과제5

\begin{align*} &P\big(L(\underline{X}) < \mu < U(\underline{X}) \big) = 1-\alpha. \\[10pt] &\hat\mu = \overline{X}. \kern{10pt} X_i \sim N(\mu,~100). \kern{10pt} \text{i.i.d} \\[25pt] &P\big(a < \hat\mu <b \big) = P\bigg({a-\mu \over 4/\sqrt{5}} < {\overline{X}-\mu \over 4/\sqrt{5}} < {b-\mu \over 4/\sqrt{5}} \bigg) \\[20pt] &=P\bigg( -Z_{0.025} < {\mu-\overline{X} \over 4/\sqrt{5}} < Z_{0.025} \bigg) \\[20pt] &=P\bigg( \overline{X}-Z_{0.025} \cdot{\sqrt{5}\over 4} < \mu < \overline{X} + Z_{0.025}\cdot{\sqrt{5}\over 4} \bigg) \end{align*}

\bigg( \overline{X}+Z_{0.025} \cdot{\sqrt{5}\over 4} \bigg) - \bigg(\overline{X}-Z_{0.025} \cdot{\sqrt{5}\over 4} \bigg) \\[15pt] = Z_{0.025} \cdot{\sqrt{5}\over 4} \times 2 = 1.96 \cdot{\sqrt{5}\over 4} = 1.096

\begin{align*} &P(Y_{1} < m < Y_{4}) = P(m \le Y_4) - P(m \le Y_1) \\[10pt] &= \sum_{x=1}^{3} \binom{4}{x} \bigg({1 \over 2}\bigg)^4 = {1 \over 16} \sum_{x=1}^{3} \binom{4}{x} \\[15pt] &= {1 \over 16}~(4 + 6 + 4) = {14 \over 16} = 0.875 \end{align*}