# Statistics - Mean Deviation of Continuous Data Series

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

 Items Frequency 0-5 5-10 10-20 20-30 30-40 2 5 1 3 12

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:

 Items Frequency 0-10 10-20 20-30 30-40 2 5 1 3

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.