Question

If \( x_1, x_2, x_3, \dots, x_n \) are \( n \) observations such that \( \sum (x_i + 2)^2 = 28n \) and \( \sum (x_i - 2)^2 = 12n \), then the variance is:

• A. \( 12 \)
• B. \( 14 \)
• C. \( 16 \)
• D. \( 20 \)

Hint:

Variance is computed using the squared deviations from the mean.

Answer 1

The Correct Option is A

We are given the following two conditions: \[ \sum (x_i + 2)^2 = 28n \] \[ \sum (x_i - 2)^2 = 12n \] Step 1: Expanding the Given Expressions Expanding the first condition: \[ \sum (x_i + 2)^2 = \sum x_i^2 + 4\sum x_i + 4n \] From the given condition, \[ \sum x_i^2 + 4\sum x_i + 4n = 28n \] Expanding the second condition: \[ \sum (x_i - 2)^2 = \sum x_i^2 - 4\sum x_i + 4n \] From the given condition, \[ \sum x_i^2 - 4\sum x_i + 4n = 12n \] Step 2: Adding the Two Equations \[ \sum x_i^2 + 4\sum x_i + 4n + \sum x_i^2 - 4\sum x_i + 4n = 28n + 12n \] Combining like terms: \[ 2\sum x_i^2 + 8n = 40n \] \[ \sum x_i^2 = 16n \] Step 3: Subtracting the Two Equations \[ \sum x_i^2 + 4\sum x_i + 4n - (\sum x_i^2 - 4\sum x_i + 4n) = 28n - 12n \] Simplifying: \[ 8\sum x_i = 16n \] \[ \sum x_i = 2n \] Step 4: Finding Variance The variance formula is: \[ \text{Variance} = \frac{1}{n} \sum x_i^2 - \left(\frac{1}{n} \sum x_i \right)^2 \] Substituting the known values: \[ \text{Variance} = \frac{1}{n} (16n) - \left( \frac{2n}{n} \right)^2 \] \[ \text{Variance} = 16 - 4 \] \[ \text{Variance} = 12 \] Step 5: Final Answer 

\[Correct Answer: (1) \ 12\]