Statistics
Variance and Standard Deviation
Grade None
Question:
<p>Given \(\sum x_i^2 = 400\), \(\sum x_i = 80\). We know that mean of squares \(\geq\) square of mean, i.e., \(\dfrac{\sum x_i^2}{n} \geq \left(\dfrac{\sum x_i}{n}\right)^2\). What is the minimum value of \(n\)?</p>
<p>14</p>
<p>16</p>
<p>18</p>
<p>20</p>
Step-by-Step Solution
Key Concept: Use the inequality relationship between mean of squares and square of mean to find when equality holds, then apply the constraint that n must be a positive integer to find the minimum value.
<p><strong>Step 1:</strong> Write the given inequality with our values:</p><p>$$\frac{\sum x_i^2}{n} \geq \left(\frac{\sum x_i}{n}\right)^2$$</p><p><strong>Step 2:</strong> Substitute $\sum x_i^2 = 400$ and $\sum x_i = 80$:</p><p>$$\frac{400}{n} \geq \left(\frac{80}{n}\right)^2$$</p><p><strong>Step 3:</strong> Simplify the right side:</p><p>$$\frac{400}{n} \geq \frac{6400}{n^2}$$</p><p><strong>Step 4:</strong> Multiply both sides by $n^2$ (positive):</p><p>$$400n \geq 6400$$</p><p><strong>Step 5:</strong> Divide by 400:</p><p>$$n \geq 16$$</p><p><strong>Step 6:</strong> Verify: When $n = 16$, mean = $\frac{80}{16} = 5$ and mean of squares = $\frac{400}{16} = 25 = 5^2$ ✓</p><p>This occurs when all $x_i = 5$ (equality condition of Cauchy-Schwarz).</p><p>∴ Answer: <strong>C (n = 16)</strong></p>
Correct Answer: C