If the variance of 10 natural numbers 1,1,1,…,1,k is less than 10, then the maximum possible value of k is ..........
Answer: 11
Solutions · 1
Official solution
\[\sigma^2 = \frac{9+k^2}{10}-\left(\frac{9+k}{10}\right)^2 < 10\]
\[10(9+k^2)-(81+k^2+18k) < 1000 \implies 9k^2-18k+9 < 1000 \implies (k-1)^2 < \frac{1000}{9}\]
\[k-1 < \frac{10\sqrt{10}}{3} \approx 10.54 \implies k < 11.54\]
Maximum possible integral (natural number) value of k is 11.