studying set I of population

Question

• let x be the mean and € the standard deviation of the statistics ×1,,,,,,, xn.

• let I=(x-3€, x+3×€) and the number of items not in I is k.

• prove that n greater or equal to 9k.
• prove that the percentage of items in I more than 80percent.

Answer 1

Note that for $x_i \not\in I$ we have $|x_i -x| \ge 3 \sigma$. Hence, \begin{align*} \sigma^2 &= \frac 1n \sum_{i=1}^n (x_i- x)^2\\ &\ge \frac 1n \sum_{x_i \not\in I} (x_i - x)^2\\ &\ge \frac 1n k \cdot 9\sigma^2\\ \iff n &\ge 9k \end{align*} The percentage of items in $I$, is by definition, $$ \frac {n-k}n \cdot 100\% = 100\% -\frac kn \cdot 100\% \ge 100\% - \frac{100}9 \% \approx 88.89\% $$