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Question

Consider the following grouped frequency distribution :

Class0-1010-2020-3030-4040-5050-60
Frequency124643

What is the mean deviation about the mean ?

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

10.65

Calculate Mean Deviation About the Mean for Grouped Data

The question asks for the mean deviation about the mean for a given grouped frequency distribution. To find this, we need to follow several steps:

  1. Find the midpoint (\(x_i\)) for each class interval.
  2. Calculate the product of the frequency (\(f_i\)) and the midpoint (\(x_i\)) for each class (\(f_i x_i\)).
  3. Calculate the mean (\(\bar{x}\)) of the distribution using the formula \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\).
  4. Calculate the absolute deviation of each midpoint from the mean, i.e., \(|x_i - \bar{x}|\).
  5. Calculate the product of the frequency (\(f_i\)) and the absolute deviation \(|x_i - \bar{x}|\) for each class, i.e., \(f_i |x_i - \bar{x}|\).
  6. Calculate the sum of \(f_i |x_i - \bar{x}|\) over all classes.
  7. Calculate the mean deviation about the mean using the formula \(\text{Mean Deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i}\).

Step-by-Step Calculation

First, let's create a table to organize the calculations:

ClassFrequency (\(f_i\))Midpoint (\(x_i\))\(f_i x_i\)\(|x_i - \bar{x}|\)\(f_i |x_i - \bar{x}|\)
0-101\(\frac{0+10}{2} = 5\)\(1 \times 5 = 5\)$|5 - 34.5| = 29.5$\(1 \times 29.5 = 29.5\)
10-202\(\frac{10+20}{2} = 15\)\(2 \times 15 = 30\)$|15 - 34.5| = 19.5$\(2 \times 19.5 = 39.0\)
20-304\(\frac{20+30}{2} = 25\)\(4 \times 25 = 100\)$|25 - 34.5| = 9.5$\(4 \times 9.5 = 38.0\)
30-406\(\frac{30+40}{2} = 35\)\(6 \times 35 = 210\)$|35 - 34.5| = 0.5$\(6 \times 0.5 = 3.0\)
40-504\(\frac{40+50}{2} = 45\)\(4 \times 45 = 180\)$|45 - 34.5| = 10.5$\(4 \times 10.5 = 42.0\)
50-603\(\frac{50+60}{2} = 55\)\(3 \times 55 = 165\)$|55 - 34.5| = 20.5$\(3 \times 20.5 = 61.5\)
Total\(\sum f_i = 20\) \(\sum f_i x_i = 690\)  \(\sum f_i |x_i - \bar{x}| = 213.0\)


 

1. Calculate the Mean (\(\bar{x}\))

Using the sums from the table:

\(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{690}{20} = 34.5\)

The mean of the distribution is 34.5.

2. Calculate the Mean Deviation About the Mean

Using the sum of $f_i |x_i - \bar{x}|$ from the table and the total frequency:

\(\text{Mean Deviation} = \frac{\sum f_i |x_i - \bar{x}|}{\sum f_i} = \frac{213.0}{20} = 10.65\)

The mean deviation about the mean is 10.65.

Conclusion

Based on the calculations, the mean deviation about the mean for the given grouped frequency distribution is 10.65.

Revision Table: Mean Deviation Calculation

Let's quickly recap the key values calculated:

  • Total Frequency ($\sum f_i$): 20
  • Sum of $f_i x_i$ ($\sum f_i x_i$): 690
  • Mean ($\bar{x}$): 34.5
  • Sum of $f_i |x_i - \bar{x}|$ ($\sum f_i |x_i - \bar{x}| $): 213.0
  • Mean Deviation about the Mean: $\frac{213.0}{20} = 10.65$

Additional Information: Measures of Dispersion

Mean deviation is a measure of dispersion that indicates how much the observations in a dataset deviate from a central value (like the mean, median, or mode). For grouped data, we use the midpoints of the classes as representative values for the observations within that class.

Other common measures of dispersion include:

  • Range: The difference between the highest and lowest values in the data. Simple but sensitive to outliers.
  • Variance: The average of the squared differences from the Mean. Provides a measure of how spread out the data is.
  • Standard Deviation: The square root of the Variance. It is widely used because it is in the same units as the original data.
  • Quartile Deviation (Semi-Interquartile Range): Half the difference between the third and first quartiles. Based on the middle 50% of the data, less affected by extreme values.

Mean deviation is useful as it considers all observations, unlike the range or quartile deviation. However, the use of absolute values in its calculation makes it less suitable for further mathematical treatments compared to variance or standard deviation.

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Similar Questions

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

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

  4. 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

  5. If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the value of d is

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