5 |5| = 5
16 16 16 - 16 = 0 |0| = 0
24 16 24 - 16 = 8 |8| = 8

The sum of the absolute differences (\(\Sigma |x_i - \bar{x}|\)) = 6 + 7 + 5 + 0 + 8 = 26.

Step 3: Calculate the Mean Deviation from the Mean

The mean deviation is the average of the absolute differences.

Mean Deviation = \(\frac{\Sigma |x_i - \bar{x}|}{n} = \frac{26}{5}\)

Mean Deviation = 5.2

The mean deviation from the mean of the given numbers is 5.2.

Revision Table: Key Concepts in Mean Deviation

Concept Description Formula (from Mean)
Mean (\(\bar{x}\)) Average of the dataset. \(\bar{x} = \frac{\Sigma x_i}{n}\)
Deviation Difference between a data point and the mean. \(x_i - \bar{x}\)
Absolute Deviation Positive value of the difference. \(|x_i - \bar{x}|\)
Mean Deviation Average of the absolute deviations. MD = \(\frac{\Sigma |x_i - \bar{x}|}{n}\)

Additional Information: Measures of Dispersion

Mean deviation is one type of measure of dispersion. Measures of dispersion quantify how spread out the data points are. Other common measures include:

  • Range: The difference between the maximum and minimum values in a dataset.
  • Variance: The average of the squared differences from the mean.
  • Standard Deviation: The square root of the variance. It is widely used and has the same units as the original data.
  • Quartile Deviation (Semi-Interquartile Range): Half the difference between the third and first quartiles.

Each measure of dispersion provides a different perspective on the spread of the data. Mean deviation is easy to understand but is less frequently used in inferential statistics compared to standard deviation because it uses absolute values, which are mathematically harder to handle in further calculations.

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Similar Questions

  1. What is the mean deviation about the mean ?

  2. What is the coefficient of mean deviation of 21, 34, 23, 39, 26, 37, 40, 20, 33, 27 (taken from mean)?

  3. What is the mean deviation of first 10 even natural numbers?

  4. The sum of deviations of n number of observations measured from 2.5 is 50. The sum of deviations of the same set of observations measured from 3.5 is -50. What is the value of n?


Important Questions from Mean Deviation

  1. What is the mean deviation about the mean ?

  2. The mean deviation about median of 10 observations is 15. If each observation is multiplied by $-3$, then find the new mean deviation about median of resulting observations.
  3. Let xi, i = 1, 2, ..., n be n observations and wi = pxi + k, i = 1, 2, ..., n where p and k are constants. If the mean of xi's is 48 and standard deviation is 12, whereas the mean of wi's is 55 and standard deviation is 15, then the value of p and k should be

  4. If the mean deviation 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to

  5. The mean of 5 observation is 5 and their variance is 124. If three of the observations are 1, 2, 6, then the mean deviation from the mean of the data is

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