If the mean deviation about median for the number 3, 5, 7, 2k, 12,...

Median is the central value (or middle observation) of statistical data if it is arranged in ascending or descending order.

Thus, if there are $n$ observations (variates) ${x}_{1}, {x}_{2}, {x}_{3}, \ldots , {x}_{n}$ arranged in ascending or descending order, then

median $= {\left(\dfrac{n + 1}{2}\right)}$^{th}$observation, if$n$is odd</p><p>and median$= \dfrac{\dfrac{n}{2}\text{th observation} + \left(\dfrac{n}{2} + 1\right)\text{th observation}}{2}$, if$n$is even.</p><p>Here we can see there are even number of terms,</p><p>So, median$= \dfrac{2k + 12}{2} = k + 6$</p><p>Mean deviation$= ∑ \dfrac{\left|{x}_{i} - M\right|}{n} = 6$</p><p>$⇒ \dfrac{\left(k + 3\right) + \left(k + 1\right) + \left(k - 1\right) + \left(6 - k\right) + \left(6 - k\right) + \left(10 - k\right) + \left(15 - k\right) + \left(18 - k\right)}{8} = 6$</p><p>$⇒ \dfrac{58 - 2k}{8} = 6$</p><p>$⇒ k = 5$</p><p>So, median$= \dfrac{2 \times 5 + 12}{2} = 11