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Question

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

This question was previously asked in
NDA I 2022 GAT Previous Year Paper (10-Apr-2022)
The correct answer is 0.22

Calculating the Coefficient of Mean Deviation from Mean

The question asks us to find the coefficient of mean deviation for a given set of data points, using the mean as the reference point for deviation.

The data set is: 21, 34, 23, 39, 26, 37, 40, 20, 33, 27.

There are 10 data points, so $n = 10$.

Understanding Coefficient of Mean Deviation

The coefficient of mean deviation (CMD) is a relative measure of dispersion. It is used to compare the variability or dispersion of different data sets, especially when their means are significantly different or they are measured in different units. It is calculated as the ratio of the Mean Deviation (MD) to an average (like mean, median, or mode).

The formula for the coefficient of mean deviation from the mean is:

$$ \text{Coefficient of Mean Deviation (CMD)} = \frac{\text{Mean Deviation (MD)}}{\text{Mean} (\bar{x})} $$

The Mean Deviation (MD) from the mean is the average of the absolute deviations of each data point from the mean:

$$ \text{Mean Deviation (MD)} = \frac{\sum |x - \bar{x}|}{n} $$

Where:

  • $x$ represents each data point.
  • $\bar{x}$ represents the mean of the data.
  • $|x - \bar{x}|$ is the absolute deviation of $x$ from the mean.
  • $\sum |x - \bar{x}|$ is the sum of the absolute deviations.
  • $n$ is the number of data points.

Step-by-Step Calculation

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

First, we need to find the mean of the given data points. The mean is the sum of all data points divided by the number of data points.

Sum of data points ($\sum x$) = $21 + 34 + 23 + 39 + 26 + 37 + 40 + 20 + 33 + 27 = 300$

Number of data points ($n$) = 10

Mean ($\bar{x}$) = $$ \frac{\sum x}{n} = \frac{300}{10} = 30 $$

The mean of the data is 30.

Step 2: Calculate the Absolute Deviations from the Mean

Next, we calculate the absolute difference between each data point and the mean (30). We can organize this in a table.

Data Point (x) Mean ($\bar{x}$) Deviation ($x - \bar{x}$) Absolute Deviation ($|x - \bar{x}|$)
21 30 21 - 30 = -9 |-9| = 9
34 30 34 - 30 = 4 |4| = 4
23 30 23 - 30 = -7 |-7| = 7
39 30 39 - 30 = 9 |9| = 9
26 30 26 - 30 = -4 |-4| = 4
37 30 37 - 30 = 7 |7| = 7
40 30 40 - 30 = 10 |10| = 10
20 30 20 - 30 = -10 |-10| = 10
33 30 33 - 30 = 3 |3| = 3
27 30 27 - 30 = -3 |-3| = 3

Step 3: Calculate the Sum of Absolute Deviations

Sum of the absolute deviations ($\sum |x - \bar{x}|$) = $9 + 4 + 7 + 9 + 4 + 7 + 10 + 10 + 3 + 3 = 66$

Step 4: Calculate the Mean Deviation (MD) from the Mean

MD = $$ \frac{\sum |x - \bar{x}|}{n} = \frac{66}{10} = 6.6 $$

The mean deviation from the mean is 6.6.

Step 5: Calculate the Coefficient of Mean Deviation (CMD) from the Mean

Now, we can calculate the coefficient of mean deviation using the MD and the mean.

CMD = $$ \frac{\text{MD}}{\bar{x}} = \frac{6.6}{30} $$

CMD = 0.22

The coefficient of mean deviation of the given data from the mean is 0.22.

Final Answer

The calculated coefficient of mean deviation is 0.22.

Revision Table: Measures of Dispersion

Measure Type Definition/Formula Example (from Mean) Use Case
Range Absolute Maximum value - Minimum value Quickest, but uses only two values.
Quartile Deviation Absolute $Q_3 - Q_1 / 2$ Used for skewed data, less affected by extreme values.
Mean Deviation Absolute $\frac{\sum |x - \text{Average}|}{n}$ (Average can be Mean, Median, Mode) Considers all values, but ignores sign of deviations.
Standard Deviation Absolute $\sqrt{\frac{\sum (x - \bar{x})^2}{n-1}}$ (Sample) or $\sqrt{\frac{\sum (x - \bar{x})^2}{n}}$ (Population) Most common, used in many statistical tests, considers squared deviations.
Coefficient of Range Relative $\frac{\text{Max} - \text{Min}}{\text{Max} + \text{Min}}$ Comparing variability of different datasets.
Coefficient of Quartile Deviation Relative $\frac{Q_3 - Q_1}{Q_3 + Q_1}$ Comparing variability of different datasets (skewed).
Coefficient of Mean Deviation Relative $\frac{\text{MD}}{\text{Average}}$ (Average can be Mean, Median, Mode) Comparing variability of different datasets.
Coefficient of Variation Relative $\frac{\text{Standard Deviation}}{\text{Mean}} \times 100$ Comparing variability relative to the mean, expressed as percentage.

Additional Information on Measures of Dispersion

Measures of dispersion, also known as measures of variability, describe the spread or scatter of data points in a distribution. They tell us how much individual data points differ from the average and from each other.

  • Absolute Measures: These measures are expressed in the same units as the data. They are useful for comparing the dispersion within a single dataset or between datasets with similar averages and units. Examples include Range, Quartile Deviation, Mean Deviation, and Standard Deviation.
  • Relative Measures: These measures are unitless and are expressed as a ratio or percentage. They are useful for comparing the dispersion of two or more datasets that have different means or are measured in different units. Examples include Coefficient of Range, Coefficient of Quartile Deviation, Coefficient of Mean Deviation, and Coefficient of Variation.

The choice of which measure of dispersion to use depends on the nature of the data and the purpose of the analysis. For example, the mean deviation is simple to calculate but is not suitable for further mathematical analysis because it ignores the signs of deviations. Standard deviation is widely used because it is amenable to algebraic treatment and forms the basis for many advanced statistical concepts.

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  1. What is the mean deviation about the mean ?

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

  3. 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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