Question

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

Answer 1

${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2$
$\Rightarrow {\sigma }^2=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$
$\Rightarrow 10(9+k^2)-(81+k^2+18k)<1000$
$\Rightarrow 90+10k^2-k^2-18k-81<1000$
$\Rightarrow 9k^2-18k+9<1000$
$\Rightarrow (k-1{)}^2<\frac{1000}{9}$
$\Rightarrow k-1<\frac{10\sqrt{10}}{3}$
$\Rightarrow k<\frac{10\sqrt{10}}{3}+1$
$\therefore$ Maximum possible integral value of $k$ is $11.$