표본평균 표준편차

확률분포표에서 확률의 합은 $1$이므로 $\displaystyle a + b = \frac{8}{9}$ $\cdots$$\cdots$ ㉠ $\displaystyle {\mathrm{E}} \left( X \right) = \frac{7}{3}$이므로 $\displaystyle 1 \times a + 3 \times \frac{1}{9} + 5 \times b = \frac{7}{3}$, $a + 5 b = 2$ $\cdots$$\cdots$ ㉡ ㉠, ㉡을 연립하면 $\displaystyle a = \frac{11}{18}$, $\displaystyle b = \frac{5}{18}$ $\mathrm{V} \mathit{\left( X \right)}$$= \mathrm{E} \mathit{\left( X ^{2} \right)} - m ^{2}$ $\displaystyle = 1 \times \frac{11}{18} + 9 \times \frac{1}{9} + 25 \times \frac{5}{18} - \left( \frac{7}{3} \right) ^{2}$ $\displaystyle = \frac{28}{9}$ $\therefore$ $\displaystyle \sigma \left( X \right) = \sqrt{\frac{28}{9}} = \frac{2 \sqrt{7}}{3}$ $\displaystyle \sigma \left( 2 {\overline{X}} + 1 \right) \leq \frac{1}{3}$에서 $\displaystyle \sigma \left( \overline{X} \right) = \frac{\sigma \left( X \right)}{\sqrt{n}}$이므로 $\displaystyle 2 \sigma \left( {\overline{X}} \right) \leq \frac{1}{3}$, $\displaystyle 2 \times \frac{\sigma \left( X \right)}{\sqrt{n}} \leq \frac{1}{3}$ $\displaystyle 2 \times \frac{\frac{2 \sqrt{7}}{3}}{\sqrt{n}} \leq \frac{1}{3}$, $\displaystyle \sqrt{n} \geq 4 \sqrt{7}$ $\therefore$ $n \geq 112$ 따라서 조건을 만족하는 자연수 $n$의 최솟값은 $112$이다.