Question

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 1

Step 1. Form the inequality :

The series of $10$ terms contains $9$ term which is $1$ and one term $k$ .

The sum of the terms is given as:

$\begin{array}{rcl}\sum _{}x& =& 1+1+1+...+1+k\\ & =& 9+k\end{array}$

And the sum of the square of the terms is given as:

$\begin{array}{rcl}\sum _{}{x}^2& =& 1^2+1^2+1^2+...+1^2+k^2\\ & =& 9+k^2\end{array}$

The variance is given by the formula: $\frac{\sum _{}{x}^2}{n}-{\left(\frac{\sum _{}x}{n}\right)}^2$.

The variance of the $10$ terms is less than $10$. So form the inequality and substitute the values found:

$\begin{array}{rcl}\frac{\sum _{}{x}^2}{n}-{\left(\frac{\sum _{}x}{n}\right)}^2& <& 10\\ & \Rightarrow & \frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10\end{array}$

Step 2. Solve the inequality.

The inequality can be simplified as:

$$\begin{gathered} \frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10 \\ \Rightarrow \frac{9+k^2}{10}-\frac{{\left(9+k\right)}^2}{100}<10 \\ \Rightarrow \frac{10\left(9+k^2\right)-{\left(9+k\right)}^2}{100}<10 \\ \Rightarrow 90+10k^2-81-18k-k^2<1000 \\ \Rightarrow 9k^2-18k+9<1000 \\ \Rightarrow 9\left(k^2-2k+1\right)<1000 \\ \Rightarrow {\left(k-1\right)}^2<\frac{1000}{9} \end{gathered}$$

Now writing it in terms of compound inequality we get:

$\begin{array}{rcl}-\sqrt{\frac{1000}{9}}& <& k-1<\sqrt{\frac{1000}{9}}\\ & \Rightarrow & -\frac{10\sqrt{10}}{3}+1<k<\frac{10\sqrt{10}}{3}+1\end{array}$

So, the value of $k$ should be less than $\frac{10\sqrt{10}}{3}+1\approx 11.54$.

As $k$ is a natural number so the maximum value of $k$ is $11$.

Hence, the maximum possible value of $k$ is $11$.