Mean deviation

Why absolute value is essential. For any data set, $\sum (x_i - \bar x) = 0$ always — this is a basic property of the mean, not a coincidence of the example. So a raw average of signed deviations from the mean is always zero, useless as a measure of spread. Taking $|x_i - a|$ before averaging removes the sign and keeps every deviation's contribution positive, so the result genuinely reflects how spread out the data is.

Ungrouped data. For $n$ observations $x_1, x_2, \ldots, x_n$ and a chosen reference point $a$ (mean or median):

where: $x_i$ are the individual observations, $a$ is the mean $\bar x$ or the median $M$, and $n$ is the number of observations.

Discrete and continuous frequency distributions. When each value $x_i$ occurs with frequency $f_i$ (or, for grouped data, $x_i$ is the mid-point of a class interval with frequency $f_i$), each deviation must be weighted by how often it occurs:

where: $f_i$ is the frequency of $x_i$ (or of the class with mid-point $x_i$), and $N$ is the total frequency, $\sum f_i$. The formula is identical for discrete and grouped continuous data — the only extra step for grouped data is computing each class mid-point first. $a = \bar x$ gives mean deviation about the mean; $a = M$ gives mean deviation about the median. NCERT theory notes that, among all possible choices of $a$, the median makes $\frac{1}{n}\sum|x_i-a|$ smallest — which is why the median is sometimes the preferred reference, although the mean is used just as often in problems.

Find the mean deviation about the median for this discrete frequency distribution.

Step 1. Find the total frequency. $N = 2+3+4+3+3 = 15$.

Step 2. Locate the median. Since $x_i$ is already sorted, the median is the value at position $\frac{N+1}{2} = \frac{16}{2} = 8$ in the ordered list of all 15 observations. Cumulative frequencies are: up to $x=2$ it is 2, up to $x=5$ it is 5, up to $x=6$ it is 9, up to $x=8$ it is 12, up to $x=10$ it is 15. Position 8 falls in the range covered by $x=6$ (positions 6 to 9), so $M = 6$.

Step 3. Compute $|x_i - 6|$ and multiply by $f_i$ for each row.

Step 4. Sum the weighted deviations and divide by $N$. $\sum f_i|x_i-M| = 8+3+0+6+12 = 29$, so $\text{MD}(M) = \dfrac{29}{15} \approx 1.93$. The mean deviation about the median is approximately 1.93.

Competitive stretch

If the same distribution's mean deviation were instead required about the mean, the method would not change — only the reference point would. First compute $\bar x = \dfrac{\sum f_i x_i}{N} = \dfrac{2(2)+5(3)+6(4)+8(3)+10(3)}{15} = \dfrac{4+15+24+24+30}{15} = \dfrac{97}{15} \approx 6.47$, then rebuild the $f_i|x_i-\bar x|$ column with this new $a$ and divide by $N$ again. This is exactly why exam questions specify "about the mean" or "about the median" — the two answers are computed by the identical formula but are generally different numbers.

• Skipping the absolute value and using signed deviations — this always sums to (approximately) zero around the mean and is a red flag that the step was missed.
• For grouped data, forgetting to use class mid-points as $x_i$, or forgetting to multiply each $|x_i - a|$ by its frequency $f_i$ before summing.
• Confusing the median's position $\frac{N+1}{2}$ with $\frac{N}{2}$, or misreading the cumulative-frequency column when locating the median in a table.
• Dividing by the wrong total — always $n$ for raw ungrouped data or $N = \sum f_i$ for frequency data, never the number of distinct $x_i$ values.

Ask the tutor: if you're stuck locating the median in a frequency table or setting up mean deviation for grouped data, ask the AI tutor to walk through your specific data set step by step.