Animated Solution for Mathematics - Statistics: Let $a_1,a_2,\dots ,a_{10}$ be 10 observations such that ${\sum }_{k=1}^{10}{a}_k=50$ and ${\sum }_{\forall k<j}{a}_k\cdot a_j=1100$. Then the standard deviation of $a_1,a_2,\dots ,a_{10}$ is equal to :

Statistics is often seen as a dry collection of data, but at its heart, it is a beautiful geometry of numbers. We are given $10$ observations, $a_1,a_2,\dots ,a_{10}$, with a total sum of ${\sum }_{k=1}^{10}{a}_k=50$ and a sum of pairwise products ${\sum }_{k<j}{a}_k{a}_j=1100$.

Our goal is to find the standard deviation. To get there, we need to build a bridge between these sums and the variance formula.

The bridge is a classic algebraic identity:

This identity is the secret key. It connects the sum of the numbers, the sum of their squares, and the sum of their pairwise products.

Now, let's substitute our known values into this identity. We have the sum of observations as $50$, so the left side becomes $(50{)}^2=2500$.

On the right side, we have $2\times 1100=2200$. Our equation now looks like this:

By simply transposing the $2200$, we find that the sum of the squares of our observations is ${\sum }_{i=1}^{10}{a}_i^2=300$. This is a massive breakthrough!

With this value in hand, we are ready to use the statistical engine: the variance formula. The variance, denoted by ${\sigma }^2$, is defined as the mean of the squares minus the square of the mean:

We have everything we need. We know $n=10$, $\sum a_i^2=300$, and $\sum a_i=50$. Let's plug these into our formula:

Simplifying this, we get ${\sigma }^2=30-(5{)}^2$, which results in $30-25=5$.

Finally, the standard deviation $\sigma$ is the square root of the variance. Therefore, the final answer is:

It is truly elegant how the algebra and statistics dance together to give us such a clean result. Remember, whenever you see sums of observations and pairwise products, look for that algebraic identity—it is your most powerful tool.