정규분포 m+σ

[출제의도] 정규분포의 특성을 이해하고 있는가? 확률변수 $X$가 정규분포 $\mathrm{N} \left( \mathit{m} , \sigma ^{2} \right)$을 따르므로 $\mathrm{P} \left( \mathit{X} \leq 3 \right)$$\displaystyle = \mathrm{P} \left( \frac{X - m}{\sigma} \leq \frac{3 - m}{\sigma} \right)$ $\displaystyle = \mathrm{P} \left( \mathit{Z} \leq \frac{3 - m}{\sigma} \right) = 0.3$ 즉, $\displaystyle 0.5 - \mathrm{P} \left( \mathit{0} \leq Z \leq \frac{m - 3}{\sigma} \right) = 0.3$ 이다. 따라서 $\displaystyle \mathrm{P} \left( \mathit{0} \leq Z \leq \frac{m - 3}{\sigma} \right) = 0.2$ 이므로 $\displaystyle \frac{m - 3}{\sigma} = 0.52$ 따라서 $m = 3 + 0.52 \sigma$ $\cdots \cdots$ ㉠ 이때 $\mathrm{P} \left( \mathit{3} \leq X \leq 80 \right)$ $\displaystyle = \mathrm{P} \left( \frac{\mathit{3 - m}}{\sigma} \leq \frac{X - m}{\sigma} \leq \frac{80 - m}{\sigma} \right)$ $\displaystyle = \mathrm{P} \left( \frac{\mathit{3 - m}}{\sigma} \leq Z \leq \frac{80 - m}{\sigma} \right)$ $\displaystyle = \mathrm{P} \left( \frac{\mathit{3 - m}}{\sigma} \leq Z \leq 0 \right) + \mathrm{P} \left( \mathit{0} \leq Z \leq \frac{80 - m}{\sigma} \right)$ $\displaystyle = 0.2 + \mathrm{P} \left( \mathit{0} \leq Z \leq \frac{80 - m}{\sigma} \right)$ $= 0.3$ 에서 $\displaystyle \mathrm{P} \left( \mathit{0} \leq Z \leq \frac{80 - m}{\sigma} \right) = 0.1$ 이므로 $\displaystyle \frac{80 - m}{\sigma} = 0.25$ 즉, $m = 80 - 0.25 \sigma$ $\cdots \cdots$ ㉡ ㉠, ㉡에서 $3 + 0.52 \sigma = 80 - 0.25 \sigma$ $0.77 \sigma = 77$ $\sigma = 100$ 따라서 $m = 3 + 0.52 \times 100 = 55$이므로 $m + \sigma = 55 + 100 = 155$