108年高考三級-統計學-詳解

108年公務人員高等考試三級考試

類 科 ：統計
科 目：統計學

解：
(一)$$\begin{cases} f_X(x)=2e^{-2x},x\ge 0\\ f_Z(z)=3e^{-3z},z\ge 0 \end{cases} \Rightarrow \begin{cases} F_X(x)=\int_0^x{2e^{-2t}\,dt}=1-e^{-2x}\\ F_Z(z)=\int_0^z{3e^{-3t}\,dt}=1-e^{-3z} \end{cases} \Rightarrow \begin{cases} P(X\le t)=1-e^{-2t}\\ P(Z\le t)=1-e^{-3t} \end{cases} \\\Rightarrow \begin{cases} P(X>t)=1-(1-e^{-2t})=e^{-2t}\\ P(Z> t)=1-(1-e^{-3t})=e^{-3t} \end{cases}\\
F_Y(y)=P(Y\le y)=P(min(X,Z)\le y)=1-P(min(X,Z)>y) =1-P(X>y)P(Z>y)\\=1-e^{-2y}\cdot e^{-3y}=1-e^{-5y} \Rightarrow f_Y(y)=F'_Y(y)=5e^{-5y}\\\Rightarrow Y\text{的pdf: }\,\bbox[red,2pt]{f_Y(y)=5e^{-5y},y\ge 0}$$
(二)$$f_Y(y)=5e^{-5y},y\ge 0 \Rightarrow E(Y)= \int_0^\infty {5ye^{-5y}\,dy}=\left. \left[-ye^{-5y}-{1\over 5}e^{-5y} \right]  \right|_0^\infty ={1\over 5}\\ \Rightarrow E(Y^2)=\int_0^\infty {5y^2e^{-5y}\,dy} =5\left. \left[-{1\over 5}y^2e^{-5y} -{2\over 25}ye^{-5y}-{2\over 125}e^{-5y} \right]  \right|_0^\infty ={2\over 25}\\ Var(Y)=E(Y^2)-(E(Y))^2= {2\over 25}-\left({1\over 5}\right)^2 = {1\over 25}\\ \Rightarrow \begin{cases} U=E(Y)+ 2\sqrt{V(Y)}={1\over 5}+2\sqrt{1\over 25} ={3\over 5}=0.6 \\L= E(Y)- 2\sqrt{V(Y)}={1\over 5}-2\sqrt{1\over 25} =-{1\over 5}=-0.2 \end{cases} \\ \Rightarrow P(L< Y< U) = P(-0.2 < Y < 0.6) = P(0  < Y < 0.6)= \int_0^{0.6}{5e^{-5y}\,dy}\\ =\left. \left[-e^{-5y} \right]  \right|_0^{0.6} = \bbox[red, 2pt]{1-e^{-3}}$$

(三)$$五取三的次數為C^5_3，其中三個正常，二個異常，機率為C^5_3p^3(1-p)^2 =10p^3(1-p)^2 \\=10(1-e^{-3})^3(e^{-3})^2 = {10(1-e^{-3})^3 \over e^{6}} \approx \bbox[red,2pt]{0.021}$$

---

解：

(一)

$$\int_A{f}=1 \Rightarrow \int_0^1{\int_0^{1-y}{f(x,y)\,dx}\,dy}=1\Rightarrow 0.5d\int_0^1{\int_0^{1-y}{xy\,dx}\,dy}=1\\ \Rightarrow 0.5d\int_0^1{\left. \left[ \frac{1}{2}x^2y\right]\right|_0^{1-y}\,dy}=1 \Rightarrow {d\over 4}\int_0^1{(1-y)^2y\,dy}= {d\over 4}\int_0^1{(y^3-2y^2+y)\,dy}\\= {d\over 4}\left. \left[{1\over 4}y^4-{2\over 3}y^3+ {1\over 2}y^2 \right]\right|_0^1 = {d\over 4}\times {1\over 12}=1 \Rightarrow \bbox[red,2pt]{d=48}$$(二) $$f_X=\int{f\,dy}=\int_0^{1-x}24xy\,dy=\left. \left[ 12xy^2 \right]\right|_0^{1-x}= 12x(1-x)^2\\ \Rightarrow \bbox[red,2pt] {f_X(x)= 12x(1-x)^2,0\le x\le 1}\\ P(X\le 0.25)=\int_0^{0.25}{f_X\,dx} =\int_0^{0.25}{12x(1-x)^2\,dx} = 12\int_0^{0.25}{ x^3-2x^2+x\, dx}\\ =12\left. \left[ {1\over 4}x^4-{2\over 3}x^3+{1\over 2}x^2 \right]\right|_0^{0.25} = 12\times {67\over 3072}={67 \over 256} \Rightarrow \bbox[red,2pt]{P(X\le 0.25)={67 \over 256}}$$

(三)$$E(3X)=\int_0^1{3xf_X(x)\,dx}= \int_0^1{36x^2(1-x)^2\,dx} =36 \int_0^1{x^2-2x^3+x^4\,dx}\\ =36 \left. \left[ {1\over 3}x^3-{1\over 2}x^4+{1\over 5}x^5 \right]\right|_0^1 =36\times {1\over 30}={6\over 5}=1.2 \Rightarrow \bbox[red,2pt] {E(3X)=1.2}$$(四)$$E(XY)=\int_0^1{\int_0^{1-y}{xyf(x,y)dx}dy}= \int_0^1{\int_0^{1-y}{24x^2y^2dx}dy} =24 \int_0^1{\left. \left[{1 \over 3}x^3y^2 \right] \right|_0^{1-y}\,dy}\\ =8 \int_0^1{y^2(1-y)^3dy} =8\int_0^1{y^2-3y^3+3y^4-y^5\,dy}= 8\left. \left[ {1\over 3}y^3-{3\over 4}y^4+ {3\over 5}y^5 -{1\over 6}y^6 \right] \right|_0^1 \\ =8\times {1\over 60}={2\over 15} \Rightarrow E(5+2XY)=5+2E(XY)= 5+2\times {2\over 15}= \frac{79}{15}\\\Rightarrow \bbox[red,2pt] {E(5+2XY)=\frac{79}{15}}$$

---

解：
(一)$$H_0:\,銷售數量符合常態分配\\ H_1:\,銷售數量不符合常態分配$$(二)$$\begin{cases} 樣本平均數\bar{x}={\sum_{i=0}^{32}{x_i}\over 32}={1515 \over 32}=47.34\\樣本標準差s=\sqrt{\sum_{i=1}^{i=32}{x_i^2}-32(\bar{x})^2\over 32-1} =14.59\end{cases} \\\Rightarrow 組界分別為\begin{cases} Q1=\bar{x}+ z_{1/4}\cdot s=\bar{x}-z_{3/4}\cdot s=37.49\\ \bar{x}=47.34\\ Q_3=\bar{x}+z_{3/4}\cdot s=47.34+0.675\cdot 14.59=57.19\end{cases}\\
此外,理論上每組均有32\div 4=8筆資料;將觀值依組界分組及理論值可得以下分配表:\\
\begin{array}{|c|c|c|c|c|}\hline
組界& 最小-37.49&37.49-47.34&47.34-57.19 &57.19-最大\\\hline
o_i & 8 & 7 & 8 & 9\\\hline
e_i & 8 & 8 & 8 & 8\\\hline
\end{array}\\
卡方檢定統計量\chi^2=\sum_{i=1}^4{(e_i-o_i)^2\over e_i} =2/8=\bbox[red,2pt]{0.25},自由度df=4-2-1=\bbox[red,2pt]{1}
$$

(三)$$查表\chi_{\alpha=0.1,df=1}^2= 2.705541 > \chi^2 \Rightarrow 不能拒絕H_0\\ 檢定結果: \bbox[red,2pt]{銷售數量為常態分配}$$

---

解：
(一)$$實驗設計方法：完全隨機設計(Completely Ramdomized Design, CRD)；\\隨機抽取15位病人並統計所使用假牙的壽命，計算統計量檢定其差異性；$$(二)$$效應模式:y_{ij}=\mu_i+\varepsilon_{ij},其中\varepsilon_{ij}為試驗誤差,i=1-3,j=1-5\\並假設\varepsilon_{ij}\sim N(0,\sigma^2) $$(三)$$\begin{cases} H_0: \mu_A =\mu_B = \mu_C  \\ H_1: \mu_A \ne \mu_B 或  \mu_A \ne \mu_C 或  \mu_B \ne \mu_C \end{cases}\\

\begin{array}{ccc|ccc}

A & B & C & A^2 &B^2 &C^2\\\hline

8 &10 & 7 & 64 & 100 &49\\

8 &12 & 8 & 64 & 144 &64\\

8 &11 & 7 & 64 & 121 &49\\

9 &12 & 6 & 81 & 144 &36\\

9 &10 & 7 & 81 & 100 &49\\\hline

42&55&35&354 &609 & 247\\

\sum{A}&\sum{B} &\sum{C} &\sum{A^2}&\sum{B^2}&\sum{C^2}

\end{array} \\\Rightarrow

\begin{cases} \bar{A}=42\div 5=8.4 \\ \bar{B}=55\div 5=11 \\ \bar{C}=35\div 5 =7\end{cases} \Rightarrow \begin{cases} s_A^2=(254-5\cdot 8.4^2)\div 4=0.3 \\ s_B^2=(609-5\cdot 11^2)\div 4=1 \\ s_C^2=(247-5\cdot 7^2)\div 4 =0.5\end{cases} \\資料總平均\bar{x}={\bar{A}+\bar{B} +\bar{C}\over 3}={26.4\over 3}=8.8 \Rightarrow SS_B=5((\bar{A}-\bar{x})^2 +(\bar{B}-\bar{x})^2 +(\bar{C}-\bar{x})^2 )\\ =5\times(0.4^2+2.2^2+1.8^2)=41.2\\ SS_W=4(S_A^2+S_B^2+S_C^2)=4(0.3+1+0.5)=7.2 \\\Rightarrow \begin{cases} SS_B=41.2\\SS_W=7.2\end{cases} \Rightarrow \begin{cases} MS_B=SS_B\div (G-1)=41.2\div (3-1)=20.6\\MS_W=SS_W\div (N-G)=7.2\div (15-3)=0.6\end{cases}\\ \Rightarrow

F=MS_B/MS_W = 20.6/0.6=34.33\\
\text{可求得 ANOVA table}:\begin{array}{|c|c|c|c|c|}\hline

& SS & df & MS &F\\\hline

組間&41.2 & 2 & 20.6 &34.33\\\hline

組內&7.2 & 12 & 0.6\\\hline

合計&48.4 & 14\\\hline

\end{array}\\
查表F_{0.05}(2,12)=3.89 \Rightarrow 拒絕區域R=\{F \mid F>3.89\} \Rightarrow 34.33\in R\\ \Rightarrow 拒絕H_0,即不同品牌材質的假牙對使用壽命\bbox[red, 2pt]{有}影響$$

(四)$$LSD=t_{\alpha/2}(N-G)\times\sqrt{MSE\left({1\over n_i}+{1\over n_j} \right)},其中自由度為N-G=15-3=12\\ \Rightarrow LSD=t_{0.025}(12) \times\sqrt{0.6\left({1\over 5}+{1\over 5} \right)}=t_{0.025}(12) \times\sqrt{6}/5\\ 查表可得t_{0.025}(12)=2.179 \Rightarrow LSD=2.179\times\sqrt{6}/5= 1.067\\ \Rightarrow \begin{cases} \begin{cases}H_0: \mu_A=\mu_B\\ H_1:\mu_A\ne \mu_B\end{cases} \Rightarrow \left| \bar{A}-\bar{B}\right|=|8.4-11|=2.6>LSD \Rightarrow 拒絕H_0 \\ \begin{cases}H_0: \mu_B=\mu_C\\ H_1:\mu_B\ne \mu_C\end{cases} \Rightarrow\left| \bar{B}-\bar{C}\right|=|11-7|=4>LSD \Rightarrow 拒絕H_0\\ \begin{cases}H_0: \mu_A=\mu_C\\ H_1:\mu_A\ne \mu_C\end{cases} \Rightarrow\left| \bar{A}-\bar{C}\right|=|8.4-7|=1.4>LSD \Rightarrow 拒絕H_0\end{cases}\\ \Rightarrow \begin{cases}品牌A與品牌B有顯著差異\\品牌B與品牌C有顯著差異\\品牌A與品牌C有顯著差異 \end{cases} ,此結果與(三)一致;$$

---

考選部未公布答案，解題僅供參考