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Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}<\mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the $5$ observations $9$,$25$, $a$, $b$, $c$ be $18$,$4$ and $\frac{136}{5}$, respectively. Then $2 \mathrm{a}+\mathrm{b}-\mathrm{c}$ is equal to ..............
$39$
$18$
$35$
$33$
Solution
$ a, b, c \in N \quad a < b < c $
$ \bar{x}=\text { mean }=\frac{9+25+a+b+c}{5}=18 $
$ a+b+c=56 $
$ \text { Mean deviation }=\frac{\sum\left|x_i-\bar{x}\right|}{n}=4 $
$ =9+7+|18-a|+|18-b|+|18-c|=20 $
$ =|18-a|+|18-b|+|18-c|=4 $
$ \text { Variance }=\frac{\Sigma\left|x_i-\bar{x}\right|^2}{n}=\frac{136}{5} $
$ =81+49+|18-a|^2+|18-b|^2+|18-c|^2=136 $
$ =(18-a)^2+(18-b)^2+(18-c)^2=6 $
$ \text { Possible values }(18-a)^2=1, \quad(18-b)^2=1 \quad(18-c)^2=4 $
$ \mathrm{a}<\mathrm{b}<\mathrm{c} \quad 18-\mathrm{a}=1 \quad 18-b=-1 \quad 18-\mathrm{c}=-2 $
$ \text { so } \quad \quad \quad \mathrm{a}=17 \quad \mathrm{~b}=19 \quad \mathrm{c}=20 $
$ \mathrm{a}+\mathrm{b}+\mathrm{c}=56 \quad 2 \mathrm{a}+\mathrm{b}-\mathrm{c} \quad 34=19-20=33$
