1. Introduction to Mean Deviation

While the range gives a basic idea of data spread, it's limited because it only uses the two most extreme values. A more robust measure is the Mean Deviation (M.D.), which considers every data point's distance from a central value (usually the mean or median).

The deviation of an observation xix_i from a central value 'a' is xi−ax_i - a. If we simply try to average these deviations from the mean, the positive and negative deviations cancel each other out, and their sum is always zero.

Example: For the data {2, 4, 6}, the mean is xˉ=4\bar{x}=4. The deviations are (2−4)=−2(2-4)=-2, (4−4)=0(4-4)=0, and (6−4)=2(6-4)=2. The sum of these deviations is −2+0+2=0-2+0+2=0.

To overcome this, we use the absolute value of each deviation, which measures the distance of each point from the center, regardless of direction. The Mean Deviation is the average of these absolute distances. 📜

2. Mean Deviation for Ungrouped Data

Ungrouped data is a simple list of individual observations.

A. Mean Deviation about the Mean

This measures the average distance of each data point from the mean of the dataset.

Steps:

  1. Calculate the Mean (xˉ\bar{x}): Sum all observations and divide by the number of observations, nn. xˉ=∑i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}
  2. Calculate Absolute Deviations: For each observation xix_i, find its distance from the mean: ∣xi−xˉ∣|x_i - \bar{x}|.
  3. Calculate Mean Deviation: Find the average of these absolute deviations. M.D.(xˉ)=∑i=1n∣xi−xˉ∣n\text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}

Example: Find the M.D. about the mean for the data: {6, 7, 10, 12, 13, 4, 8, 12}

  1. Mean: xˉ=6+7+10+12+13+4+8+128=728=9\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9.
  2. Absolute Deviations: The distances from the mean (9) are: ∣6−9∣=3|6-9|=3, ∣7−9∣=2|7-9|=2, ∣10−9∣=1|10-9|=1, ∣12−9∣=3|12-9|=3, ∣13−9∣=4|13-9|=4, ∣4−9∣=5|4-9|=5, ∣8−9∣=1|8-9|=1, ∣12−9∣=3|12-9|=3.
  3. Mean Deviation: M.D.(xˉ\bar{x}) = 3+2+1+3+4+5+1+38=228=2.75\frac{3+2+1+3+4+5+1+3}{8} = \frac{22}{8} = \mathbf{2.75}.

B. Mean Deviation about the Median

This measures the average distance of each data point from the median of the dataset.

Steps:

  1. Calculate the Median (M): Arrange the data in ascending order and find the middle value.
  2. Calculate Absolute Deviations: For each observation xix_i, find its distance from the median: ∣xi−M∣|x_i - M|.
  3. Calculate Mean Deviation: Find the average of these absolute deviations. M.D.(M)=∑i=1n∣xi−M∣n\text{M.D.}(M) = \frac{\sum_{i=1}^{n} |x_i - M|}{n}

Example: Find the M.D. about the median for the data: {3, 9, 5, 3, 12, 10, 18, 4, 7}

  1. Median: First, sort the data: {3, 3, 4, 5, 7, 9, 10, 12, 18}. There are n=9n=9 observations. The median is the middle (5th5^{th}) value, so M=7M=7.
  2. Absolute Deviations: The distances from the median (7) are: ∣3−7∣=4,∣3−7∣=4,∣4−7∣=3,∣5−7∣=2,∣7−7∣=0,∣9−7∣=2,∣10−7∣=3,∣12−7∣=5,∣18−7∣=11|3-7|=4, |3-7|=4, |4-7|=3, |5-7|=2, |7-7|=0, |9-7|=2, |10-7|=3, |12-7|=5, |18-7|=11.
  3. Mean Deviation: M.D.(MM) = 4+4+3+2+0+2+3+5+119=349≈3.78\frac{4+4+3+2+0+2+3+5+11}{9} = \frac{34}{9} \approx \mathbf{3.78}.

3. Mean Deviation for Grouped Data

Grouped data is presented in a frequency table.

A. Discrete Frequency Distribution

Here, each observation xix_i has a corresponding frequency fif_i.

  • Mean Deviation about the Mean: M.D.(xˉ)=∑i=1nfi∣xi−xˉ∣N,where N=∑fi and xˉ=∑fixiN\text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{n} f_i |x_i - \bar{x}|}{N}, \quad \text{where } N=\sum f_i \text{ and } \bar{x} = \frac{\sum f_i x_i}{N}

  • Mean Deviation about the Median: M.D.(M)=∑i=1nfi∣xi−M∣N,where M is the median\text{M.D.}(M) = \frac{\sum_{i=1}^{n} f_i |x_i - M|}{N}, \quad \text{where M is the median}

Example: Find the M.D. about the mean for the following distribution:

xix_i 5 10 15 20
fif_i 2 3 4 1
  1. Calculate Mean: N=2+3+4+1=10N = 2+3+4+1=10. The sum ∑fixi=(2⋅5)+(3⋅10)+(4⋅15)+(1⋅20)=10+30+60+20=120\sum f_i x_i = (2 \cdot 5) + (3 \cdot 10) + (4 \cdot 15) + (1 \cdot 20) = 10+30+60+20=120. So, xˉ=120/10=12\bar{x} = 120/10=12.
  2. Calculate Deviations:
xix_i fif_i ∣xi−12∣|x_i - 12| fi∣xi−12∣f_i |x_i - 12|
5 2 7 14
10 3 2 6
15 4 3 12
20 1 8 8
  1. Calculate M.D.: The sum ∑fi∣xi−xˉ∣=14+6+12+8=40\sum f_i|x_i - \bar{x}| = 14+6+12+8=40. M.D.(xˉ\bar{x}) = 4010=4\frac{40}{10} = \mathbf{4}.

B. Continuous Frequency Distribution

For data in class intervals, we use the mid-point (xix_i) of each class as the representative value.

  • M.D. about the Mean: The calculation is the same as for discrete data, but xix_i is the mid-point.

  • M.D. about the Median: The median must first be calculated using the formula: M=l+N2−Cf×hM = l + \frac{\frac{N}{2} - C}{f} \times h where ll is the lower limit of the median class, NN is the total frequency, CC is the cumulative frequency of the class before the median class, ff is the frequency of the median class, and hh is the class width.

Example: Find the M.D. about the median for the data:

Class 0-10 10-20 20-30 30-40
Frequency 5 8 15 12
  1. Find the Median: First, create a table with mid-points (xix_i) and cumulative frequency (cfcf).
Class Freq (fif_i) Mid-point (xix_i) cf
0-10 5 5 5
10-20 8 15 13
20-30 15 25 28
30-40 12 35 40 
N=40N=40, so N/2=20N/2=20. The median class is the class where the cf first exceeds 20, which is **20-30**. 
l=20,C=13,f=15,h=10l=20, C=13, f=15, h=10.
M=20+20−1315×10=20+7015≈24.67M = 20 + \frac{20-13}{15} \times 10 = 20 + \frac{70}{15} \approx 24.67.
  1. Calculate Deviations and M.D.:
xix_i fif_i ∣xi−24.67∣|x_i - 24.67| fi∣xi−24.67∣f_i |x_i - 24.67|
5 5 19.67 98.35
15 8 9.67 77.36
25 15 0.33 4.95
35 12 10.33 123.96 
The sum ∑fi∣xi−M∣=98.35+77.36+4.95+123.96=304.62\sum f_i|x_i - M| = 98.35+77.36+4.95+123.96 = 304.62.
M.D.(MM) = 304.6240≈7.62\frac{304.62}{40} \approx \mathbf{7.62}.