Question

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are:

• A. 2 and 9
• B. 3 and 8
• C. 4 and 7
• D. 5 and 6

Hint:

When solving problems involving mean and variance, use the relationships between sum, squares, and products of the observations to set up equations that can be solved simultaneously.

Answer 1

The Correct Option is C

We are given the following information: - The mean of the five observations is 4. - The variance of the five observations is 5.2. - Three of the observations are 1, 2, and 6. Step 1: Use the mean formula The mean of the five observations is given by: \[ \frac{1 + 2 + 6 + x + y}{5} = 4 \] Simplifying: \[ \frac{9 + x + y}{5} = 4 \] Multiplying both sides by 5: \[ 9 + x + y = 20 \] Thus: \[ x + y = 11 \] Step 2: Use the variance formula The variance is given by: \[ \frac{1}{5} \left[ (1 - 4)^2 + (2 - 4)^2 + (6 - 4)^2 + (x - 4)^2 + (y - 4)^2 \right] = 5.2 \] This simplifies to: \[ \frac{1}{5} \left[ 9 + 4 + 4 + (x - 4)^2 + (y - 4)^2 \right] = 5.2 \] Simplifying further: \[ \frac{17 + (x - 4)^2 + (y - 4)^2}{5} = 5.2 \] Multiplying both sides by 5: \[ 17 + (x - 4)^2 + (y - 4)^2 = 26 \] Thus: \[ (x - 4)^2 + (y - 4)^2 = 9 \] Step 3: Solve the system of equations We now have the system of equations: 1. \( x + y = 11 \) 2. \( (x - 4)^2 + (y - 4)^2 = 9 \) Expanding the second equation: \[ (x - 4)^2 + (y - 4)^2 = (x^2 - 8x + 16) + (y^2 - 8y + 16) = 9 \] Simplifying: \[ x^2 + y^2 - 8x - 8y + 32 = 9 \] \[ x^2 + y^2 - 8x - 8y = -23 \] Substitute \( x + y = 11 \) into the equation: \[ x^2 + y^2 - 8(11) = -23 \] \[ x^2 + y^2 - 88 = -23 \] \[ x^2 + y^2 = 65 \] Now, use the identity \( (x + y)^2 = x^2 + y^2 + 2xy \) to find \( xy \): \[ (11)^2 = 65 + 2xy \] \[ 121 = 65 + 2xy \] \[ 2xy = 56 \] \[ xy = 28 \] Step 4: Solve for \( x \) and \( y \) Now, we solve the quadratic equation: \[ t^2 - (x + y)t + xy = 0 \] Substitute \( x + y = 11 \) and \( xy = 28 \): \[ t^2 - 11t + 28 = 0 \] Solving this quadratic equation: \[ t = \frac{11 \pm \sqrt{11^2 - 4 \times 1 \times 28}}{2} = \frac{11 \pm \sqrt{121 - 112}}{2} = \frac{11 \pm \sqrt{9}}{2} = \frac{11 \pm 3}{2} \] Thus, the solutions are: \[ t = \frac{11 + 3}{2} = 7 \quad {and} \quad t = \frac{11 - 3}{2} = 4 \] Thus, the other two observations are 4 and 7. Therefore, the correct answer is Option C.