If the mean of $4,7,2,8,6$ and a is 7 , then the mean deviation from the median of these observations is

Given observations are 4, 7, 2, 8, 6, $a$ and mean is 7. We know $$ \begin{aligned} & \text { Mean }=\frac{4+7+2+8+6+a}{6} \\ & \Rightarrow 7=\frac{4+7+2+8+6+a}{6} \Rightarrow a=15 \end{aligned} $$ Now, given observations can be written in ascending order which is $2,4,6,7,8,15$ Since, No. of observation is even $\therefore$ Median $=\frac{\left(\frac{6}{2}\right) \text { th observation }+\left(\frac{6}{2}+1\right) \text { th observation }}{2}$ $=\frac{\text { 3rd observation }+4 \text { th observation }}{2}$ $=\frac{6+7}{2}=\frac{13}{2}$ $$ \begin{aligned} & \text { Now, Mean deviation }=\frac{\sum_{i=1}^6\left|x_i-\frac{13}{2}\right|}{6} \\ & =\frac{\left|4-\frac{13}{2}\right|+\left|7-\frac{13}{2}\right|+\left|2-\frac{13}{2}\right|+\left|8-\frac{13}{2}\right|+\left|6-\frac{13}{2}\right|+\left|15-\frac{13}{2}\right|}{6} \\ & =\frac{\frac{5}{2}+\frac{1}{2}+\frac{9}{2}+\frac{3}{2}+\frac{1}{2}+\frac{17}{2}}{6}=\frac{18}{6}=3 \\ & \end{aligned} $$