Question

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

Answer 1

Answer: 11

${\sigma }^2=\frac{\Sigma x^2}{n}-{\left(\frac{\Sigma x}{n}\right)}^2$
$=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$
$90+10k^2-81-k^2-18k<1000$
$9k^2-18k-991<0$
$k^2-2k<\frac{991}{9}$
$(k-1{)}^2<\frac{1000}{9}$
$\frac{-10\sqrt{10}}{3}<k-1<\frac{10\sqrt{10}}{3}$
$k<\frac{10\sqrt{10}}{3}+1$
$k\le 11$
Maximum value of $k$ is 11 .