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Statistics - Limitations of mean deviation

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Mean Deviation is the arithmetic mean of the absolute values of the deviations of individual values from a central value (usually the mean or median).

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The most significant limitation of Mean Deviation is that it ignores the algebraic signs of the deviations. By taking absolute values ∣xi−a∣|x_i - a|, we treat negative deviations as positive, which makes the measure less suitable for further algebraic treatment and advanced statistical analysis.

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The sum of deviations of items from their arithmetic mean is always zero, i.e., ∑(xi−xˉ)=0\sum (x_i - \bar{x}) = 0. Mean deviation avoids this by using absolute values, but this lack of sign sensitivity is mathematically inconvenient in calculus and probability theory.

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Mean Deviation is not as stable as Standard Deviation when samples are drawn from the same population. It varies more from sample to sample.

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Mean Deviation is minimum when calculated about the Median. However, for many distributions, the Mean is a more common representative value, leading to a conflict in choosing the 'best' central tendency for this measure.

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It is not capable of further mathematical manipulation. For instance, you cannot calculate the combined mean deviation of two series if you know their individual mean deviations, sizes, and means (unlike Standard Deviation and Variance).

📐Formulae

M.D.(xˉ)=∑i=1n∣xi−xˉ∣nM.D.(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}

M.D.(M)=∑i=1n∣xi−M∣nM.D.(M) = \frac{\sum_{i=1}^{n} |x_i - M|}{n}

M.D.(xˉ)=∑i=1nfi∣xi−xˉ∣N, where N=∑fiM.D.(\bar{x}) = \frac{\sum_{i=1}^{n} f_i |x_i - \bar{x}|}{N}, \text{ where } N = \sum f_i

Coefficient of M.D.=M.D.Average used (Mean or Median)\text{Coefficient of } M.D. = \frac{M.D.}{\text{Average used (Mean or Median)}}

💡Examples

Problem 1:

Given the data set x:4,8,12x: 4, 8, 12, show why we use absolute deviations instead of simple deviations and calculate the Mean Deviation about the mean.

Solution:

  1. Find the Mean (xˉ\bar{x}): xˉ=4+8+123=243=8\bar{x} = \frac{4 + 8 + 12}{3} = \frac{24}{3} = 8

  2. Calculate simple deviations (xi−xˉ)(x_i - \bar{x}): 4−8=−48−8=012−8=4Sum=0\begin{array}{r} 4 - 8 = -4 \\ 8 - 8 = 0 \\ 12 - 8 = 4 \\ \hline \text{Sum} = 0 \end{array} This sum is always zero, making it useless as a measure of dispersion.

  3. Calculate absolute deviations ∣xi−xˉ∣|x_i - \bar{x}|: ∣4−8∣=4|4 - 8| = 4 ∣8−8∣=0|8 - 8| = 0 ∣12−8∣=4|12 - 8| = 4

  4. Mean Deviation: M.D.(xˉ)=4+0+43=83≈2.67M.D.(\bar{x}) = \frac{4 + 0 + 4}{3} = \frac{8}{3} \approx 2.67

Explanation:

This example illustrates the primary limitation: if we do not use the absolute value (modulus), the sum of deviations from the mean is always zero. However, by using ∣xi−xˉ∣|x_i - \bar{x}|, we lose the 'direction' of the data points, which prevents the use of this formula in more complex algebraic proofs.