이항분포 분산 최솟값

$\mathrm{X}$가 이항분포 $\displaystyle \mathrm{B} \left( 120 , \frac{1}{121} \right)$을 따르므로 $\displaystyle m = \mathrm{E} ( \mathrm{X} ) = \sum\limits _{{k} = 0} ^{120} k \cdot \mathrm{P} ( \mathrm{X} = k ) = 120 \times \frac{1}{121} = \frac{120}{121}$ $\displaystyle \mathrm{V} ( \mathrm{X} ) = \sum\limits _{{k} = 0} ^{120} k ^{2} \mathrm{P} ( \mathrm{X} = k ) - m ^{2} = 120 \times \frac{1}{121} \times \frac{120}{121} = \frac{120 ^{2}}{121 ^{2}}$ $\displaystyle \begin{aligned} f ( x ) & = \sum\limits _{k = 0} ^{120} ( x ^{2} - 2 ax k + a ^{2} k ^{2} ) \mathrm{P} ( \mathrm{X} = k ) \end{aligned}$ $\displaystyle = x ^{2} - 2 ax \sum\limits _{k = 0} ^{120} k \cdot \mathrm{P} ( \mathrm{X} = k ) + a ^{2} \sum\limits _{k = 0} ^{120} k ^{2} \cdot \mathrm{P} ( \mathrm{X} = k )$ $\displaystyle = x ^{2} - 2 am x + a ^{2} \sum\limits _{k = 0} ^{120} k ^{2} \cdot \mathrm{P} ( \mathrm{X} = k )$ $\displaystyle = ( x - am ) ^{2} + a ^{2} \left( \sum\limits _{k = 0} ^{120} k ^{2} \cdot \mathrm{P} ( \mathrm{X} = k ) - m ^{2} \right)$ $\displaystyle = ( x - am ) ^{2} + \frac{120 ^{2}}{121 ^{2}} a ^{2}$ 함수$f ( x )$는 $x = am$일 때, 최솟값 $\displaystyle \frac{120 ^{2}}{121 ^{2}} a ^{2}$ 이다. $\displaystyle \therefore \frac{120 ^{2}}{121 ^{2}} a ^{2} = 1$에서 $\displaystyle a = \frac{121}{120} ( a > 0 )$ $\therefore 120 a = 121$