Question

The mean of $5$ observations is $5$ and their variance is $124$. If three of the observations are $1, 2$ and $6$ ; then the mean deviation from the mean of the data is :

• A. 2.4
• B. 2.8
• C. 2.5
• D. 2.6

Answer 1

Answer: (B)

2.8

Answer 2

$\overline{ x }=\frac{ x _{1}+ x _{2}+ x _{3}+ x _{4}+ x _{5}}{5}=5$
$\displaystyle\sum_{ i =1}^{5} x _{ i }=25 \ldots \ldots \ldots . .( i )$
Also $\sigma^{2}=124$
$\Rightarrow \frac{\sum x _{1}^{2}}{5}-(\overline{ x })^{2}=124$
$\Rightarrow \frac{\sum x _{1}^{2}}{5}=124+25=149$
$\Rightarrow \left(x_{1}^{2}+x_{2}^{2}+\ldots . .+x_{5}^{2}\right)=745$
$\Rightarrow x_{1}^{2}+x_{2}^{2}=704 \ldots \ldots \ldots (ii)$
by (i) } $x_{1}+x_{2}=16 \ldots \ldots \ldots \ldots . (iii)$
$2 x_{1} x_{2}+704=256$
$x_{1} x_{2}=\frac{256-704}{2}$
$x_{1} x_{2}=128-352=-224 \ldots \ldots \ldots \ldots$ (iv)
Now $\frac{\sum\left|x_{1}-5\right|}{5}-\frac{\left|x_{1}-5\right|+\left|x_{2}-5\right|+4+3+1}{5}$
$=\frac{8+\left|x_{1}-5\right|+\left|11-x_{1}\right|}{5}$
$=\frac{8+6}{5}=2.8$ Ans