Solve this following


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



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


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

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

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


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

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

$\mathrm{k} \leq 11$

Maximum value of $\mathrm{k}$ is 11 .


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