정규분포 조건과 확률

(가)에서 $\mathrm{P} ( \mathit{X} \geq 64 ) = \mathrm{P} ( \mathit{X} \leq 56 )$이므로 $\displaystyle m = \frac{64 + 56}{2} = 60$ 이때, $\mathrm{V} ( \mathit{X} ) = \mathrm{E} ( \mathit{X} ^{2} ) - \left\{ \mathrm{E} ( \mathit{X} ) \right\} ^{2}$이므로 $\mathrm{V} ( \mathit{X} ) = 3616 - 60 ^{2} = 16$ $\therefore \sigma ( X ) = 4$ $\displaystyle \therefore \mathrm{P} ( \mathit{X} \leq 68 ) = \mathrm{P} \left( \mathit{Z} \leq \frac{68 - 60}{4} \right)$$= \mathrm{P} ( \mathit{Z} \leq 2 )$ $= 0.5 + \mathrm{P} ( 0 \leq \mathit{Z} \leq 2 )$$= 0.5 + 0.4772$ $( \because \mathrm{P} ( \mathit{0} \leq Z \leq 2 ) = \mathrm{P} ( \mathit{m} \leq X \leq m + 2 \sigma ) )$$= 0.9772$ [다른 풀이] (가)조건에서

그래프의 대칭의 중심은 $60$이므로 $m = 60$ (나)조건에서 $E ( X ^{2} ) = V ( X ) + \left\{ E ( X ) \right\} ^{2} = 6 ^{2} + 60 ^{2} = 3616$ $\therefore \sigma ^{2} = 16 , \sigma = 4$ $\mathrm{P} ( \mathit{X} \leq 68 ) = \mathrm{P} ( \mathit{X} \leq m + 2 \sigma )$ $= 0.5 + \mathrm{P} ( 0 \leq \mathit{X} \leq m + 2 \sigma ) = 0.9772$