Question

The mean deviation about median of variates $13,14,15,....99,100$ is

• A. $1936$
• B. $21.5$
• C. $23.5$
• D. $22$

Answer 1

The correct option is D $22$

$13,14,15,.....99,100$

Total number of observation $=88$

Median $=\frac{{\left(\frac{88}{2}\right)}^{th}term+{(\frac{88}{2}+1)}^{th}term}{2}$

Median $=\frac{56+57}{2}$

Median $M=56.5$

$x_i-M$, we get

$-43.5,-42.5,-41.5,............42.5,43.5$

Now, $|x_i-M|$

$43.5,42.5,41.5,.....,1.5,0.5,0.5,\mathrm{1.5.......42.5},43.5$

Mean deviation $=\frac{{\sum }_{i=1}^{88}|x_i-M|}{88}$

Mean deviation $=\frac{22(43.5+0.5)+22(0.5+43.5)}{88}$ [Using Sum formula of $AP$ when first and last terms are given]

Mean deviation $=22$.