The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2,4,10,12 and 14, then find the remaining two observations.

Options

A. 5,7
B. 3,5
C. 6,8
D. 4,2

Answer

To solve the problem, we need to find the two remaining observations given the mean and variance of seven observations. Let's break it down step by step.

### Step 1: Understand the Given Information
- We have 7 observations.
- The mean of these observations is 8.
- The variance of these observations is 16.
- Five of the observations are: 2, 4, 10, 12, and 14.

### Step 2: Calculate the Sum of All Observations
The mean is calculated as:
\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \]
Given that the mean is 8 and there are 7 observations:
\[ 8 = \frac{S}{7} \implies S = 8 \times 7 = 56 \]
Where \( S \) is the sum of all observations.

### Step 3: Calculate the Sum of the Known Observations
Now, let's calculate the sum of the known observations:
\[ 2 + 4 + 10 + 12 + 14 = 42 \]

### Step 4: Set Up the Equation for the Remaining Observations
Let the two remaining observations be \( x \) and \( y \). Therefore, we can write:
\[ 42 + x + y = 56 \]
This simplifies to:
\[ x + y = 56 - 42 = 14 \quad \text{(Equation 1)} \]

### Step 5: Calculate the Variance
The variance formula is given by:
\[ \text{Variance} = \frac{\sum (x_i^2)}{n} - \left(\frac{\sum x_i}{n}\right)^2 \]
Where \( n \) is the number of observations. We know:
- Variance = 16
- \( n = 7 \)
- Mean = 8, so \( \left(\frac{\sum x_i}{n}\right)^2 = 8^2 = 64 \)

Now, substituting the values into the variance formula:
\[ 16 = \frac{(2^2 + 4^2 + 10^2 + 12^2 + 14^2 + x^2 + y^2)}{7} - 64 \]
Calculating the squares of the known observations:
\[ 2^2 = 4, \quad 4^2 = 16, \quad 10^2 = 100, \quad 12^2 = 144, \quad 14^2 = 196 \]
Adding these:
\[ 4 + 16 + 100 + 144 + 196 = 460 \]
Now substituting back into the variance equation:
\[ 16 = \frac{460 + x^2 + y^2}{7} - 64 \]
Rearranging gives:
\[ 16 + 64 = \frac{460 + x^2 + y^2}{7} \]
\[ 80 = \frac{460 + x^2 + y^2}{7} \]
Multiplying both sides by 7:
\[ 560 = 460 + x^2 + y^2 \]
Thus:
\[ x^2 + y^2 = 560 - 460 = 100 \quad \text{(Equation 2)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( x + y = 14 \)
2. \( x^2 + y^2 = 100 \)

Using the identity \( x^2 + y^2 = (x + y)^2 - 2xy \):
\[ 100 = 14^2 - 2xy \]
Calculating \( 14^2 \):
\[ 100 = 196 - 2xy \]
Rearranging gives:
\[ 2xy = 196 - 100 = 96 \implies xy = 48 \quad \text{(Equation 3)} \]

### Step 7: Solve for \( x \) and \( y \)
Now we can solve the equations:
1. \( x + y = 14 \)
2. \( xy = 48 \)

Let \( x \) and \( y \) be the roots of the quadratic equation:
\[ t^2 - (x+y)t + xy = 0 \implies t^2 - 14t + 48 = 0 \]
Using the quadratic formula:
\[ t = \frac{14 \pm \sqrt{14^2 - 4 \cdot 48}}{2} \]
Calculating the discriminant:
\[ 14^2 - 192 = 196 - 192 = 4 \]
Thus:
\[ t = \frac{14 \pm 2}{2} \]
Calculating the roots:
1. \( t = \frac{16}{2} = 8 \)
2. \( t = \frac{12}{2} = 6 \)

### Conclusion
The two remaining observations are \( x = 8 \) and \( y = 6 \). Therefore, the two observations are:
\[ \boxed{6 \text{ and } 8} \]