두 정규분포 조건

[출제의도] 정규분포를 표준화하여 확률을 구할 수 있는가를 묻는 문제이다. $\displaystyle \mathrm{P} \mathit{(} X \geq x ) = \mathrm{P} \mathit{\left( \frac{X - m}{\sigma} \geq \frac{x - m}{\sigma} \right)} = \mathrm{P} \left( \mathit{Z} \geq \frac{x - m}{\sigma} \right)$, $\displaystyle \mathrm{P} \mathit{(} Y \geq y ) = \mathrm{P} \mathit{\left( \frac{Y - am}{\sqrt{b} \sigma} \geq \frac{y - am}{\sqrt{b} \sigma} \right)} = \mathrm{P} \left( \mathit{Z} \geq \frac{y - am}{\sqrt{b} \sigma} \right)$이므로 (가)에서 $\displaystyle - \frac{m}{\sigma} = - \frac{am}{\sqrt{b} \sigma}$ $\therefore$ $\displaystyle a = \sqrt{b}$$\cdots \cdots$① (나)에서 $\mathrm{1} = P \mathit{(} X \leq 1 ) + \mathrm{P} \mathit{(} Y \geq 2 )$ $\displaystyle = \mathrm{P} \mathit{\left( \frac{X - m}{\sigma} \leq \frac{1 - m}{\sigma} \right)} + \mathrm{P} \mathit{\left( \frac{Y - am}{\sqrt{b} \sigma} \geq \frac{2 - am}{\sqrt{b} \sigma} \right)}$ $\displaystyle = \mathrm{P} \mathit{\left( Z \leq \frac{1 - m}{\sigma} \right)} + \mathrm{P} \mathit{\left( Z \geq \frac{2 - am}{\sqrt{b} \sigma} \right)}$ 이므로 $\displaystyle \frac{1 - m}{\sigma} = \frac{2 - am}{\sqrt{b} \sigma}$$\cdots \cdots$ ② ①, ②를 연립하여 풀면, $a = 2 , b = 4$이다.