Question

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

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

Answer 1

The correct option is A $1936$

Here, $n=88$

$Median=\frac{{44}^{th}obs+{45}^{th}obs}{2}$

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

$\therefore$ $Median=56.5$

$MD=\frac{{\sum }_{i=1}^{88}|x_i-56.5|}{88}$

$=\frac{43.5+42.5+......+0.5+0.5+..+43.5}{88}$

$=\frac{1+3+5+...+87}{88}$

$=22$

Now, $1,3,5,..,87$ are in $AP$

$a=1,d=2$

$a_x=a+(n-1)d$

$87=1+(n-1)2$

$86=(n-1)2$

$n-1=43$

$\therefore$ $n=44$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{44}{2}\left[2\right(1)+(44-1\left)2\right]$

$=22[2+86]$

$=22\left[88\right]$

$=1936$