The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 1

Q: Wait, do I really need to find the individual values of $x$ and $y$ to get the answer?

That is the beauty of this problem! You absolutely do not. By using algebraic identities, we can find the difference directly without ever knowing the specific values of $x$ and $y$. It saves time and reduces the chance of calculation errors.

Q: Why does the variance formula have that minus $(\bar{x})^2$ term? It looks weird.

That term is the 'correction factor' that centers the variance around the mean. Without it, you'd be calculating the raw second moment, not the spread. Think of it as shifting your coordinate system to the mean.

Q: Is there a risk of getting a negative value for the variance?

Variance is a measure of spread, so it must always be non-negative. If you ever calculate a negative variance, check your arithmetic—you've likely swapped the mean and the sum of squares in the formula.

Question

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

Options:

(A) $2$

(B) $4$

(C) $3$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a dataset of seven observations: $2, 4, 10, 12, 14, x, y$. We are provided with the mean $\bar{x} = 8$ and the variance $\sigma^2 = 16$.

Our objective is to determine the absolute difference between the two missing values, denoted as $|x - y|$.

### The Power of the Mean

The mean is defined as the sum of all observations divided by the total count $n=7$. We express this as:

$$\bar{x} = \frac{\sum x_i}{n} = 8$$

Substituting our known values into the equation:

$$\frac{2 + 4 + 10 + 12 + 14 + x + y}{7} = 8$$

Summing the known constants yields $42 + x + y = 56$. This provides our first fundamental constraint:

$$x + y = 14$$

### The Variance—The Measure of Spread

To handle the variance efficiently, we utilize the computational formula:

$$	ext{Var} = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

Given $	ext{Var} = 16$ and $\bar{x} = 8$, we substitute these values:

$$16 = \frac{\sum x_i^2}{7} - 64$$

Rearranging the terms, we find:

$$\frac{\sum x_i^2}{7} = 80 \Rightarrow \sum x_i^2 = 560$$

### The Algebraic Bridge

First, we calculate the sum of the squares of the known observations:

$$2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 4 + 16 + 100 + 144 + 196 = 460$$

Since the total sum of squares is $560$, the sum of the squares of the unknowns must be:

$$x^2 + y^2 = 560 - 460 = 100$$

We now possess a system of two equations:
1) $x + y = 14$
2) $x^2 + y^2 = 100$

### Final Calculation

To find $|x - y|$, we utilize the algebraic identity $(x - y)^2 = x^2 + y^2 - 2xy$. First, we determine $2xy$ using the expansion of $(x + y)^2$:

$$(x + y)^2 = x^2 + y^2 + 2xy$$
$$14^2 = 100 + 2xy$$
$$196 = 100 + 2xy \Rightarrow 2xy = 96$$

Now, substitute these values back into the identity for the difference:

$$(x - y)^2 = 100 - 96 = 4$$

Taking the square root of both sides, we find the absolute difference:

$$|x - y| = \sqrt{4} = 2$$

The absolute difference between the two missing numbers is **2**.

Animated Solution for Mathematics - Statistics: The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

Select Answer:

Visualized Solution

Visualizing the Data

Defining the Unknowns $x$ and $y$

Applying the Mean Formula

Summing the Known Values

Finding the Sum of Unknowns

The Variance Formula

Setting up the Variance Equation

Calculating Sum of Squares

Finding Sum of Squares of Unknowns

Using Algebraic Identity

Calculating the Product Term

The Final Difference

Summary and Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of the Mean

The Variance—The Measure of Spread

The Algebraic Bridge

Final Calculation

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