Key Concepts and Formulas

• Variance ($\sigma^2$): A statistical measure that quantifies the spread or dispersion of a set of data points around their mean. For $n$ observations $x_1, x_2, \dots, x_n$, the computational formula for variance is: $\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$
• Non-Negativity of Variance: A fundamental property of variance is that it can never be negative. Since variance is calculated from squared differences, and squares of real numbers are always non-negative, their sum and average must also be non-negative. Therefore: $\sigma^2 \ge 0$
• Cauchy-Schwarz Inequality (Alternative Insight): For any real numbers $a_1, \dots, a_n$ and $b_1, \dots, b_n$, the inequality states: $\left(\sum_{i=1}^n a_i b_i\right)^2 \le \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right)$ This inequality can be used to derive the same constraint on $n$.

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Step-by-Step Solution

We are given $n$ observations $x_1, x_2, \dots, x_n$ with the following sums:

1. Sum of squares of observations: $\sum x_i^2 = 400$
2. Sum of observations: $\sum x_i = 80$

Our goal is to find a possible value of $n$ from the given options.

Step 1: Apply the Principle of Non-Negativity of Variance

The most crucial concept here is that the variance of any set of real numbers must be non-negative. We will use the formula for variance and set it greater than or equal to zero.

• Why this step? The variance formula connects the sum of observations, the sum of squares of observations, and the number of observations ($n$). By applying the non-negativity constraint ($\sigma^2 \ge 0$), we can form an inequality involving $n$ that will help us find its possible values.

$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \ge 0$

Step 2: Substitute the Given Values into the Inequality

Now, we will substitute the provided numerical values for $\sum x_i^2$ and $\sum x_i$ into the inequality derived in Step 1.

• Given values:

  • $\sum x_i^2 = 400$
  • $\sum x_i = 80$

• First, calculate the term $\left(\sum x_i\right)^2$: $\left(\sum x_i\right)^2 = (80)^2 = 6400$

• Substitute these into the variance inequality: $\frac{400}{n} - \frac{6400}{n^2} \ge 0$

  • Explanation: The term $\left(\frac{\sum x_i}{n}\right)^2$ correctly expands to $\frac{(\sum x_i)^2}{n^2}$, which gives $\frac{6400}{n^2}$. It's important not to confuse this with $\frac{\sum x_i^2}{n^2}$.

Step 3: Solve the Inequality for $n$}

Our next task is to algebraically manipulate the inequality $\frac{400}{n} - \frac{6400}{n^2} \ge 0$ to find the possible range for $n$.

• Find a Common Denominator: To combine the fractions, we use $n^2$ as the common denominator. We multiply the first term $\frac{400}{n}$ by $\frac{n}{n}$: $\frac{400n}{n^2} - \frac{6400}{n^2} \ge 0$

• Combine the Terms: Now, we can combine the numerators over the common denominator: $\frac{400n - 6400}{n^2} \ge 0$

• Analyze the Denominator: Since $n$ represents the number of observations, it must be a positive integer ($n \ge 1$). This means $n^2$ will always be strictly positive ($n^2 > 0$).

• Determine the Sign of the Numerator: For the entire fraction $\frac{400n - 6400}{n^2}$ to be greater than or equal to zero, and knowing that the denominator $n^2$ is positive, the numerator $(400n - 6400)$ must also be greater than or equal to zero. $400n - 6400 \ge 0$

• Isolate $n$: Add $6400$ to both sides of the inequality: $400n \ge 6400$ Divide both sides by $400$. Since $400$ is a positive number, the direction of the inequality remains unchanged: $n \ge \frac{6400}{400}$ $n \ge 16$

Step 4: Interpret the Result and Choose the Correct Option

The inequality $n \ge 16$ tells us that the number of observations, $n$, must be an integer greater than or equal to 16.

Let's examine the given options: (A) 18 (B) 15 (C) 12 (D) 9

Out of these options, only $n = 18$ satisfies the condition $n \ge 16$. The other options (15, 12, 9) are all less than 16, which would imply a negative variance, an impossible scenario.

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Common Mistakes & Tips

• Don't Forget Non-Negativity: The principle that variance ($\sigma^2$) must be $\ge 0$ is the cornerstone of this problem. Always remember this fundamental statistical property.
• Correct Squaring of the Mean: A frequent error is to incorrectly square the mean term. Remember that $\left(\frac{\sum x_i}{n}\right)^2$ means squaring the entire mean, resulting in $\frac{(\sum x_i)^2}{n^2}$, not $\frac{\sum x_i^2}{n^2}$ (which is the mean of squares).
• Nature of $n$: Always keep in mind that $n$ represents the number of observations, so it must be a positive integer ($n \in \{1, 2, 3, \dots\}$). This allows us to confidently state that $n > 0$ and $n^2 > 0$, simplifying the inequality solving process.
• Cauchy-Schwarz as an Alternative: As shown in the "Key Concepts" section, the problem can also be solved elegantly using the Cauchy-Schwarz inequality by setting $a_i = x_i$ and $b_i = 1$. This directly leads to $(\sum x_i)^2 \le n (\sum x_i^2)$, which upon substituting values gives $80^2 \le n \cdot 400 \Rightarrow 6400 \le 400n \Rightarrow n \ge 16$. This provides a quick verification or an alternative solution path for those familiar with it.

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Summary

This problem demonstrates how a fundamental statistical principle – the non-negativity of variance – can be used to deduce constraints on unknown parameters. By setting the variance formula greater than or equal to zero and substituting the given sums of observations and squares of observations, we established an inequality for $n$. Solving this inequality yielded $n \ge 16$. Comparing this result with the provided options, we found that only $n=18$ is a possible value.

The final answer is $\boxed{\text{18}}$ which corresponds to option (A).