Statistics - Mean Deviation of Continuous Data Series


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When data is given based on ranges alongwith their frequencies. Following is an example of continous series:

Items0-55-1010-2020-3030-40
Frequency251312

In case of continous series, a mid point is computed as $\frac{lower-limit + upper-limit}{2}$ and Mean Deviation is computed using following formula.

Formula

${MD} =\frac{\sum{f|x-Me|}}{N} = \frac{\sum{f|D|}}{N}$

Where −

  • ${N}$ = Number of observations.

  • ${f}$ = Different values of frequency f.

  • ${x}$ = Different values of mid points for ranges.

  • ${Me}$ = Median.

The Coefficient of Mean Deviation can be calculated using the following formula.

${Coefficient\ of\ MD} =\frac{MD}{Me}$

Example

Problem Statement:

Let's calculate Mean Deviation and Coefficient of Mean Deviation for the following continous data:

Items0-1010-2020-3030-40
Frequency2513

Solution:

Based on the given data, we have:

ItemsMid-pt
${x_i}$
Frequency
${f_i}$
${f_ix_i}$ ${|x_i-Me|}$${f_i|x_i-Me|}$
0-10521014.5429.08
10-20155754.5422.7
20-30251256.545.46
30-4035310514.5446.38
  ${N=11}$${\sum f=215}$ ${\sum {f_i|x_i-Me|} = 103.62}$

Median

${Me} = \frac{215}{11} \\[7pt] \, = {19.54}$

Based on the above mentioned formula, Mean Deviation ${MD}$ will be:

${MD} = \frac{\sum{f|D|}}{N} \\[7pt] \, = \frac{103.62}{11} \\[7pt] \, = {9.42}$

and, Coefficient of Mean Deviation ${MD}$ will be:

${=\frac{MD}{Me}} \, = \frac{9.42}{19.54} \\[7pt] \, = {0.48}$

The Mean Deviation of the given numbers is 9.42.

The coefficient of mean deviation of the given numbers is 0.48.

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