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

The Correct Option is B

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