If x_1, x_2, ldots, x_n are n observations su | vedclass

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(D) We know that the variance of a set of observations is always non-negative,i.e.,$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^

Vedclass · Accepted Answer

$16$

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(D) We know that the variance of a set of observations is always non-negative,i.e.,$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \geq 0$. Given $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$. Substituting these values into the inequality: $\frac{400}{n} - \left(\frac{80}{n}\right)^2 \geq 0$ $\frac{400}{n} - \frac{6400}{n^2} \geq 0$ Multiplying by $n^2$ (since $n > 0$): $400n - 6400 \geq 0$ $400n \geq 6400$ $n \geq \frac{6400}{400}$ $n \geq 16$. Thus,the least value of $n$ is $16$.

If $x_1, x_2, \ldots, x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 80$,then the least value of $n$ is

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