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

To find the standard deviation of the observations \( a_1, a_2, \ldots, a_{10} \), we are given:

• \(\sum_{k=1}^{10} a_k = 50\) 
• \(\sum_{1 \leq k < j \leq 10} a_k \cdot a_j = 1100\)

Let's use these conditions to find the standard deviation.

1. The formula for the variance of a data set is given by: \(\sigma^2 = \frac{1}{n} \left( \sum_{k=1}^{n} a_k^2 - \frac{1}{n} \left( \sum_{k=1}^{n} a_k \right)^2 \right)\) where \( n = 10 \) in this case.
2. We need to determine \(\sum_{k=1}^{10} a_k^2\). We know:
   • \((\sum_{k=1}^{10} a_k )^2 = \sum_{k=1}^{10} a_k^2 + 2 \sum_{1 \leq k < j \leq 10} a_k \cdot a_j\)
   • Plugging in the given values: \(50^2 = \sum_{k=1}^{10} a_k^2 + 2 \cdot 1100\)
   • Therefore: \(2500 = \sum_{k=1}^{10} a_k^2 + 2200\)
   • Calculate \(\sum_{k=1}^{10} a_k^2\): \(\sum_{k=1}^{10} a_k^2 = 2500 - 2200 = 300\)
3. Substitute into the variance formula:
   • \( \sigma^2 = \frac{1}{10} \left( 300 - \frac{1}{10} \times 2500 \right) \)
   • \( \sigma^2 = \frac{1}{10} (300 - 250) \)
   • \( \sigma^2 = \frac{1}{10} \times 50 = 5 \)
4. The standard deviation is: \(\sigma = \sqrt{5}\)

Hence, the standard deviation of the observations \( a_1, a_2, \ldots, a_{10} \) is \(\sqrt{5}\), which matches the option given.

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