두 정규분포 확률 비교

[출제의도] 정규분포의 표준화 성질 추론하기 확률변수 $X , Y$는 각각 정규분포$N \mathit{(} m , ( 2 \sigma ) ^{2} )$, $\mathrm{N} \mathit{(} 2 m , \sigma ^{2} )$을 따른다. ㄱ. $\displaystyle {\mathrm{P}} ( X \leq 0 ) = {\mathrm{P}} \left( Z \leq - \frac{m}{2 \sigma} \right)$ $\displaystyle {\mathrm{P}} \left( Y \geq \frac{5}{2} m \right) = {\mathrm{P}} \left( Z \geq \frac{m}{2 \sigma} \right)$ (참) ㄴ. $\displaystyle {\mathrm{P}} ( m \leq X \leq 2 m ) = {\mathrm{P}} \left( 0 \leq Z \leq \frac{m}{2 \sigma} \right)$ $\displaystyle \frac{1}{2} {\mathrm{P}} ( 2 m \leq Y \leq 3 m ) = \frac{1}{2} {\mathrm{P}} \left( 0 \leq Z \leq \frac{m}{\sigma} \right)$ $\displaystyle {\mathrm{P}} ( m \leq X \leq 2 m ) \ne \frac{1}{2} {\mathrm{P}} ( 2 m \leq Y \leq 3 m )$ (거짓) ㄷ. $\displaystyle {\mathrm{P}} \left( Z \geq \frac{a - m}{2 \sigma} \right) + {\mathrm{P}} \left( Z \leq \frac{b - 2 m}{\sigma} \right) = 1$이 되려면 $\displaystyle \frac{a - m}{2 \sigma} = \frac{b - 2 m}{\sigma}$ 이다. $\displaystyle \therefore b = \frac{a + 3 m}{2}$ (참)