Question

Find the mean deviation about the median for the data given below.

45, 36, 50, 60, 53, 46, 51, 48, 72, 42

Answer 1

Arranging the given data in an ascending order, we get:

$36,42,45,46,48,50,51,53,60,72$

Here n = 10, which is even.

$\therefore$ $Median$ $=\frac{1}{2}.\{\frac{n}{2}th\text{observation}+(\frac{n}{2}+1)th\text{observation}\}$

$=\frac{1}{2}$ $\text{(5th observation + 6th observation)}$

$=\frac{1}{2}(48+50)=\frac{98}{2}=49$

Thus, M = 49.

The values of $(x_i-M)$ are

$-13,-7,-4,-3,-1,1,2,4,11,23$

$\therefore {\sum }_{i=1}^{10}|x_i-M|$ $=(13+7+4+3+1+1+2+4+11+23)=69$

$\Rightarrow MD\left(M\right)=\frac{{\sum }_{i=1}^{10}|x_i-M|}{10}=\frac{69}{10}=6.9$