Question

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

Answer 1

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