Let the mean and the variance of 20 observations x₁, x₂,...x₂₀ be 15 and 9, respectively. For α ∈ R,

Q: Why do we need to calculate $\sum x_i^2$? Can't we just use the variance formula directly?

Great question! The variance formula gives us the relationship between the mean and the sum of squares. Without $\sum x_i^2$, we cannot expand the term $\sum (x_i + \alpha)^2$ because it contains the $x_i^2$ term. Think of $\sum x_i^2$ as the 'energy' of the dataset that we must extract first.

Q: Wait, why is the sum of $\alpha^2$ equal to $20\alpha^2$?

Remember that $\alpha$ is a constant, not a variable like $x_i$. When you sum a constant $\alpha^2$ for $n$ observations, you are essentially adding $\alpha^2$ to itself $n$ times. Since $n=20$, the sum is simply $20 \times \alpha^2$.

Q: How do I know which $\alpha$ is the maximum? Is it $-2$ or $-28$?

This is a classic trap! On the number line, $-2$ is to the right of $-28$. Therefore, $-2$ is greater than $-28$. Always visualize the number line when comparing negative values to avoid this common mistake.

Question

Let the mean and the variance of 20 observations $x_1, x_2, \dots, x_{20}$ be 15 and 9, respectively. For $\alpha \in \mathbb{R}$, if the mean of $(x_1 + \alpha)^2, (x_2 + \alpha)^2, \dots, (x_{20} + \alpha)^2$ is 178, then the square of the maximum value of $\alpha$ is equal to

Sigma Percentile · Accepted Answer

### The Hidden Architecture of Data

Statistics is not just a collection of numbers; it is a language of patterns. When you look at a set of observations $x_1, x_2, \dots, x_{20}$, you are looking at a system with its own unique DNA—its mean and its variance.

In this problem, we are given a mean of $15$ and a variance of $9$. We are asked to transform this data by adding a constant $\alpha$ to each term and squaring the result. Beneath this transformation lies a beautiful algebraic structure waiting to be revealed.

### Phase 1: Unlocking the DNA

Before we can tackle the transformation, we must understand the original dataset. We know the mean $\bar{x} = 15$ and the number of observations $n = 20$.

The definition of the mean is $\bar{x} = \frac{\sum x_i}{n}$. By simple multiplication, we find the sum of all observations:
$$\sum x_i = 15 	imes 20 = 300$$

Next, we look at the variance. The variance $\sigma^2$ is defined as $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. This formula is the bridge between the average value of the data and the average of the squares.

Substituting our known values, we get:
$$9 = \frac{\sum x_i^2}{20} - (15)^2$$

Solving for the sum of squares, we find $\frac{\sum x_i^2}{20} = 9 + 225 = 234$, which leads us to:
$$\sum x_i^2 = 4680$$

### Phase 2: The Algebraic Transformation

Now, we introduce the transformation: $(x_i + \alpha)^2$. The problem states that the mean of these new terms is $178$.

Mathematically, this is expressed as:
$$\frac{\sum (x_i + \alpha)^2}{20} = 178$$

Multiplying by $20$, we get the sum of the new terms:
$$\sum (x_i + \alpha)^2 = 3560$$

We expand the expression using the binomial identity $(a+b)^2 = a^2 + 2ab + b^2$. The summation becomes:
$$\sum (x_i^2 + 2\alpha x_i + \alpha^2) = 3560$$

By the linearity of summation, we distribute this:
$$\sum x_i^2 + 2\alpha \sum x_i + \sum \alpha^2 = 3560$$

Since $\alpha$ is a constant, $\sum \alpha^2$ is simply $20\alpha^2$. We now have a clean, manageable equation:
$$\sum x_i^2 + 2\alpha \sum x_i + 20\alpha^2 = 3560$$

### Phase 3: The Quadratic Resolution

We substitute the values we found earlier: $\sum x_i^2 = 4680$ and $\sum x_i = 300$. The equation becomes:
$$4680 + 2\alpha(300) + 20\alpha^2 = 3560$$

Simplifying this, we get:
$$20\alpha^2 + 600\alpha + 1120 = 0$$

Dividing the entire equation by $20$ yields the elegant quadratic:
$$\alpha^2 + 30\alpha + 56 = 0$$

Factorizing this is straightforward. We look for two numbers that multiply to $56$ and add to $30$, which are $28$ and $2$. Thus:
$$(\alpha + 28)(\alpha + 2) = 0$$

This gives us two possible values for $\alpha$: $-28$ and $-2$. The question asks for the square of the **maximum** value of $\alpha$.

Comparing $-28$ and $-2$, it is clear that $-2$ is the maximum. Finally, the result is:
$$(-2)^2 = \mathbf{4}$$

Enter Numerical Value:

Visualized Solution

Identify Given Parameters

Calculate Sum of Observations $\sum x_{i}$

Calculate Sum of Squares $\sum x_{i}^{2}$

Analyze the New Mean Condition

Expand the Algebraic Expression

Substitute Known Values

Form the Quadratic Equation

Solve for $α$

Find the Final Answer

The Sigma Insight: Measures of Dispersion

The Hidden Architecture of Data

Phase 1

Phase 2

Phase 3

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