확률분포표에서 표본평균의 분산

크기가 $n$인 표본을 복원추출하였을 때, 표본평균 $\displaystyle \overline{X}$의 분산은 $\displaystyle \mathrm{V} \mathit{(} \overline{X} ) = \frac{\mathrm{V} ( \mathit{X} )}{n}$이고, 문제에서 $\displaystyle \mathrm{V} \mathit{(} \overline{X} ) = \frac{7}{12}$이므로 $\displaystyle \frac{\mathrm{V} ( \mathit{X} )}{3} = \frac{17}{12}$ $\displaystyle \therefore \mathrm{V} ( \mathit{X} ) = \frac{17}{4}$ 한편, $\mathrm{V} ( \mathit{X} ) = \mathrm{E} ( \mathit{X} ^{2} ) - \left\{ \mathrm{E} ( \mathit{X} ) \right\} ^{2}$에서 $\mathrm{E} ( \mathit{X} ^{2} )$$\displaystyle = 0 ^{2} \times \frac{1}{3} + 3 ^{2} \times a + 6 ^{2} \times \left( \frac{2}{3} - a \right)$ $= 9 a + 24 - 36 a = 24 - 27 a$ $\mathrm{E} ( \mathit{X} )$$\displaystyle = 0 \times \frac{1}{3} + 3 \times a + 6 \times \left( \frac{2}{3} - a \right)$$= 3 a + 4 - 6 a = 4 - 3 a$ 이므로 $\displaystyle \frac{17}{4} = ( 24 - 27 a ) - ( 4 - 3 a ) ^{2}$, $\displaystyle \frac{17}{4} = 24 - 27 a - ( 16 - 24 a + 9 a ^{2} )$ $\displaystyle \frac{17}{4} = - 9 a ^{2} - 3 a + 8$ $36 a ^{2} + 12 a - 15 = 0$, $12 a ^{2} + 4 a - 5 = 0$, $( 6 a + 5 ) ( 2 a - 1 ) = 0$ $\displaystyle \therefore a = \frac{1}{2} ( \because a \geq 0 )$