The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is

Options

A. 1216
B. 1456
C. 1072
D. 1792

Answer

To solve the problem, we need to find the sum of the cubes of the two remaining observations given the mean and variance of five observations, three of which are already known.

### Step-by-Step Solution:

1. **Understanding the Mean**:
The mean of the five observations is given as 5. The mean is calculated as:
\[ \text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}} \]
Let the two unknown observations be \( \alpha \) and \( \beta \). The three known observations are 1, 3, and 5. Therefore, we can express the sum of the observations as:
\[ 1 + 3 + 5 + \alpha + \beta = 5 \times 5 \]
Simplifying this:
\[ 9 + \alpha + \beta = 25 \]
Thus,
\[ \alpha + \beta = 25 - 9 = 16 \quad \text{(Equation 1)} \]

2. **Understanding the Variance**:
The variance is given as 8. The formula for variance is:
\[ \text{Variance} = \frac{\sum (x_i^2)}{n} - \left(\frac{\sum x_i}{n}\right)^2 \]
Here, \( n = 5 \) and the mean \( \mu = 5 \). Therefore:
\[ 8 = \frac{1^2 + 3^2 + 5^2 + \alpha^2 + \beta^2}{5} - 5^2 \]
Calculating the known squares:
\[ 1^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35 \]
Thus, substituting back:
\[ 8 = \frac{35 + \alpha^2 + \beta^2}{5} - 25 \]
Rearranging gives:
\[ 8 + 25 = \frac{35 + \alpha^2 + \beta^2}{5} \]
\[ 33 = \frac{35 + \alpha^2 + \beta^2}{5} \]
Multiplying through by 5:
\[ 165 = 35 + \alpha^2 + \beta^2 \]
Thus,
\[ \alpha^2 + \beta^2 = 165 - 35 = 130 \quad \text{(Equation 2)} \]

3. **Using the Equations**:
We have two equations:
- \( \alpha + \beta = 16 \) (Equation 1)
- \( \alpha^2 + \beta^2 = 130 \) (Equation 2)

We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting Equation 1 into this:
\[ 130 = 16^2 - 2\alpha\beta \]
\[ 130 = 256 - 2\alpha\beta \]
Rearranging gives:
\[ 2\alpha\beta = 256 - 130 = 126 \]
Thus,
\[ \alpha\beta = \frac{126}{2} = 63 \quad \text{(Equation 3)} \]

4. **Finding the Sum of Cubes**:
The sum of cubes can be calculated using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) \]
Substituting the known values:
\[ \alpha^3 + \beta^3 = 16 \left( 130 - 63 \right) \]
\[ = 16 \times 67 = 1072 \]

### Final Answer:
The sum of the cubes of the remaining two observations is \( \boxed{1072} \).