Question

Consider $x_{1},x_{2},...,x_{n}$ observations such that $\sum_{i=1}^{n}{x_{i}}^{2}=500$ and $\sum_{i=1}^{n}x_{i}=50$. Then a minimum number of observations required is

• A. 25
• B. 5
• C. 10
• D. 15

Hint:

The Cauchy-Schwarz inequality for a vector of 1s and the vector $x$ provides the same result: $(\sum x_i)^2 \le n (\sum x_i^2)$. Plugging in the numbers: $2500 \le n(500) \implies n \ge 5$. It's a faster way to remember the relationship.

Answer 1

The Correct Option is B

This problem uses the fundamental statistical property that the variance of a real-valued set of data must be non-negative ($Var(X) \ge 0$).

Step 1: \color{redRecall the Formula for Variance
The variance of $n$ observations is given by:
$Var(X) = \frac{1}{n}\sum x_i^2 - (\frac{1}{n}\sum x_i)^2$.

Step 2: \color{redApply the Non-Negativity Condition
Since variance cannot be negative:
$\frac{1}{n}\sum x_i^2 - (\frac{1}{n}\sum x_i)^2 \ge 0$
$\frac{\sum x_i^2}{n} \ge \frac{(\sum x_i)^2}{n^2}$.

Step 3: \color{redSubstitute the Given Values
Substitute $\sum x_i^2 = 500$ and $\sum x_i = 50$:
$\frac{500}{n} \ge \frac{50^2}{n^2}$
$\frac{500}{n} \ge \frac{2500}{n^2}$.

Step 4: \color{redSolve the Inequality for n
Assuming $n > 0$, multiply both sides by $n^2$:
$500n \ge 2500$
$n \ge \frac{2500}{500}$
$n \ge 5$.

Step 5: \color{redConclusion
The calculation shows that for the given sums to be mathematically possible for real numbers, the number of observations $n$ must be at least 5.
Therefore, the minimum number is 5.
This corresponds to Option (2).