Let $$a_1, a_2, \ldots, a_{10}$$ 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 $$a_1, a_2, \ldots, a_{10}$$ is equal to :

Given: $$\sum_{k=1}^{10} a_k = 50$$ and $$\sum_{k < j} a_k a_j = 1100$$.

We know that $$\left(\sum_{k=1}^{10} a_k\right)^2 = \sum_{k=1}^{10} a_k^2 + 2\sum_{k < j} a_k a_j$$

$$50^2 = \sum a_k^2 + 2(1100)$$

$$2500 = \sum a_k^2 + 2200$$

$$\sum a_k^2 = 300$$

Variance: $$\sigma^2 = \frac{\sum a_k^2}{n} - \left(\frac{\sum a_k}{n}\right)^2 = \frac{300}{10} - \left(\frac{50}{10}\right)^2 = 30 - 25 = 5$$

Standard deviation: $$\sigma = \sqrt{5}$$

The answer is $$\sqrt{5}$$, which corresponds to Option (2).