Question

If \( x_1, x_2, \dots, x_{25} \) be 25 observations such that \( \sum_{i=1}^{25} (x_i + 5)^2 = 2500 \) and \( \sum_{i=1}^{25} (x_i - 5)^2 = 1000 \). Then, the ratio of Mean and Standard deviation of the given observations is:

• A. \( \dfrac{1}{4} \)
• B. \( \dfrac{1}{2} \)
• C. \( \dfrac{2}{3} \)
• D. \( \dfrac{1}{5} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given two summations involving deviations from \( \pm 5 \). By expanding these squares and subtracting/adding the equations, we can solve for \( \sum x_i \) (to find the Mean) and \( \sum x_i^2 \) (to find the Variance and Standard Deviation).
Step 2: Key Formula or Approach:
1. \( (x_i + 5)^2 - (x_i - 5)^2 = 20x_i \)
2. \( (x_i + 5)^2 + (x_i - 5)^2 = 2x_i^2 + 50 \)
3. \( \text{Mean } (\bar{x}) = \frac{\sum x_i}{n} \); \( \text{Variance } (\sigma^2) = \frac{\sum x_i^2}{n} - (\bar{x})^2 \).
Step 3: Detailed Explanation:
1. Subtracting the two given equations: \[ \sum [(x_i + 5)^2 - (x_i - 5)^2] = 2500 - 1000 = 1500 \] \[ \sum (20x_i) = 1500 \implies 20 \sum x_i = 1500 \implies \sum x_i = 75 \] \[ \bar{x} = \frac{75}{25} = 3 \] 2. Adding the two given equations: \[ \sum [(x_i + 5)^2 + (x_i - 5)^2] = 2500 + 1000 = 3500 \] \[ \sum (2x_i^2 + 50) = 3500 \implies 2 \sum x_i^2 + (50 \times 25) = 3500 \] \[ 2 \sum x_i^2 + 1250 = 3500 \implies 2 \sum x_i^2 = 2250 \implies \sum x_i^2 = 1125 \] 3. Find Variance (\( \sigma^2 \)): \[ \sigma^2 = \frac{1125}{25} - (3)^2 = 45 - 9 = 36 \implies \sigma = 6 \] 4. Ratio of Mean to S.D.: \[ \text{Ratio} = \frac{\bar{x}}{\sigma} = \frac{3}{6} = \frac{1}{2} \]
Step 4: Final Answer:
The ratio of Mean and Standard deviation is \( \frac{1}{2} \).