Question

Let $a_1, a_2, \ldots, a_{10}$ be 10 observations such that $\sum_{k=1}^{10} a_k = 50 \quad \text{and} \quad \sum_{1 \leq k < j \leq 10} a_k \cdot a_j = 1100$. Then the standard deviation of $a_1, a_2, \ldots, a_{10}$ is equal to:

• A. $5$
• B. $\sqrt{5}$
• C. $10$
• D. $\sqrt{15}$

Answer 1

Answer: (B)

$\sqrt{5}$

Answer 2

Step 1. Use the Formula for Standard Deviation:

$\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{10} a_i^2 - \left( \frac{1}{n} \sum_{i=1}^{10} a_i \right)^2}$

where $n = 10$.

Step 2. Calculate $\sum_{i=1}^{10} a_i^2$: We know:

$\sum_{i=1}^{10} a_i = 50, \quad \sum_{1 \leq k < j \leq 10} a_k \cdot a_j = 1100$
Expanding:

$\left( \sum_{i=1}^{10} a_i \right)^2 = \sum_{i=1}^{10} a_i^2 + 2 \sum_{1 \leq k < j \leq 10} a_k \cdot a_j$

$2500 = \sum_{i=1}^{10} a_i^2 + 2200 \Rightarrow \sum_{i=1}^{10} a_i^2 = 300$

Step 3. Compute Standard Deviation:

$\sigma = \sqrt{\frac{300}{10} - \left( \frac{50}{10} \right)^2} = \sqrt{30 - 25} = \sqrt{5}$