If the mean of the data p ,   6 , 6 , 7 , 8 , 11 , 15 , 16 , is 3 times p , then the mean deviation of…

We know that mean deviation for ungrouped data:

For '$n$' observation $x_1,x_2\dots \dots \dots \dots ..x_n$, the mean deviation about their mean $\overline{x}$ is given by $M.D=\frac{{\sum }_{i=1}^n\left|x_i-\overline{x}\right|}{N}$. where $N$ is the number of observations

Now given data of numbers are $p, 6,6,7,8,11,15$, and $16 .$

So, mean $\overline{x}=\frac{p+6+6+7+8+11+15+16}{8}=3p$

$\Rightarrow 23p=69\Rightarrow p=3$

So given data is $3,6,6,7,8,11,15,16$ and mean is $3p=3\times 3=9$.
$\therefore$ Mean deviation $=\frac{\left|3-9\right|+\left|6-9\right|+\left|6-9\right|+\left|7-9\right|+\left|8-9\right|+\left|11-9\right|+\left|15-9\right|+\left|16-9\right|}{8}$
Mean deviation $=\frac{\left|x_i-\overline{x}\right|}{n}=\frac{30}{8}=3.75$