Calculate mean deviation from median for: 3, 5, 7, 9, 11, 13

Question

Calculate the mean deviation from the median for the data set: 3, 5, 7, 9, 11, 13.

Solution — Step by Step

The data is already in ascending order: 3, 5, 7, 9, 11, 13.

Number of observations: $n = 6$ (even number).

For an even number of observations, the median is the average of the $\frac{n}{2}$ th and $\left(\frac{n}{2}+1\right)$ th values.

$\frac{n}{2} = 3$ rd value = 7

$\frac{n}{2} + 1 = 4$ th value = 9

\text{Median} = \frac{7 + 9}{2} = 8

For each observation $x_i$ , find $|x_i - \text{Median}| = |x_i - 8|$ :

| $x_i$ | $x_i - 8$ | $|x_i - 8|$ | |---|---|---| | 3 | −5 | 5 | | 5 | −3 | 3 | | 7 | −1 | 1 | | 9 | +1 | 1 | | 11 | +3 | 3 | | 13 | +5 | 5 |

\sum |x_i - \text{Median}| = 5 + 3 + 1 + 1 + 3 + 5 = 18

\text{Mean Deviation (MD)} = \frac{\sum |x_i - M|}{n} = \frac{18}{6} = \mathbf{3}

The mean deviation from the median is 3. This means, on average, each data point is 3 units away from the median value of 8.

A mean deviation of 3 relative to a median of 8 represents a coefficient of mean deviation of $\frac{3}{8} = 0.375$ — this can be used to compare the spread of different data sets with different medians.

Why This Works

Mean deviation measures the average spread of data around a central value. We use absolute values $|x_i - M|$ because if we didn’t, the positive and negative deviations would cancel out (the sum of deviations from the mean = 0 always; sum from the median = 0 or close to 0).

For this particular data set (an arithmetic sequence: 3, 5, 7, 9, 11, 13 with common difference 2), the deviations are symmetric: 5, 3, 1, 1, 3, 5. The symmetry arises because the data is symmetric about its median (8). The mean deviation equals the average of these symmetric values.

Mean deviation from the median is always ≤ mean deviation from the mean in general. The median minimises the sum of absolute deviations — a useful property for robust statistics.

Alternative Method — Formula Shortcut for Arithmetic Sequences

For a symmetric data set like this one, the mean deviation can be computed by averaging just the positive deviations:

Positive deviations: 1, 3, 5 (the values 7, 5, 3 are below the median by 1, 3, 5 respectively)

Their average = $\frac{1 + 3 + 5}{3} = 3$

Since the deviations are symmetric, the overall mean deviation = average of all |deviations| = 3. ✓

Common Mistake

The most common error is using the signed deviations instead of absolute values. If you sum $(-5) + (-3) + (-1) + 1 + 3 + 5 = 0$ and divide by 6, you get 0 — which is a trivially useless result. Mean deviation always uses absolute values $|x_i - M|$ .

Another error is finding the median incorrectly for an even number of data points. With 6 values, the median is the average of the 3rd and 4th values, NOT the 3rd value alone. Here, median = (7+9)/2 = 8, not 7.

For CBSE Class 11, the mean deviation formula is the same for both mean and median as the central measure — only the value you subtract changes. Always check: is the question asking for MD from the mean or MD from the median? They give different answers for the same data set.

Question

Solution — Step by Step

Why This Works

Alternative Method — Formula Shortcut for Arithmetic Sequences

Common Mistake

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