Mean deviation about mean

Two class sections score the same average of 60 out of 100 on a test. In one section every student is bunched between 55 and 65; in the other, marks swing wildly from 20 to 100. The average alone cannot tell these two classes apart — it hides exactly the information a teacher, or a statistician, actually wants: how spread out the data is. Mean deviation about the mean fixes this by measuring the average distance of every observation from the mean, using absolute values so that deviations above and below the mean do not cancel out. It works the same way for a plain list of numbers (ungrouped data) and for data organised into a frequency table (grouped data), and it is the natural stepping stone toward the more famous measure of spread, standard deviation, which you meet next.

The core insight

Mean deviation about the mean averages the absolute distance of every observation from $\bar x$ , giving a single number, in the same units as the data, that tells you how tightly or loosely the data is clustered around its average.

The core idea

A crude measure of spread like range (largest value minus smallest) only looks at two extreme observations and ignores everyone else. Mean deviation about the mean is better because it uses every observation: it finds each value's distance from the mean, then averages those distances. There is one subtlety. If you just average the signed deviations $(x_i - \bar x)$ , you always get zero — this is a basic algebraic property of the mean (the positive and negative deviations exactly cancel). So before averaging, every deviation is converted to a positive number by taking its absolute value $|x_i - \bar x|$ , which discards the sign but keeps the size. For ungrouped data $x_1, x_2, \dots, x_n$ with mean $\bar x$ , the mean deviation about the mean is:

\text{MD} = \frac{1}{n}\sum_{i=1}^{n} |x_i - \bar x|

where: $x_i$ is the $i$ -th observation, $\bar x$ is the mean of all $n$ observations, and $|x_i - \bar x|$ is the absolute (unsigned) deviation of that observation from the mean. For grouped or discrete-frequency data, where the value $x_i$ occurs with frequency $f_i$ (for class-interval data, $x_i$ is taken as the mid-point of the class), the formula becomes a frequency-weighted average of the absolute deviations:

\text{MD} = \frac{\sum f_i |x_i - \bar x|}{\sum f_i}, \qquad \text{where } \bar x = \frac{\sum f_i x_i}{\sum f_i}

where: $f_i$ is the frequency of value (or class) $i$ , $\sum f_i$ is the total number of observations (which for class-interval data is the total frequency, not the number of rows in the table), and $\bar x$ is the frequency-weighted mean. A small mean deviation says the data huddles close to the mean; a large one says the data is scattered widely. Because absolute deviations are added rather than squared, mean deviation carries the same units as the original data — marks, hours, rupees — which makes it directly interpretable, unlike variance (whose units are squared) or standard deviation, which you will meet as a refinement of this same idea.

Figure

Number line showing seven marks scattered around a mean with the absolute deviation of each point marked as a bracketed distance back to the mean

Derivation AWhy absolute value, not the signed deviation

Show that using signed deviations $(x_i - \bar x)$ to measure spread always collapses to zero regardless of how scattered the data actually is, so replacing them with absolute deviations $|x_i - \bar x|$ is not a stylistic choice but the only way to get a non-trivial, always-positive measure of dispersion out of the deviations from the mean.

Step 1. Start from the definition of the mean: $\bar x = \dfrac{1}{n}\sum_{i=1}^n x_i$ , which rearranges to $\sum_{i=1}^n x_i = n\bar x$ .

Step 2. Consider the sum of the signed deviations, $\sum_{i=1}^n (x_i - \bar x)$ . Splitting the sum term by term gives $\sum_{i=1}^n x_i - \sum_{i=1}^n \bar x = \sum_{i=1}^n x_i - n\bar x$ (since $\bar x$ is a constant added $n$ times).

Step 3. Substitute $\sum_{i=1}^n x_i = n\bar x$ from Step 1: $\sum_{i=1}^n (x_i - \bar x) = n\bar x - n\bar x = 0$ . This holds for every data set, however scattered or however tightly clustered — it is an identity, not a coincidence, so the average signed deviation $\frac{1}{n}\sum(x_i - \bar x)$ is always exactly zero and can never be used to compare the spread of two data sets.

Step 4. Since the sign cancellation is unavoidable, the only way to keep the "distance from the mean" information while destroying the cancellation is to strip the sign before averaging. Taking $|x_i - \bar x|$ makes every term non-negative, so $\sum |x_i - \bar x| = 0$ only when every observation equals the mean exactly; any real scatter makes the sum strictly positive, and dividing by $n$ (or by $\sum f_i$ for frequency data) turns that sum into the mean deviation, a genuine, non-trivial measure of dispersion.

Worked example

The marks (out of 20) obtained by 7 students in a class test are $12,\ 15,\ 9,\ 18,\ 11,\ 14,\ 13$ . Find the mean deviation about the mean.

Step 1. Find the mean: $\bar x = \dfrac{12+15+9+18+11+14+13}{7} = \dfrac{92}{7}$ . We keep this as a fraction rather than rounding early, to avoid arithmetic drift in later steps.

Step 2. Find $|x_i - \bar x|$ for each value: for $12$ it is $\left|12 - \frac{92}{7}\right| = \frac{8}{7}$ ; for $15$ it is $\frac{13}{7}$ ; for $9$ it is $\frac{29}{7}$ ; for $18$ it is $\frac{34}{7}$ ; for $11$ it is $\frac{15}{7}$ ; for $14$ it is $\frac{6}{7}$ ; for $13$ it is $\frac{1}{7}$ .

Step 3. Sum the absolute deviations: $\sum |x_i - \bar x| = \dfrac{8+13+29+34+15+6+1}{7} = \dfrac{106}{7}$ .

Step 4. Divide by $n = 7$ : $\text{MD} = \dfrac{106/7}{7} = \dfrac{106}{49} \approx 2.16$ marks.

A typical student's score deviates from the class mean by about $2.16$ marks.

Competitive stretch

For grouped data, the same logic just adds a frequency weight. Forty students' weekly study hours are grouped into classes $0$ to $10$ (frequency $4$ ), $10$ to $20$ (frequency $8$ ), $20$ to $30$ (frequency $14$ ), $30$ to $40$ (frequency $10$ ), and $40$ to $50$ (frequency $4$ ), using mid-points $5, 15, 25, 35, 45$ . The mean works out to $\bar x = \frac{1020}{40} = 25.5$ hours, and summing $f_i|x_i - 25.5|$ across all five classes gives a total of $346.0$ , so $\text{MD} = \dfrac{346}{40} = 8.65$ hours — a much larger spread than the marks example, because study habits vary far more than test scores here.

Common traps

Forgetting to recompute the mean for the given data set. Mean deviation is always taken about the mean of this data set — never reuse a mean from an earlier, unrelated problem.
Skipping the absolute value. Adding signed deviations and dividing by $n$ always gives (approximately) zero, as the derivation above shows — a zero or near-zero "mean deviation" is a red flag that absolute values were forgotten.
Using class boundaries instead of the mid-point for grouped data. For class-interval data, $x_i$ must be $\frac{\text{lower limit}+\text{upper limit}}{2}$ , not the interval's endpoints.
Dividing by the number of classes instead of $\sum f_i$ . For grouped data the denominator is the total number of observations (total frequency), not the number of rows in the frequency table.

Ask the tutor: if you are unsure how to set up the deviation table for a grouped-data problem, or want to check your absolute-value column, ask the AI tutor to walk through mean deviation about the mean with your own numbers.

Why do we use absolute deviations instead of just squaring the deviations like standard deviation does?

Both tricks solve the same cancellation problem, but they are not identical in spirit. Absolute value is the simplest possible fix — it keeps mean deviation easy to compute by hand and keeps the answer in the original units of the data. Squaring, used for variance and standard deviation, has better mathematical properties for further statistical theory (it is differentiable everywhere, which matters for later calculus-based work), but it changes the units to "squared marks" or "squared hours," which is harder to interpret directly. Mean deviation is usually taught first because it is the more intuitive, direct measure.

Do I always need to find the mean first before I can find the mean deviation?

Yes — the name says it plainly: it is the mean deviation about the mean, so $\bar x$ must be computed first, from the same data set, before you can find a single deviation $|x_i - \bar x|$ . Skipping straight to deviations without computing $\bar x$ is not possible; every step in the worked example depends on knowing $\bar x = 92/7$ before the deviation table can even be started.

What is the difference between mean deviation about the mean and mean deviation about the median?

They use the same idea — average absolute distance from a central value — but centre the deviations on a different statistic. Mean deviation about the mean uses $\bar x$ ; mean deviation about the median uses the median $M$ instead, so the formula becomes $\text{MD} = \frac{1}{n}\sum|x_i - M|$ (or the frequency-weighted version for grouped data). Both are valid measures of spread; ISC Class 11 typically introduces the mean version first since the mean is already familiar from earlier classes.

Mean deviation about mean

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads