정규분포표 활용

[출제의도] 표준정규분포표를 이용하여 주어진 문제를 해결한다. $\displaystyle \mathit{F} \left( x \right) = \mathrm{P} \left( \mathit{X} \leq x \right) = \mathrm{P} \left( \mathit{Z} \leq \frac{x - m}{\sigma} \right)$ $\displaystyle \mathit{F} \left( \frac{13}{2} \right) = \mathrm{0.8413}$, $\mathrm{P} ( \mathit{0} \leq Z \leq 1 ) = 0.3413$이므로 $\displaystyle \frac{\frac{13}{2} - m}{\sigma} = 1$, $\displaystyle \sigma = \frac{13}{2} - m$ $\cdots$$\cdots$ ㉠ $\displaystyle 0.5 \leq F \left( \frac{\mathit{11}}{2} \right) \leq 0.6915$이므로 $\displaystyle 0 \leq \frac{\frac{11}{2} - m}{\sigma} \leq 0.5$ $\cdots$$\cdots$ ㉡ ㉡에 ㉠을 대입하여 정리하면 $\displaystyle \frac{9}{2} \leq m \leq \frac{11}{2}$이고 $m$이 자연수이므로 $m = 5$, $\displaystyle \sigma = \frac{3}{2}$ $\mathit{F} \left( k \right) = \mathrm{0.9772}$이므로 $\displaystyle \frac{k - 5}{\frac{3}{2}} = 2$, $k = 8$