Let x₁, x₂, ..., xₙ be n observations such that Σxᵢ² = 400 and Σxᵢ = 80. Then the possible value of

Q: Wait, why is variance always non-negative? Can't it be negative if the data is weird?

That is a fantastic question! Think about the formula for variance: $\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$. Because we are squaring the difference $(x_i - \bar{x})$, every single term in that sum is either positive or zero. Adding up non-negative numbers will always result in a non-negative sum. Therefore, variance is fundamentally a measure of 'spread' and can never be negative. If you ever calculate a negative variance, it is a sign that there is a calculation error!

Q: Why did we multiply by $n^2$ without flipping the inequality sign?

Great catch on the inequality rules! When we multiply an inequality by a variable, we must be careful. If the variable is negative, the sign flips. However, here $n$ represents the number of observations, which must be a positive integer ($n > 0$). Since $n^2$ is always positive for any $n > 0$, multiplying both sides by $n^2$ is perfectly safe and does not require us to flip the inequality sign.

Q: Is there another way to solve this besides using variance?

Yes, you can use the Cauchy-Schwarz inequality! The Cauchy-Schwarz inequality states that $(\sum a_i b_i)^2 \le (\sum a_i^2)(\sum b_i^2)$. If you let $a_i = x_i$ and $b_i = 1$, you get $(\sum x_i)^2 \le (\sum x_i^2)(\sum 1^2)$. Substituting the values gives $(80)^2 \le (400)(n)$, which simplifies to $6400 \le 400n$, leading to $n \ge 16$. Both methods lead to the same beautiful result!

Question

Let $x_1, x_2, \dots, x_n$ be n observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then the possible value of n among the following is

Options:

(A) 15

(B) 18

(C) 9

(D) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given $n$ observations, $x_1, x_2, \dots, x_n$, with two specific constraints: the sum of their squares is $\sum x_i^2 = 400$ and the sum of the observations is $\sum x_i = 80$.

Our objective is to determine the possible value of $n$ from a given set of options. This problem relies on understanding the fundamental constraints that govern the distribution of any set of real numbers.

### The Variance Insight

To solve this, we utilize the concept of variance, denoted as $\sigma^2$. Variance measures the spread of data points around their mean and is defined by the formula:

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

Because variance represents the average of squared deviations from the mean, it is a physical impossibility for the spread of data to be negative. Therefore, we must satisfy the fundamental constraint:

$$\sigma^2 \ge 0$$

### The Algebraic Dance

Using the constraint $\sigma^2 \ge 0$, we can rewrite the inequality as:

$$\frac{\sum x_i^2}{n} \ge \left(\frac{\sum x_i}{n}\right)^2$$

Substituting our known values $\sum x_i^2 = 400$ and $\sum x_i = 80$ into the inequality, we obtain:

$$\frac{400}{n} \ge \left(\frac{80}{n}\right)^2$$

Expanding the right side of the expression yields:

$$\frac{400}{n} \ge \frac{6400}{n^2}$$

Since $n$ represents the number of observations, it must be a positive integer ($n > 0$). We can safely multiply both sides by $n^2$ without reversing the inequality sign:

$$400n \ge 6400$$

Dividing both sides by 400, we arrive at the final constraint:

$$n \ge 16$$

### The Final Verdict

We have determined that for this dataset to exist, the number of observations $n$ must be at least 16.

Comparing this to the provided options (15, 18, 9, and 12), we observe that 9, 12, and 15 are all less than 16, which violates our derived constraint.

Consequently, only **18** satisfies the condition $n \ge 16$. By respecting the non-negative nature of variance, we have successfully identified the only valid possibility.

Animated Solution for Mathematics - Statistics: Let $x_{1}, x_{2}, \dots, x_{n}$ be n observations such that $\sum x_{i}^{2} = 400$ and $\sum x_{i} = 80$ . Then the possible value of n among the following is

Select Answer:

Visualized Solution

Identify Given Data

The Variance Constraint

Substitution of Values

Algebraic Simplification

Solving for $n$

Number Line Visualization

Final Conclusion

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Variance Insight

The Algebraic Dance

The Final Verdict

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