The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4,5 and 7, then remaining two observations are :

Options

A. 10,11
B. 3,18
C. 8,13
D. 1,20

Answer

To solve the problem, we need to find the two remaining observations \( x \) and \( y \) given the mean and variance of six distinct observations. The four known observations are 2, 4, 5, and 7.

### Step-by-Step Solution:

1. **Calculate the Mean:**
The mean of the observations is given as 6.5. The formula for the mean is:
\[ \text{Mean} = \frac{\text{Sum of observations}}{n} \]
where \( n \) is the number of observations. Here, \( n = 6 \).

Therefore, we can write:
\[ 6.5 = \frac{2 + 4 + 5 + 7 + x + y}{6} \]
Simplifying the left side:
\[ 6.5 \times 6 = 2 + 4 + 5 + 7 + x + y \]
\[ 39 = 18 + x + y \]
Rearranging gives us:
\[ x + y = 39 - 18 = 21 \quad \text{(Equation 1)} \]

2. **Calculate the Variance:**
The variance is given as 10.25. The formula for variance is:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \]
Substituting the known values:
\[ 10.25 = \frac{2^2 + 4^2 + 5^2 + 7^2 + x^2 + y^2}{6} - (6.5)^2 \]
Calculating \( (6.5)^2 \):
\[ 6.5^2 = 42.25 \]
Now substituting the squares of the known values:
\[ 10.25 = \frac{4 + 16 + 25 + 49 + x^2 + y^2}{6} - 42.25 \]
Summing the squares:
\[ 10.25 = \frac{94 + x^2 + y^2}{6} - 42.25 \]
Rearranging gives:
\[ 10.25 + 42.25 = \frac{94 + x^2 + y^2}{6} \]
\[ 52.5 = \frac{94 + x^2 + y^2}{6} \]
Multiplying both sides by 6:
\[ 315 = 94 + x^2 + y^2 \]
Rearranging gives:
\[ x^2 + y^2 = 315 - 94 = 221 \quad \text{(Equation 2)} \]

3. **Solve the System of Equations:**
We now have two equations:
- Equation 1: \( x + y = 21 \)
- Equation 2: \( x^2 + y^2 = 221 \)

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 21 - x \]
Substituting this into Equation 2:
\[ x^2 + (21 - x)^2 = 221 \]
Expanding the square:
\[ x^2 + (441 - 42x + x^2) = 221 \]
Combining like terms:
\[ 2x^2 - 42x + 441 - 221 = 0 \]
Simplifying gives:
\[ 2x^2 - 42x + 220 = 0 \]
Dividing the entire equation by 2:
\[ x^2 - 21x + 110 = 0 \]
Now we can factor this quadratic:
\[ (x - 10)(x - 11) = 0 \]
Thus, we have:
\[ x = 10 \quad \text{or} \quad x = 11 \]
Using \( y = 21 - x \):
- If \( x = 10 \), then \( y = 11 \).
- If \( x = 11 \), then \( y = 10 \).

### Final Result:
The remaining two observations are **10 and 11**.