표본평균 확률

확률변수 $X$가 정규분포 ${\mathrm{N}} ( m , \sigma ^{2} )$을 따른다고 하면 $\displaystyle \mathrm{P} ( \mathit{X} \geq 3.4 ) = \frac{1}{2}$이므로 $m = 3.4$ 또, ${\mathrm{P}} ( Z \leq - 1 ) = {\mathrm{P}} ( Z \geq 1 )$이므로 $\mathrm{P} ( \mathit{X} \leq 3.9 ) + \mathrm{P} ( \mathit{Z} \geq 1 ) = 1$에서 $\mathrm{P} ( \mathit{X} \leq 3.9 ) = \mathrm{P} ( \mathit{Z} \geq 1 )$ $\displaystyle \mathrm{P} ( \mathit{X} \leq 3.9 ) = {\mathrm{P}} \left( Z \geq \frac{3.9 - 3.4}{\sigma} \right) = {\mathrm{P}} \left( Z \geq \frac{0.5}{\sigma} \right)$ 이므로 $\displaystyle \frac{0.5}{\sigma} = 1$, $\sigma = 0.5$ 따라서 확률변수 $X$가 정규분포 ${\mathrm{N}} ( 3.4 , 0.5 ^{2} )$을 따르므로 확률변수 $\displaystyle \overline{X}$는 정규분포 ${\mathrm{N}} ( 3.4 , 0.1 ^{2} )$을 따른다. 즉, $\displaystyle \mathrm{P} ( \mathit{\overline{X}} \geq 3.55 )$$\displaystyle = {\mathrm{P}} \left( Z \geq \frac{3.55 - 3.4}{0.1} \right)$ $= {\mathrm{P}} ( Z \geq 1.5 )$$= 0.5 - {\mathrm{P}} ( 0 \leq Z \leq 1.5 )$ $= 0.5 - 0.4332$$= 0.0668$