Mathematics - Statistics Question with Solution | TestHub

${\sigma }^2=\frac{\Sigma x^2}{n}-{\left(\frac{\Sigma x}{n}\right)}^2$

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

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

$9k^2-18k-991<0$

$k^2-2k<\frac{991}{9}$

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

$\frac{-10\sqrt{10}}{3}<k-1<\frac{10\sqrt{10}}{3}$

$k<\frac{10\sqrt{10}}{3}+1$

$k\le 11$

Maximum value of $k$ is $11.$