Key Concepts and Formulas

1. Mean ($\bar{x}$): The mean is the arithmetic average of a set of $n$ observations, $x_1, x_2, \dots, x_n$. It represents the central tendency of the data. $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$ This formula allows us to find the sum of all observations if the mean and number of observations are known.

2. Variance ($\sigma^2$): Variance quantifies the spread or dispersion of data points around their mean. A larger variance indicates greater spread. The most computationally efficient formula for variance, especially when dealing with sums of squares, is: $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$ Here, $\sum_{i=1}^{n} x_i^2$ is the sum of the squares of all observations. This formula helps establish a relationship involving the squares of the unknown observations.

Step-by-Step Solution

Step 1: Understand the Problem and Define Variables

We are given information about a set of 6 distinct observations and need to find two missing observations.

• Total number of observations, $n = 6$.
• Mean of the observations, $\bar{x} = 6.5$.
• Variance of the observations, $\sigma^2 = 10.25$.
• Four known observations: $2, 4, 5, 7$.

Let the two unknown distinct observations be $a$ and $b$. Our goal is to determine the values of $a$ and $b$.

Step 2: Formulate the First Equation using the Mean

We use the mean formula to establish a relationship between the sum of the unknown observations ($a+b$) and the known values.

The mean formula is $\bar{x} = \frac{\sum x_i}{n}$. We know $\bar{x}=6.5$ and $n=6$. The sum of all observations ($\sum x_i$) is the sum of the four known observations plus $a$ and $b$. $\sum x_i = 2 + 4 + 5 + 7 + a + b$ $\sum x_i = 18 + a + b$ Now, substitute these into the mean formula: $6.5 = \frac{18 + a + b}{6}$ To find the sum $18+a+b$, multiply both sides by $6$: $6.5 \times 6 = 18 + a + b$ $39 = 18 + a + b$ Subtract $18$ from both sides to find the sum of the unknown observations: $a + b = 39 - 18$ $\boldsymbol{a + b = 21 \quad \text{(Equation 1)}}$ This is our first key equation, relating the sum of the two unknown observations.

Step 3: Formulate the Second Equation using the Variance

Next, we use the variance formula to establish a second, independent relationship between the unknown observations, specifically involving their squares.

The computational formula for variance is $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. We know $\sigma^2 = 10.25$ and $\bar{x} = 6.5$. First, let's calculate the sum of squares of the known observations: $2^2 + 4^2 + 5^2 + 7^2 = 4 + 16 + 25 + 49 = 94$ The sum of squares of all observations ($\sum x_i^2$) will be the sum of squares of known observations plus $a^2$ and $b^2$: $\sum x_i^2 = 94 + a^2 + b^2$ Now, substitute these values, along with $n=6$ and $\bar{x}=6.5$, into the variance formula: $10.25 = \frac{94 + a^2 + b^2}{6} - (6.5)^2$ Calculate $(6.5)^2$: $(6.5)^2 = 42.25$ Substitute this value back into the equation: $10.25 = \frac{94 + a^2 + b^2}{6} - 42.25$ This equation now relates $a^2+b^2$ to the given statistical measures.

Step 4: Solve the Variance Equation for $a^2+b^2$

We will algebraically manipulate the equation from Step 3 to isolate $a^2+b^2$. Add $42.25$ to both sides of the equation: $10.25 + 42.25 = \frac{94 + a^2 + b^2}{6}$ $52.5 = \frac{94 + a^2 + b^2}{6}$ Multiply both sides by $6$: $52.5 \times 6 = 94 + a^2 + b^2$ $315 = 94 + a^2 + b^2$ Subtract $94$ from both sides to isolate $a^2+b^2$: $a^2 + b^2 = 315 - 94$ $\boldsymbol{a^2 + b^2 = 221 \quad \text{(Equation 2)}}$ We now have two algebraic equations relating $a$ and $b$.

Step 5: Solve the System of Equations for $a$ and $b$

We have the following system of equations:

1. $a + b = 21$
2. $a^2 + b^2 = 221$

From Equation 1, express $b$ in terms of $a$: $b = 21 - a$ Substitute this expression for $b$ into Equation 2: $a^2 + (21 - a)^2 = 221$ Expand $(21 - a)^2$ using the identity $(X-Y)^2 = X^2 - 2XY + Y^2$: $a^2 + (21^2 - 2 \times 21 \times a + a^2) = 221$ $a^2 + (441 - 42a + a^2) = 221$ Combine like terms: $2a^2 - 42a + 441 = 221$ Move all terms to one side to form a standard quadratic equation: $2a^2 - 42a + 441 - 221 = 0$ $2a^2 - 42a + 220 = 0$ Divide the entire equation by $2$ to simplify: $a^2 - 21a + 110 = 0$ Factor this quadratic equation. We need two numbers that multiply to $110$ and add up to $-21$. These numbers are $-10$ and $-11$. $(a - 10)(a - 11) = 0$ This gives two possible values for $a$: $a = 10 \quad \text{or} \quad a = 11$ Now, find the corresponding values for $b$ using $b = 21 - a$:

• If $a=10$, then $b = 21 - 10 = 11$.
• If $a=11$, then $b = 21 - 11 = 10$. In both cases, the two remaining observations are $10$ and $11$.

Step 6: Verify Distinctness and Final Observations

The problem states that the 6 observations are distinct. The given observations are: $2, 4, 5, 7$. The calculated observations are: $10, 11$. Combining all observations: $\{2, 4, 5, 7, 10, 11\}$. All these numbers are unique, satisfying the distinctness condition.

Therefore, the remaining two observations are $10$ and $11$.

Common Mistakes & Tips

• Arithmetic Precision: Be extremely careful with basic calculations, especially squaring numbers and multiplications (e.g., $6.5 \times 6 = 39$, $21^2 = 441$). Errors here are common.
• Variance Formula Choice: Always prefer the computational formula for variance, $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$, as it generally simplifies calculations and avoids working with deviations from the mean, which can introduce fractions or decimals.
• Checking Conditions: Don't forget to check all conditions mentioned in the problem, such as "distinct observations." This ensures the validity of your solution.
• Option Checking (for MCQs): Once you establish the system of equations ($a+b=21$ and $a^2+b^2=221$), you can often quickly check the given options. For option (A) (10, 11): $10+11=21$ (correct) and $10^2+11^2 = 100+121=221$ (correct). This rapid verification can save time.

Summary

We systematically solved this problem by first using the definition of the mean to find the sum of the two unknown observations. Then, we used the computational formula for variance to establish a relationship involving the sum of their squares. This led to a system of two algebraic equations, which was solved by substitution, resulting in a quadratic equation. Solving the quadratic equation yielded the values for the two missing observations, which were then verified against the distinctness condition.

The final answer is $\boxed{\text{10, 11}}$ which corresponds to option (A).