Question

Find the mean deviation about the median for the data
$36,72,46,42,60,45,53,46,51,49$

Answer 1

Given =36,72,46,42,60,45,53,46,51,49

n = 10

median $=\frac{(\frac{n}{2}{)}^{th}obs+(\frac{n}{2}+1{)}^{th}}{2}obs$

$\frac{5^{th}obs+6^{th}obs}{2}$

$=\frac{46+49}{2}=\frac{95}{2}=47.5$

mean deviation above median =$\frac{ϵ|x:-m|}{10}$

= $\frac{(136-47.51+142-47.51+145-47.51+146-47.51+146-47.51+149-47-51+151-47.51+153-47.51+160-47.51+172-47.51)}{10}$

= $\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

= $\frac{70}{10}=7$