1. Interpret the data

There are $10$ natural numbers: 1,1,1,ots,1,k So we have $9$ numbers equal to $1$ and one number equal to $k$.

2. Find the mean

The sum of all observations is: $9\cdot 1 + k = 9+k$ Hence the mean is: $\bar{x} = \frac{9+k}{10}$

3. Use the formula for variance

Variance of $n$ observations is: $\sigma^2 = \frac{1}{n}\sum (x_i-\bar{x})^2$

Here,

• $9$ observations are $1$
• $1$ observation is $k$

So, $\sigma^2 = \frac{1}{10}\left[9\left(1-\frac{9+k}{10}\right)^2 + \left(k-\frac{9+k}{10}\right)^2\right]$

Now simplify each term: $1-\frac{9+k}{10} = \frac{10-(9+k)}{10} = \frac{1-k}{10}$ $k-\frac{9+k}{10} = \frac{10k-(9+k)}{10} = \frac{9k-9}{10} = \frac{9(k-1)}{10}$

Thus, $\sigma^2 = \frac{1}{10}\left[9\left(\frac{k-1}{10}\right)^2 + \left(\frac{9(k-1)}{10}\right)^2\right]$

$= \frac{1}{10}\left[9\cdot \frac{(k-1)^2}{100} + \frac{81(k-1)^2}{100}\right]$

$= \frac{1}{10}\cdot \frac{90(k-1)^2}{100}$

$= \frac{9(k-1)^2}{100}$

4. Apply the condition that variance is less than $10$

$\frac{9(k-1)^2}{100} < 10$

Multiply by $100$: $9(k-1)^2 < 1000$

$(k-1)^2 < \frac{1000}{9}$

Since $\frac{1000}{9} \approx 111.11$ we get $(k-1)^2 < 111.11$

So, $k-1 < \sqrt{111.11} \approx 10.54$

Since $k$ is a natural number, $k-1 \le 10 \Rightarrow k \le 11$

Thus the maximum possible value is: $\boxed{11}$

5. Verification by substitution

For $k=11$: $\sigma^2 = \frac{9(11-1)^2}{100} = \frac{9\cdot 100}{100} = 9 < 10$

For $k=12$: $\sigma^2 = \frac{9(12-1)^2}{100} = \frac{9\cdot 121}{100} = 10.89 > 10$

So $11$ is indeed the greatest possible value.

6. Comparison with stored answer

Stored correct answer: $11$

Our derived answer is also $11$, so they agree.