107年公務人員高等考試三級考試
類 科 ：經建行政、工業行政、農業行政、交通技術
科 目：統計學
科 目：統計學
(一)$$2500\times P\left(X>360 \right)=47 \Rightarrow P\left(X>360 \right)={47\over 2500} \Rightarrow P\left({X-\mu \over \sigma}> {360-\mu\over \sigma }\right)={47\over 2500}\\ \Rightarrow P\left(Z>{360-256\over \sigma}\right)= P\left(Z>{104\over \sigma}\right)=0.0188 \\ 查表可得P(Z<-2.08)=0.0188 \Rightarrow P(Z>2.08)=0.0188 \Rightarrow {104\over \sigma}=2.08 \Rightarrow \sigma={104\over 2.08}=\bbox[red, 2pt]{50}$$
(二)$$ P\left(X\ge a \right)={177 \over 2500}\Rightarrow P\left(Z\ge {a-\mu\over \sigma} \right)=0.0708 \Rightarrow P\left(Z\ge {a-256\over 50} \right)=0.0708\\ 查表可得P(Z<-1.47)=0.0708 \Rightarrow P(Z>1.47)=0.0708 \Rightarrow {a-256\over 50}=1.47 \\\Rightarrow a={50\times 1.47+256}=\bbox[red, 2pt]{329.5}$$
解：
解：
(一)$$\begin{cases} X_1\sim N(\mu_1,\sigma_1^2=1)\\ X_2\sim N(\mu_2,\sigma_2^2=4)\end{cases} \Rightarrow \bar{X_1}-\bar{X_2}\sim N\left( \mu_1-\mu_2,{1\over n_1}+{4\over n_2}\right) \\ \Rightarrow P\left( -z_{0.05/2}\le
{ (\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)\over \sqrt{{1\over n_1}+{4\over n_2}}}\le z_{0.05/2}\right)=0.95\\ \Rightarrow P\left(-z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}\le (\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2) \le z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}\right)=0.95 \\\Rightarrow P\left((\bar{X_1}-\bar{X_2})-z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}\le \mu_1-\mu_2 \le (\bar{X_1}-\bar{X_2})+z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}\right)=0.95\\ \Rightarrow \mu_1-\mu_2的95\%信賴區間為\left((\bar{X_1}-\bar{X_2})-z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}, \;(\bar{X_1}-\bar{X_2})+z_{0.025}\cdot \sqrt{{1\over n_1}+{4\over n_2}}\right)\\ = \bbox[red, 2pt]{\left( \bar{X_1}-\bar{X_2})\pm 1.96\cdot \sqrt{{1\over n_1}+{4\over n_2}}\right)}$$(二)$$信賴區間最短\Rightarrow {1\over n_1}+{4\over n_2}最小，且滿足n_1+n_2=100\\ \text{利用 Lagrange Multiplier 求解, 即令}\begin{cases} f(n_1,n_2)= {1\over n_1}+{4\over n_2}\\ g(n_1,n_2)=n_1+n_2-100\end{cases} \\ \Rightarrow \begin{cases} {\partial f\over \partial n_1}= \lambda {\partial g\over \partial n_1}\\ {\partial f\over \partial n_2}= \lambda {\partial g\over \partial n_2} \\g=0\end{cases} \Rightarrow \begin{cases} {-1\over n_1^2}= \lambda\\ {-4\over n_2^2}= \lambda \\n_1+n_2=100\end{cases} \Rightarrow \begin{cases} 4n_1^2= n_2^2\Rightarrow n_2=2n_1\\n_1+n_2=100\end{cases} \Rightarrow n_1+2n_1=100 \\\Rightarrow \begin{cases}n_1=100/3\\n_2= 200/3\end{cases}\xrightarrow{取整數} \bbox[red, 2pt]{\begin{cases}n_1=33\\n_2= 67\end{cases}}$$
解：
(一)$$E(X)=\int{xf(x)\,dx}=\int_{\theta_1}^{\theta_2}{{x\over \theta_2-\theta_1}dx} =\left. \left[{x^2\over 2(\theta_2-\theta_1)} \right] \right|_{\theta_1}^{\theta_2} ={\theta_2^2 -\theta_1^2\over 2(\theta_2-\theta_1)}=\frac{1}{2}(\theta_1+\theta_2)\\ E(X^2)=\int{x^2f(x)\,dx}=\int_{\theta_1}^{\theta_2}{{x^2\over \theta_2-\theta_1}dx} =\left. \left[{x^3\over 3(\theta_2-\theta_1)} \right] \right|_{\theta_1}^{\theta_2} ={\theta_2^3 -\theta_1^3\over 3(\theta_2-\theta_1)}=\frac{1}{3}(\theta_1^2+ \theta_1\theta_2+\theta_2^2)\\ \Rightarrow Var(X)=E(X^2)-(E(X))^2= \frac{1}{3}(\theta_1^2+ \theta_1\theta_2+\theta_2^2)-\left(\frac{1}{2}(\theta_1+\theta_2) \right)^2\\=\frac{1}{12}\theta_1^2 -\frac{1}{6}\theta_1\theta_2 +\frac{1}{12}\theta_2^2 ={1\over 12}(\theta_2-\theta_1)^2\\ \Rightarrow \bbox[red, 2pt] {\begin{cases} E(X)= \frac{1}{2}(\theta_1+\theta_2) \\V(X)= {1\over 12}(\theta_2-\theta_1)^2\end{cases}}$$(二)$$\begin{cases} E(X)=\frac{1}{2}(\theta_1+\theta_2)={X_1+X_2+\cdots+X_n\over n}=\sum{X_i}/n=\bar{X}\cdots(1) \\ E(X^2)= \frac{1}{3}(\theta_1^2+ \theta_1\theta_2+\theta_2^2)={X_1^2+X_2^2+\cdots+X_n^2\over n}=\sum{X_i^2}/n \cdots(2)\end{cases}\\ 式(1)\Rightarrow \theta_1=2\bar{X}-\theta_2\;代入式(2)\Rightarrow \left( 2\bar{X}-\theta_2 \right)^2 +\left( 2\bar{X}-\theta_2 \right)\theta_2+ \theta_2^2={3\over n}\sum{X_i^2}\\ \Rightarrow 4\bar{X}^2-2\bar{X}\theta_2 +\theta_2^2={3\over n}\sum{X_i^2} \Rightarrow \bar{X}^2-2\bar{X}\theta_2 +\theta_2^2={3\over n}\sum{X_i^2}-3\bar{X}^2 \\ \Rightarrow \left(\theta_2-\bar{X}\right)^2={3\over n}\sum{X_i^2}-3\bar{X}^2 ={3\over n}\left(\sum{X_i^2}-n\bar{X}^2\right) ={3\over n}\sum{(X_i-\bar{X})^2}\\={3(n-1)\over n}\cdot{\sum{(X_i-\bar{X})^2} \over n-1} = {3(n-1)\over n}\cdot s_X^2 \Rightarrow \theta_2=\bar{X}+{3(n-1)\over n}\cdot s_X^2\\ 同理,式(1)\Rightarrow \theta_2=2\bar{X}-\theta_1\;代入式(2)\Rightarrow \theta_1=\bar{X}-{3(n-1)\over n}\cdot s_X^2\\ 因此\theta_1的估計值\hat{\theta_1}及\theta_2的估計值\hat{\theta_2}為\bbox[red, 2pt]{\begin{cases} \hat{\theta_1}=\bar{X}-{3(n-1)\over n}\cdot s_X^2\\ \hat{\theta_2}=\bar{X}+{3(n-1)\over n}\cdot s_X^2\end{cases},其中s_X^2={\sum{(X_i-\bar{X})^2} \over n-1}}$$(三)$$L\left(X_1,X_2,\dots,X_n\mid \theta_1,\theta_2 \right)={1\over (\theta_2-\theta_1)^n}\\ L要最大化\Rightarrow \theta_2-\theta_1要最小化且滿足X_i\in [\theta_1,\theta_2],\text{for }i=1-n\\\Rightarrow \bbox[red, 2pt] {\begin{cases} \hat{\theta_1}=\min\{X_1,X_2,\dots,X_n\} \\ \hat{\theta_2}=\max\{X_1,X_2,\dots,X_n\} \end{cases}}$$
解：$$\bar{x} =(1.9+2.4+4.2+3.5+3.0)\div 5= 15\div 5=3\\
s^2=((1.9-3)^2+((2.4-3)^2+(4.2-3)^2 +(3.5-3)^2+(3-3)^2)\div (5-1)\\ = 3.26\div 4=0.815 \\
{(n-1)s^2\over \sigma^2} \sim \chi_{(n-1)}^2 \Rightarrow P\left(\chi_{0.975}^2(n-1) \le {(n-1)s^2\over \sigma^2}\le \chi_{0.025}^2(n-1)\right)=0.95 \\ \Rightarrow P\left(\chi_{0.975}^2(4) \le {4s^2\over \sigma^2}\le \chi_{0.025}^2(4)\right)=0.95 \Rightarrow P\left({4s^2\over \chi_{0.025}^2(4)} \le \sigma^2\le {4s^2 \over \chi_{0.975}^2(4)}\right)=0.95 \\ \sigma 的95\%信賴區間為\left(\sqrt{4s^2\over \chi_{0.025}^2(4)}, \sqrt{4s^2 \over \chi_{0.975}^2(4)}\right) =\left(\sqrt{4\times 0.815\over 11.14}, \sqrt{4\times 0.815 \over 0.48}\right)=(0.54,2.61)
\\\begin{cases} H_0:\sigma=1 \\ H_1:\sigma\ne 1 \end{cases} \Rightarrow \sigma=1 落在\sigma 的95\%信賴區間內\Rightarrow 不能拒絕 H_0 \\ \Rightarrow \bbox[red,2pt]{相信}該公司的宣稱$$
解：
數字& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9\\\hline
觀察值o_i&11 &12 &10 &9 & 9 & 10 &15 & 14 & 17 &13\\\hline
理論值e_i&12 &12 &12 &12 &12 &12 &12 &12 &12 &12
\end{array} \\ \Rightarrow \chi^2=(1^2+0^2+2^2+3^2+3^2+2^2 +3^2+2^2+5^2+1^2)\div 12=5.5\\
\begin{cases} H_0:資料服從均勻分配\\H_1: 資料不服從均勻分配\end{cases} \\\Rightarrow 拒絕區域R=\{\chi^2 \mid \chi^2>\chi_{0.05}^2(9)= 16.92(查表)\} \Rightarrow 5.5 \notin R \\\Rightarrow 不能拒絕H_0 \Rightarrow 各數字被搖出的\bbox[red, 2pt]{機率相等}$$
考選部未公布答案，解題僅供參考
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