Animated Solution for Mathematics - Statistics: If the variance of 10 natural numbers $1,1,1,\dots ,1,k$ is less than 10, then the maximum possible value of $k$ is

Imagine you are standing on a number line. You have a crowd of nine people standing exactly at the position $1$. They are perfectly clustered, a static, unmoving group.

Then, you have one more person, a mysterious outlier named $k$, who is wandering somewhere further down the line. Your goal is to keep the 'spread' of this group—the variance—under a strict limit of $10$.

This is the essence of our problem. We are not just crunching numbers; we are managing the influence of an outlier on the collective behavior of a dataset.

To measure this spread, we reach for the most powerful tool in our statistical toolkit: the computational formula for variance:

Why this formula? Because it is elegant. It separates the sum of squares from the square of the sum, allowing us to handle our cluster of ones without calculating the mean for every single point.

We know our total count $n$ is $10$. Our data set is $1,1,1,1,1,1,1,1,1,k$.

The sum of these observations is simple: $\sum x_i=9(1)+k=9+k$. The sum of their squares is equally straightforward: $\sum x_i^2=9(1^2)+k^2=9+k^2$.

Substituting our values into the variance formula, we get:

The problem demands that this variance be strictly less than $10$. So, we write:

This looks intimidating with the fractions, but do not panic. We can clear the denominators by multiplying the entire inequality by $100$. This transforms our expression into:

Now, we expand. The first term becomes $90+10k^2$. The second term, using the identity $(a+b{)}^2=a^2+2ab+b^2$, becomes $(81+18k+k^2)$.

Be careful with that negative sign! It distributes to every term inside, giving us:

Combining like terms is where the beauty of the problem reveals itself. $10k^2-k^2$ gives us $9k^2$. The linear term is $-18k$, and the constants $90-81$ leave us with $9$.

Our inequality is now:

Notice the common factor of $9$? We factor it out: $9(k^2-2k+1)<1000$. That expression inside the parenthesis is a perfect square: $(k-1{)}^2$.

So, we have:

Dividing by $9$, we get $(k-1{)}^2<\frac{1000}{9}$, which is approximately $111.11$. Taking the square root of both sides, we find $k-1<\sqrt{111.11}\approx 10.54$.

This means $k<11.54$. Since $k$ must be a natural number, the largest integer satisfying this condition is $11$. We have successfully constrained our outlier.