If the variance of 10 natural numbers $1,1,1, \ldots, 1, k$ is less than 10 , then the maximum possi

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

Professor PC Thomas Classes · Answer

(11) : Given,\ $\sigma^{2}=\frac{\sum x^{2}}{n}-\left(\frac{\sum x}{n}ight)^{2}<10$\ $\Rightarrow \frac{9+k^{2}}{10}-\left(\frac{9+k}{10}ight)^{2}<10$\ $\Rightarrow \quad 90+10 k^{2}-81-k^{2}-18 k<1000$\ $\Rightarrow \quad 9 k^{2}-18 k+9<1000$\ $\Rightarrow \quad 9(k-1)^{2}<1000$\ $\Rightarrow(k-1)^{2}<\frac{1000}{9}$\ $\Rightarrow k<\frac{10 \sqrt{10}}{3}+1 \approx 11.54$\ $	herefore \quad$ Maximum integral value of $k$ is 11 .\