Question:
If the variance of 10 natural numbers $1,1,1, \ldots, 1, k$ is less than 10 , then the maximum possible value of $k$ is
Solution:
$\sigma^{2}=\frac{\Sigma X^{2}}{n}-\left(\frac{\Sigma x}{n}\right)^{2}$
$\sigma^{2}=\frac{\left(9+k^{2}\right)}{10}-\left(\frac{9+k}{10}\right)^{2}<10$
$\left(90+k^{2}\right) 10-\left(81+k^{2}+8 k\right)<1000$
$90+10 k^{2}-k^{2}-18 k-81<1000$
$9 k^{2}-18 k+9<1000$
$(k-1)^{2}<\frac{1000}{9} \Rightarrow k-1<\frac{10 \sqrt{10}}{3}$
$k<\frac{10 \sqrt{10}}{3}+1$
Maximum integral value of $k=11$
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