# If the variance of 10 natural numbers {1,1,1.......1, k} is less than 10, then the maximum possible value of k is _______. Option: 1 9 Option: 2 11 Option: 3 13 Option: 4 15

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

$\\90+10 k^{2}-81-k^{2}-18 k<1000 \\ \\9 k^{2}-18 k-991<0$

\begin{aligned} &\mathrm{k}^{2}-2 \mathrm{k}<\frac{991}{9}\\ &(\mathrm{k}-1)^{2}<\frac{1000}{9}\\ &\frac{-10 \sqrt{10}}{3}<\mathrm{k}-1<\frac{10 \sqrt{10}}{3}\\ &\mathrm{k}<\frac{10 \sqrt{10}}{3}+1\\ &k \leq 11\\ &\text {Maximum value of } \mathrm{k} \text { is } 11 \text { . } \end{aligned}

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