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 observations are:

• A. 4, 7
• B. 2, 10
• C. 5, 6
• D. 2, 9

Hint:

To solve problems involving the mean and variance of a set of observations, use the basic definitions of the mean and variance, and set up equations accordingly.

Answer 1

The Correct Option is A

Let the five observations be \( x_1 = 1, x_2 = 2, x_3 = 6, x_4 \), and \( x_5 \). The mean of the five observations is given by: \[ \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = 4 \] Substituting the known values: \[ \frac{1 + 2 + 6 + x_4 + x_5}{5} = 4 \] This simplifies to: \[ 9 + x_4 + x_5 = 20 \] Thus, \[ x_4 + x_5 = 11 \] So the sum of the other two observations is 11. Next, the variance is given by: \[ \text{Variance} = \frac{1}{5} \sum_{i=1}^{5} (x_i - \mu)^2 \] Substituting the known values, the variance is 5.2: \[ 5.2 = \frac{1}{5} \left[ (1-4)^2 + (2-4)^2 + (6-4)^2 + (x_4 - 4)^2 + (x_5 - 4)^2 \right] \] Simplifying the terms: \[ 5.2 = \frac{1}{5} \left[ 9 + 4 + 4 + (x_4 - 4)^2 + (x_5 - 4)^2 \right] \] \[ 5.2 = \frac{1}{5} \left[ 17 + (x_4 - 4)^2 + (x_5 - 4)^2 \right] \] \[ 26 = 17 + (x_4 - 4)^2 + (x_5 - 4)^2 \] \[ 9 = (x_4 - 4)^2 + (x_5 - 4)^2 \] Now, we know that \( x_4 + x_5 = 11 \). Let's substitute \( x_4 = 4 \) and \( x_5 = 7 \), which satisfies both the conditions: \[ (4 - 4)^2 + (7 - 4)^2 = 0 + 9 = 9 \] Thus, the other two observations are \( 4 \) and \( 7 \).