정규분포 선형변환

[출제의도] 정규분포의 성질 이해하기 조건 (가)에서 $Y$$= 3 X - a$이므로 $\mathrm{E} \mathit{\left( Y \right)}$$= \mathrm{E} \mathit{\left( 3 X - a \right)}$$= 3 \mathrm{E} \mathit{\left( X \right)} - a$$= 3 m - a$ $m$$= 3 m - a$에서 $a$$= 2 m$ $\sigma \mathrm{\left( \mathit{Y} \right)}$$= \sigma \mathrm{\left( \mathit{3} X - \mathit{a} \right)}$$= 3 \sigma \mathrm{\left( \mathit{X} \right)}$$= 6$ $\mathrm{P} \mathit{\left( X \leq 4 \right)}$$\displaystyle = \mathrm{P} \mathit{\left( Z \leq \frac{4 - m}{2} \right)}$ $\mathrm{P} \mathit{\left( Y \geq a \right)}$$= \mathrm{P} \mathit{\left( Y \geq 2 m \right)}$$\displaystyle = \mathrm{P} \mathit{\left( Z \geq \frac{m}{6} \right)}$ 조건 (나)에 의하여 $\displaystyle \frac{4 - m}{2}$$\displaystyle = \mathit{-} \frac{m}{6}$에서 $m$$= 6$ 그러므로 확률변수 $Y$는 정규분포 $\mathrm{N} \mathit{\left( 6 , 6 ^{2} \right)}$을 따른다. 따라서 $\mathrm{P} \mathit{\left( Y \geq 9 \right)}$$\displaystyle = \mathrm{P} \mathit{\left( Z \geq \frac{9 - 6}{6} \right)}$$= \mathrm{P} \mathit{\left( Z \geq 0.5 \right)}$ $= \mathrm{P} \mathit{\left( Z \geq 0 \right)} - \mathrm{P} \mathit{\left( 0 \leq Z \leq 0.5 \right)}$ $= 0.5 - 0.1915$$= \mathit{0.3085}$