Mean and variance of five observations are respectively 4 and 5.2. If three observations among them are 1, 2 and 6, then remaining two observations will be:

Question

Accepted Answer

To solve this problem, we need to determine the remaining two observations given the mean and variance of the five observations, as well as three known observations.

### Explanation for Option 3 being correct:

1. **Mean Calculation:**
   - The mean of the five observations is given as 4. Therefore, the sum of all five observations is:
     $$
     \text{Mean} = \frac{\text{Sum of observations}}{5} = 4 \implies \text{Sum of observations} = 4 \times 5 = 20
     $$
   - We know three of the observations: 1, 2, and 6. Their sum is:
     $$
     1 + 2 + 6 = 9
     $$
   - Therefore, the sum of the remaining two observations is:
     $$
     20 - 9 = 11
     $$

2. **Variance Calculation:**
   - The variance of the five observations is given as 5.2. The formula for variance is:
     $$
     \text{Variance} = \frac{\sum (x_i - \text{Mean})^2}{5} = 5.2
     $$
   - This implies:
     $$
     \sum (x_i - 4)^2 = 5.2 \times 5 = 26
     $$
   - For the known observations (1, 2, and 6), we calculate:
     $$
     (1-4)^2 + (2-4)^2 + (6-4)^2 = 9 + 4 + 4 = 17
     $$
   - Therefore, the sum of the squares of deviations for the remaining two observations is:
     $$
     26 - 17 = 9
     $$

3. **Finding the Remaining Observations:**
   - Let the remaining two observations be $ x $ and $ y $. We have:
     $$
     x + y = 11
     $$
     $$
     (x-4)^2 + (y-4)^2 = 9
     $$
   - Expanding the second equation:
     $$
     (x-4)^2 + (y-4)^2 = (x^2 - 8x + 16) + (y^2 - 8y + 16) = 9
     $$
     $$
     x^2 + y^2 - 8x - 8y + 32 = 9
     $$
     $$
     x^2 + y^2 - 8x - 8y = -23
     $$
   - Using $ x + y = 11 $, we substitute $ y = 11 - x $ into the equation:
     $$
     x^2 + (11-x)^2 - 8x - 8(11-x) = -23
     $$
     $$
     x^2 + 121 - 22x + x^2 - 8x - 88 + 8x = -23
     $$
     $$
     2x^2 - 22x + 33 = -23
     $$
     $$
     2x^2 - 22x + 56 = 0
     $$
   - Solving this quadratic equation, we find $ x = 4 $ and $ y = 7 $.

### Explanation for Other Options:

- **Option 1 (5 and 8):** The sum is 13, not 11, and the variance calculation does not satisfy the condition.
- **Option 2 (8 and 3):** The sum is 11, but the variance calculation does not satisfy the condition.
- **Option 4 (6 and 4):** The sum is 10, not 11, and the variance calculation does not satisfy the condition.

Thus, Option 3 (4 and 7) is the only set of observations that satisfies both the mean and variance conditions, making it the correct answer.