Mathematics - Statistics Question with Solution | TestHub

Given${\mathrm{b}}_1,{\mathrm{b}}_2,{\mathrm{b}}_3,\dots ..{\mathrm{b}}_{11},$ are in $\mathrm{A}.\mathrm{P}.$ 

$\therefore$ Variance of $\left({\mathrm{b}}_1,{\mathrm{b}}_2,\dots ..{\mathrm{b}}_{11}\right)=$Variance of $(0,\mathrm{d},2\mathrm{d},\dots ..10\mathrm{d})=90$

$\Rightarrow \frac{{\sum }_{i=1}^{11}{b}_i^2}{11}-{\left(\frac{{\sum }_{i=1}^{11}{b}_i}{11}\right)}^2=90$

$\Rightarrow \frac{d^2}{11}·\frac{10.11.21}{6}-d^2·\frac{55}{11}·\frac{55}{11}=90$

$\Rightarrow 35d^2-25d^2=90$

$\Rightarrow d^2=9 \Rightarrow d=\pm 3$

It is an increasing $A.P. \Rightarrow d$ is positive.

$\Rightarrow d=3$