확률분포 표준편차

$\displaystyle c \sum\limits _{k = 1} ^{n} k = 1$에서 $\displaystyle c = \frac{2}{n ( n + 1 )}$ $X$의 표준편차가 $\displaystyle \sqrt{6}$이므로 $X$의 분산 $V ( X )$는 $\displaystyle V ( X ) = \sum\limits _{k = 1} ^{n} k ^{2} \cdot ( c k ) - \left( \sum\limits _{k = 1} ^{n} k \cdot c k \right) ^{2} = 6$에서 $\displaystyle c \sum\limits _{k = 1} ^{n} k ^{3} - \left( \sum\limits _{k = 1} ^{n} c k ^{2} \right) ^{2} = 6$ 위 식에 $\displaystyle c = \frac{2}{n ( n + 1 )}$를 대입하여 정리하면 $\displaystyle \frac{2}{n ( n + 1 )} \times \left\{ \frac{n ( n + 1 )}{2} \right\} ^{2} - \left\{ \frac{2}{n ( n + 1 )} \times \frac{n ( n + 1 ) ( 2 n + 1 )}{6} \right\} ^{2} = 6$ $\displaystyle \frac{n ( n + 1 )}{2} - \frac{( 2 n + 1 ) ^{2}}{9} = 6$, $n ^{2} + n - 110 = 0$ $( n + 11 ) ( n - 10 ) = 0$$\therefore n = 10$