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