표준편차와 평균 같음

평균은 $\displaystyle 0 \times \frac{1}{10} + 1 \times \frac{1}{2} + a \times \frac{2}{5} = \frac{2}{5} a + \frac{1}{2} = \mathrm{E} ( X ) = \mathit{m}$ 제곱의 평균은 $\displaystyle 0 ^{2} \times \frac{1}{10} + 1 ^{2} \times \frac{1}{2} + a ^{2} \times \frac{2}{5} = \mathrm{E} ( X ^{2} ) = \frac{\mathit{2}}{5} a ^{2} + \frac{1}{2}$ $\sigma ( \mathrm{X} ) = \mathrm{E} ( X )$이므로 $\mathrm{V} ( X ) = E ( X ) ^{2} = E ( X ^{2} ) - E ( X ) ^{2} , 2 E ( X ) ^{2} = E ( X ^{2} )$ $\displaystyle \frac{2}{5} a ^{2} + \frac{1}{2} = 2 \times \left( \frac{2}{5} a + \frac{1}{2} \right) ^{2} = \frac{8}{25} a ^{2} + \frac{4}{5} a + \frac{1}{2}$ $\displaystyle \frac{2}{25} a ^{2} - \frac{4}{5} a = \frac{2}{25} \left( a ^{2} - 10 a ) = 0 \right.$, $a = 10$ $\displaystyle \mathrm{E} ( X ) = \frac{2}{5} \times 10 + \frac{1}{2} = \frac{9}{2} , E ( X ^{2} ) = \frac{2}{5} \times 100 + \frac{1}{2} = \frac{81}{2}$ $\displaystyle \therefore \mathrm{E} ( X ^{2} ) + E ( X ) = \frac{9}{2} + \frac{81}{2} = 45$