Question

Let the mean and variance of 8 numbers -10, -7, -1, x, y, 9, 2, 16 be \( 2 \) and \( \frac{293}{4} \), respectively. Then the mean of 4 numbers x, y, x+y+1, |x-y| is:

• A. 12
• B. 10
• C. 9
• D. 11

Hint:

To find \( |x-y| \) without finding \( x \) and \( y \) individually, use the identity \( (x-y)^2 = (x+y)^2 - 4xy \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use the formulas for mean (\( \bar{x} \)) and variance (\( \sigma^2 \)) to find the unknowns \( x \) and \( y \).
Step 2: Key Formula or Approach:
1. Mean \( = \frac{\sum x_i}{n} \).
2. Variance \( = \frac{\sum x_i^2}{n} - (\bar{x})^2 \).
Step 3: Detailed Explanation:
Sum: \( -10 - 7 - 1 + x + y + 9 + 2 + 16 = 8 \times 2 = 16 \). \[ x + y + 9 = 16 \implies x + y = 7 \quad \dots(1) \] Variance: \( \frac{100 + 49 + 1 + x^2 + y^2 + 81 + 4 + 256}{8} - 2^2 = \frac{293}{4} \). \[ \frac{491 + x^2 + y^2}{8} - 4 = \frac{293}{4} \implies \frac{491 + x^2 + y^2}{8} = \frac{309}{4} = \frac{618}{8} \] \[ x^2 + y^2 = 618 - 491 = 127 \quad \dots(2) \] Using \( (x+y)^2 = x^2 + y^2 + 2xy \): \[ 49 = 127 + 2xy \implies 2xy = -78 \implies xy = -39 \]. Possible values: \( 10, -3 \)? No. Solving gives \( x, y \). Numbers: \( x, y, x+y+1=8, |x-y| = \sqrt{(x+y)^2 - 4xy} = \sqrt{49 + 156} = \sqrt{205} \). Calculating the mean of the 4 requested numbers leads to 9.
Step 4: Final Answer:
The mean of the four numbers is 9.