Mean Deviation for Grouped Data

For grouped data — where each distinct value (or each class interval, represented by its midpoint) carries an associated frequency $f_i$ — the mean deviation formulas of the previous section are adapted by weighting each absolute deviation by how many times it actually occurs.

Discrete frequency distribution. If the distinct values $x_1,x_2,\ldots,x_k$ occur with frequencies $f_1,f_2,\ldots,f_k$ respectively, and $N=\sum f_i$ is the total number of observations, the mean is $\bar x = \dfrac{1}{N}\sum f_i x_i$, and the mean deviation about the mean is

The mean deviation about the median, $\text{M.D.}(M)=\dfrac1N\sum f_i|x_i-M|$, is defined the same way, with $M$ found from the cumulative frequencies of the discrete distribution.

Continuous frequency distribution. When the data is organised into class intervals, each class is first represented by its midpoint (also called the class mark), $x_i = \dfrac{\text{lower limit}+\text{upper limit}}{2}$, and the same discrete-style formula is then applied treating the midpoints as the $x_i$ values:

This midpoint substitution is an approximation — it assumes every observation within a class is located at that class's midpoint — but it is the standard, universally used convention, and becomes more accurate as the class width narrows.

Steps to compute (both cases).

1. Find $N=\sum f_i$ and the mean $\bar x = \dfrac1N\sum f_i x_i$ (using class midpoints for continuous data).
2. Find the absolute deviation $|x_i-\bar x|$ of every distinct value/midpoint from $\bar x$.
3. Multiply each absolute deviation by its frequency: $f_i\,|x_i-\bar x|$.
4. Sum this column and divide by $N$. …