자연계열 2번

[논제 1] 표본평균 $\displaystyle \overline X$는 정규분포 $\displaystyle \mathrm{N}\left(m,\frac{\sigma^2}n\right)$을 따르므로, $\displaystyle Z=\frac{\overline X-m}{\frac\sigma{\sqrt n}}$로 표준화하면,$\displaystyle \mathrm{P}\left(\overline X\ge m+\frac1{\sqrt n}\right)=\mathrm{P}\left(\frac{\overline X-m}{\frac\sigma{\sqrt n}}\ge\frac1\sigma\right)=\mathrm{P}\left(Z\ge\frac1\sigma\right)=0.0668$따라서 $\displaystyle \mathrm{P}\left(0\le Z\le\frac1\sigma\right)=0.5-0.0668=0.4332$이며 표준정규분포표로부터 $\displaystyle \frac1\sigma=1.5$, 즉, $\displaystyle \sigma=\frac23$이다.

그러므로$\displaystyle h(x)=\mathrm{P}(\sqrt n(\overline X-m)\le-x)=\mathrm{P}\left(\frac{\overline X-m}{\frac\sigma{\sqrt n}}\le-\frac x\sigma\right)=\mathrm{P}\left(Z\le-\frac32x\right)$이다. 따라서 $\displaystyle h(x)$는 감소함수이므로$\displaystyle \begin{aligned}&\sum_{k=1}^{10}\left|h\left(\frac15-\frac k5\right)-h\left(-\frac k5\right)\right|\\&=\left(h\left(-\frac15\right)-h(0)\right)+\left(h\left(-\frac25\right)-h\left(-\frac15\right)\right)+\cdots+\left(h(-2)-h\left(-\frac95\right)\right)\\&=h(-2)-h(0)\end{aligned}$

이다. 그런데 $\displaystyle h(-2)=\mathrm{P}(Z\le3),h(0)=\mathrm{P}(Z\le0)$이므로$\displaystyle \sum_{k=1}^{10}\left|h\left(\frac15-\frac k5\right)-h\left(-\frac k5\right)\right|=h(-2)-h(0)=\mathrm{P}(0\le Z\le3)=0.4987$이다.

[논제 2] $\displaystyle \overline X$는 정규분포 $\displaystyle \mathrm{N}\left(m,\frac{\sigma^2}n\right)$, $\displaystyle Y$는 정규분포 $\displaystyle \mathrm{N}(m+2,4\sigma^2)$을 각각 따르므로, 조건$\displaystyle \mathrm{P}(\overline X\le m+1)=\mathrm{P}(Y\le m+12)$에서$\displaystyle \mathrm{P}\left(\frac{\overline X-m}{\frac\sigma{\sqrt n}}\le\frac1{\frac\sigma{\sqrt n}}\right)=\mathrm{P}\left(\frac{Y-m-2}{2\sigma}\le\frac{10}{2\sigma}\right)$이다. 따라서$\displaystyle \mathrm{P}\left(Z\le\frac{\sqrt n}\sigma\right)=\mathrm{P}\left(Z\le\frac5\sigma\right)$

이므로 $\displaystyle n=25$이다. 이로부터$\displaystyle \mathrm{P}(|\overline X-m|\ge a)=\mathrm{P}\left(\left|\frac{\overline X-m}{\frac\sigma{\sqrt{25}}}\right|\ge\frac a{\frac\sigma{\sqrt{25}}}\right)=\mathrm{P}\left(|Z|\ge\frac{5a}\sigma\right)=0.0124$이므로 $\displaystyle \mathrm{P}\left(|Z|\le\frac{5a}\sigma\right)=1-0.0124=0.9876$이다. $\displaystyle \mathrm{P}\left(0\le Z\le\frac{5a}\sigma\right)=\frac{0.9876}2=0.4938$에서 표준정규분포표로부터 $\displaystyle \frac{5a}\sigma=\frac52$이므로 $\displaystyle a=\frac13$이다.