Table of Contents

## Mean Deviation Definition

Mean deviation is defined as the mean of absolute deviations of the values from the central value.**For individual series:**\[MD\left( \overline{X} \right)=\frac{\sum{\left| \left( {{X}_{i}}-\overline{X} \right) \right|}}{N}\]**For discrete data with frequency, mean deviation is calculated as:**\[MD\left( \overline{X} \right)=\frac{\sum{f\left| \left( {{X}_{i}}-\overline{X} \right) \right|}}{\sum{f}}\]In the case of continuous series ‘X’ represents the mid-value of the class interval. Similarly, we can have mean deviation from median or mode. X’ is replaced by median or mode in the above formula. However, mean deviation from median is the least. It is known as minimal property of mean deviation.

The corresponding relative measures are coefficient of mean deviation.\[\text{Coefficient of M}D\left( \overline{X} \right)=\frac{\text{M}D\left( \overline{X} \right)}{\overline{X}}\]

In the following three articles in Statistics we will discuss various types of Absolute Measure of Dispersion in details:

1. Range (R)

2. Quartile Deviation (Q.D.)

3. Standard Deviations (S.D.)

## Merits

1. Based on all values.

2. Less affected by extreme values.

3. Not affected by sampling fluctuations

## Demerits

1. Not capable of further algebraic treatment.

2. Does not take into account negative signs.

## Uses

1. When sampling size is small.

2. In statistical analysis of certain economic, business and social phenomena.

Example 01 |

**Compute MD and Coefficient of MD from mean for the following data: 21, 32, 38, 41, 49, 54, 59, 66, 68**

**Solution:**

\[Mean\left( \overline{X} \right)=\frac{\sum\limits_{i=1}{{{X}_{i}}}}{N}=\frac{428}{9}=47.55\]

X | \[d=\left| {{X}_{i}}-\overline{X} \right|\] |

21 | 26.55 |

32 | 15.55 |

38 | 9.55 |

41 | 6.55 |

49 | 1.45 |

54 | 6.45 |

59 | 11.45 |

66 | 18.45 |

68 | 20.45 |

\[\sum{X=428}\] | \[\sum{\left| {{X}_{i}}-\overline{X} \right|}=\sum{d=116.45}\] |

\[MD=\frac{\sum{\left| {{X}_{i}}-\overline{X} \right|}}{N}=\frac{116.45}{9}=12.94\]\[\text{Coefficient of M}D\left( \overline{X} \right)=\frac{\text{M}D\left( \overline{X} \right)}{\overline{X}}\]\[\Rightarrow \text{Coefficient of M}D\left( \overline{X} \right)=\frac{12.94}{47.55}=0.272\]

Example 02 |

**Following are the wages of workers. Find mean deviation from median and its coefficient.**

X | 59 | 32 | 67 | 43 | 22 | 17 | 64 | 55 | 47 | 80 | 25 |

Wages | 17 | 22 | 25 | 32 | 43 | 47 | 55 | 59 | 64 | 67 | 80 |

**Solution:**

\[Median={{\left( \frac{11+1}{2} \right)}^{th}}item={{6}^{th}}item=47\]

X | Wages (f) | \[\left| {{X}_{i}}-Me \right|=\left| {{X}_{i}}-47 \right|\] | \[{{f}_{i}}.\left| {{X}_{i}}-Me \right|\] |

59 | 17 | 30 | 510 |

32 | 22 | 25 | 550 |

67 | 25 | 22 | 550 |

43 | 32 | 15 | 480 |

22 | 43 | 4 | 172 |

17 | 47 ← Me | 0 | 0 |

64 | 55 | 8 | 440 |

55 | 59 | 12 | 708 |

47 | 64 | 17 | 1088 |

80 | 67 | 20 | 1340 |

25 | 80 | 33 | 2640 |

\[\sum{{{f}_{i}}}=511\] | \[\sum{\left| {{X}_{i}}-Me \right|=186}\] | \[\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}=8478\] |

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}}{\sum{{{f}_{i}}}}=\frac{8478}{511}=16.591\]\[\text{Coefficient of M}D=\frac{MD}{Median}\]\[\Rightarrow \text{Coefficient of M}D=\frac{16.591}{47}=0.353\]

Example 03 |

**Compute mean deviation about its mode and its coefficient.**

X | 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |

f | 6 | 19 | 40 | 23 | 65 | 83 | 55 | 20 | 9 |

**Solution:**

The highest frequency is 83 and hence mode is 120.

X | f | \[d=\left| {{X}_{i}}-Mode \right|\] | fd |

20 | 6 | 100 | 600 |

40 | 19 | 80 | 1520 |

60 | 40 | 60 | 2400 |

80 | 23 | 40 | 920 |

100 | 65 | 20 | 1300 |

120 | 83 | 0 | 0 |

140 | 55 | 20 | 1100 |

160 | 20 | 40 | 800 |

180 | 9 | 60 | 540 |

\[N=\sum{{{f}_{i}}=320}\] | \[\sum{{{f}_{i}}.\left| {{X}_{i}}-Mode \right|}=9180\] |

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Mode \right|}}{N}=\frac{9180}{320}=28.69\]\[\text{Coefficient of M}D=\frac{MD}{Mode}=\frac{28.69}{120}=0.239\]

Example 04 |

**Find out the mean deviation from the data given below about its median.**

Salaries | 40 | 50 | 50-100 | 100-200 | 200-400 |

No. of Employees | 22 | 18 | 10 | 8 | 2 |

**Solution:**

Here N = 22+18+10+8+2 = 60\[Median={{\left( \frac{60+1}{2} \right)}^{th}}item={{30.5}^{th}}item\]It lies in 40 cumulative frequency (cf) and against 40 cf discrete value is 50.

X | No. of Employees | x( Mid value) | Cumulative frequency (cf) | \[d=\left| {{x}_{i}}-Me \right|\] | fd |

40 | 22 | 40 | 22 | 10 | 220 |

50 | 18 | 50 | 40 | 0 | 0 |

50-100 | 10 | 75 | 50 | 25 | 250 |

100-200 | 8 | 150 | 58 | 100 | 800 |

200-400 | 2 | 300 | 60 | 250 | 500 |

\[N=\sum{{{f}_{i}}}=60\] | \[\sum{{{f}_{i}}.\left| {{x}_{i}}-Me \right|}=1770\] |

\[MD=\frac{\sum{{{f}_{i}}.\left| {{X}_{i}}-Me \right|}}{\sum{{{f}_{i}}}}=\frac{1770}{60}=29.5\]\[\text{Coefficient of M}D=\frac{MD}{Median}=\frac{29.5}{50}=0.59\]

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