Question

If the mean and the variance of 6, 4, 8, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then $ a + b + ab $ is equal to:

• A. 105
• B. 103
• C. 100
• D. 106

Hint:

In solving problems involving variance and mean, use the formula for variance to derive the necessary equations. Then, use the system of equations to solve for unknowns such as \( a + b + ab \).

Answer 1

The Correct Option is B

Given: Mean \( = 9 \) and Variance \( = 9.25 \). The given numbers are \( 6, 4, 8, 8, b, 12, 10, 13 \). \[ \text{Mean} = \frac{53 + a + b}{7} = 9 \] \[ \Rightarrow 53 + a + b = 63 \quad \text{or} \quad a + b = 19 \] \[ \text{Variance:} \quad \sigma^2 = \frac{1}{7} \left[ 37 + 529 + a^2 + b^2 \right] \] \[ \Rightarrow 9.25 = \frac{37 + 529 + a^2 + b^2}{7} \] \[ \Rightarrow 648 + 74 = 529 + a^2 + b^2 \quad \Rightarrow \quad a^2 + b^2 = 193 \] Now we have the following system of equations: \[ a + b = 19 \quad \text{and} \quad a^2 + b^2 = 193 \] From this, we can solve for \( a + b + ab \): \[ (a + b)^2 = a^2 + b^2 + 2ab \] \[ 19^2 = 193 + 2ab \quad \Rightarrow \quad 361 = 193 + 2ab \] \[ \Rightarrow 2ab = 168 \quad \Rightarrow \quad ab = 84 \] Thus, \( a + b + ab = 103 \).