Question

If the mean and variance of observations \( x, y, 12, 14, 4, 10, 2 \) is \(8\) and \(16\) respectively, where \( x>y \), then the value of \( 3x - y \) is:

• A. \(18\)
• B. \(20\)
• C. \(22\)
• D. \(24\)

Hint:

For mean–variance problems:
Use mean to form sum equations
Variance helps form square relations
Combine symmetric sums: \(x+y\), \(x^2+y^2\), \(xy\)

Answer 1

The Correct Option is A

Step 1: Total number of observations \( n = 7 \). Given mean \( \bar{x} = 8 \): \[ \text{Sum of observations} = 7 \times 8 = 56 \] Sum of known observations: \[ 12 + 14 + 4 + 10 + 2 = 42 \] \[ \Rightarrow x + y = 56 - 42 = 14 (1) \]
Step 2: Variance is given by: \[ \sigma^2 = \frac{1}{n}\sum (x_i - \bar{x})^2 \] Given variance \( = 16 \): \[ \sum (x_i - 8)^2 = 7 \times 16 = 112 \]
Step 3: Compute squared deviations of known terms: \[ (12-8)^2 = 16,\; (14-8)^2 = 36,\; (4-8)^2 = 16,\; (10-8)^2 = 4,\; (2-8)^2 = 36 \] \[ \Rightarrow \text{Sum} = 108 \] Thus, \[ (x-8)^2 + (y-8)^2 = 112 - 108 = 4 (2) \]
Step 4: Expand equation (2): \[ x^2 + y^2 - 16(x+y) + 128 = 4 \] Using \( x + y = 14 \): \[ x^2 + y^2 = 100 (3) \]
Step 5: From: \[ (x+y)^2 = x^2 + y^2 + 2xy \] \[ 196 = 100 + 2xy \Rightarrow xy = 48 \]
Step 6: Solve for \( x, y \): \[ t^2 - 14t + 48 = 0 \Rightarrow t = 8, 6 \] Given \( x>y \): \[ x = 8,\; y = 6 \]
Step 7: Required value: \[ 3x - y = 3(8) - 6 = 18 \]