**Mean Deviation from the Mean for UnGrouped Data**

Suppose we have a discrete data with n observations x₁, x₂, … x_{n}. Then we adopt the following procedure for computing the mean deviation from the mean of the given data.

**Step 1:** Calculate the arithmetic mean (x̄) of the n observations. Let it be “a”.

**Step 2:** Find the deviations of each x₁ from ‘a’ i.e., x₁ – a, x₂ – a, … x_{n} – a.

**Step 3:** Find the absolute value i.e., [x₁ – a], [x₂ – a], … [x_{n} – a] of these deviations by ignoring the negative sign, if any. In the deviations computed in step 2.

**Step 4:** Find the arithmetic mean of the absolute values of the deviations.

i.e., M.D from the mean = \(\frac{\sum\limits_{i\,=\,1}^{n}{\left| {{x}_{i}}\,-\,a \right|}}{n}\).

**Example:** Find the mean deviation from the mean of the following discrete data: 6, 7, 10, 12, 13, 14, 12, 16.

**Solution:** The arithmetic mean of the given data is

x̄ = (6 + 7 + 10 + 12 + 13 + 4 + 12 + 16)/ 8 = 0

The absolute values of the deviations |x₁ – x̄| are 4, 3, 0, 2, 3, 6, 2, 6.

∴ The mean deviation from the mean = \(\frac{\sum\limits_{i\,=\,1}^{8}{\left| {{x}_{i}}\,-\,\overline{x} \right|}}{8}\),

= (4 + 3 + 0 + 2 + 3 + 6 + 2 + 6)/ 8

= 26/8

= 3.25.