If the variance of 10 natural numbers $1,1,1\dots .,1,k$ is less than 10 then the maximum possible value of $k$ is

Given observations 1,1,1,….,1,k

Variance <10

$\frac{{\displaystyle \sum {x_i}^2}}{n}-{\left(\frac{{\displaystyle \sum x_i}}{n}\right)}^2<10$

$\Rightarrow \frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$

$\Rightarrow \frac{9+k^2}{10}-\frac{81+k^2+18k}{100}<10$

$\Rightarrow 90+10k^2-81-k^2-18k<1000$

$\Rightarrow 9k^2-18k+9-1000<0$

$\Rightarrow 9k^2-18k<991$

$\Rightarrow 9k^2-18k+9<1000$

$\Rightarrow k^2-2k+1<\frac{1000}{9}$

$\Rightarrow {\left(k-1\right)}^2<\frac{1000}{9}$

$\Rightarrow \frac{-10\sqrt{10}}{3}<\left(k-1\right)<\frac{10\sqrt{10}}{3}$

$\Rightarrow -10.54<k-1<10.54$

$\Rightarrow -9.54<k<11.54$

$\therefore \max imumvalueofkis$ 11