정규분포 대칭

[출제의도] 정규분포의 성질 이해하기 확률변수 $X$가 정규분포 $N ( m , \sigma ^{2} )$을 따르므로 $\displaystyle P \left( X \geq 2 m - \frac{1}{2} a \right) = P \left( Z \geq \frac{m - \frac{1}{2} a}{\sigma} \right) = 0.1$ $P ( 0 \leq Z \leq 1.28 ) = 0.4$에서 $P ( Z \geq 1.28 ) = 0.1$ $\displaystyle \frac{m - \frac{1}{2} a}{\sigma} = 1.28$, $\displaystyle m - \frac{1}{2} a = 1.28 \sigma \cdots$㉠ $\displaystyle P \left( \frac{1}{2} a \leq X \leq a \right) = P \left( \frac{\frac{1}{2} a - m}{\sigma} \leq Z \leq \frac{a - m}{\sigma} \right)$ $\displaystyle = P \left( - 1.28 \leq Z \leq \frac{a - m}{\sigma} \right) = 0.1$ $P ( 0 \leq Z \leq 0.88 ) = 0.3$, $P ( 0 \leq Z \leq 1.28 ) = 0.4$에서 $P ( - 1.28 \leq Z \leq - 0.88 ) = 0.1$ $\displaystyle \frac{a - m}{\sigma} = - 0.88$, $a - m = - 0.88 \sigma \cdots$㉡ ㉠, ㉡을 연립하여 풀면 $a = 0.8 \sigma$, $m = 1.68 \sigma$ $\displaystyle \therefore \frac{10 m}{a} = \frac{16.8 \sigma}{0.8 \sigma} = 21$