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Step-by-Step Solution
Step 1: Represent the unknown observations
Let the remaining two unknown observations be $a$ and $b$.
Given that there are 6 observations in total, of which 4 are already known (2, 4, 5, and 7).
Hence:
$$a + b + 2 + 4 + 5 + 7 = 6 \times \text{(mean)}.$$
Step 2: Use the given mean to find the sum of observations
The mean of the 6 observations is 6.5, which means their sum must be
$$6 \times 6.5 = 39.$$
Therefore:
$$a + b + (2 + 4 + 5 + 7) = 39 \implies a + b + 18 = 39,$$
which simplifies to
$$a + b = 21.$$
Step 3: Use the variance formula
Variance $(\sigma^2)$ is defined as:
$$\sigma^2 = \text{(Mean of the squares of the observations)} \;-\; (\text{Mean of the observations})^2.$$
We are given the variance as 10.25. Thus:
$$10.25 = \frac{2^2 + 4^2 + 5^2 + 7^2 + a^2 + b^2}{6} \;-\; (6.5)^2.$$
Step 4: Compute the mean of the squares
First, compute $(6.5)^2$:
$$(6.5)^2 = 42.25.$$
So, if $M$ is the mean of the squares, we have
$$10.25 = M - 42.25 \implies M = 10.25 + 42.25 = 52.5.$$
Since $M$ is the average (mean) of the squares of the 6 observations,
$$\frac{2^2 + 4^2 + 5^2 + 7^2 + a^2 + b^2}{6} = 52.5.$$
Step 5: Write the equation for the sum of squares
Multiplying both sides by 6,
$$2^2 + 4^2 + 5^2 + 7^2 + a^2 + b^2 = 6 \times 52.5 = 315.$$
Now compute the known squares:
$$2^2 = 4,\quad 4^2 = 16,\quad 5^2 = 25,\quad 7^2 = 49.$$
So their sum is $4 + 16 + 25 + 49 = 94.$
Thus:
$$a^2 + b^2 + 94 = 315 \implies a^2 + b^2 = 221.$$
Step 6: Use the relationship $a + b = 21$
From Step 2, $a + b = 21.$ Let $b = 21 - a.$ Substitute into $a^2 + b^2 = 221$:
$$a^2 + (21 - a)^2 = 221.$$
Expanding:
$$a^2 + (441 - 42a + a^2) = 221.$$
Combine like terms:
$$2a^2 - 42a + 441 = 221.$$
Simplify further:
$$2a^2 - 42a + 220 = 0.$$
Divide everything by 2:
$$a^2 - 21a + 110 = 0.$$
Step 7: Solve the quadratic equation
We have:
$$(a - 10)(a - 11) = 0.$$
Hence:
$$a = 10 \quad \text{or} \quad a = 11.$$
Since $b = 21 - a$,
- If $a = 10$, then $b = 21 - 10 = 11.$
- If $a = 11$, then $b = 21 - 11 = 10.$
Step 8: Conclude the remaining observations
The remaining two distinct observations are $10$ and $11.$ Thus, the correct choice is (10, 11).
