110年地方三等-交通技術-統計學詳解

 110 年特種考試地方政府公務人員考試試題

等 別： 三等考試
類 科： 交通技術
科 目： 統計學

解答：

(一)$$\sum f(x)=1 \Rightarrow p+4p+ 2p+p +2p=10p=1 \Rightarrow p=1/10\\\cases{E(X)=-2p-4p+p+4p = -p=-1/10\\ E(X^2)=4p+4p +p +8p=17p= 17/10\\ E(X^3)=-8p-4p+p +16p=5p=5/10\\ E(X^4)=16p+4p+p+32p = 53p=53/10}\\ 因此E(Y^2)= E((2X-1)^4)= E(16X^4-32X^3 +24X^2 -8X+1) \\ =16E(X^4) -32E(X^3) +24E(X^2)-8E(X)+1 =848-160p+408p+8p+1\\ =1104p+1= \bbox[red,2pt]{111.4}$$(二)$$E(XY)=E(X(2X-1)^2)= E(4X^3-4X^2+X) =4E(X^3)-4E(X^2)+E(X) \\ =20p -68p-p=-49p=-4.9;\\ E(Y)=E((2X-1)^2)=E(4X^2-4X+1) =4E(X^2) -4E(X)+1 =68p+4p+1= 8.2 \\ 因此Cov(X,Y)= E(XY)-E(X)E(Y)= -4.9-(-0.1)(8.2)= \bbox[red,2pt]{-4.08}$$

解答：

$$由題意知:\cases{P(A)=a+d+f+g = 0.5\\P(B)= b+d+e+g=0.3 \\ P(C)= c+e+f+g=0.1\\ P(A\cap B)= d+g=0.15\\ P(A\cap C)= f+g= 0.05\\ P(B\cap C)= e+g = 0.03\\ P(A\cap B\cap C)= g=0.01} \Rightarrow \cases{a=0.31\\ b=0.13\\ c=0.03\\ d=0.14\\ e=0.02\\ f=0.04\\g=0.01}\\(一)僅使用過快遞A的機率=a=\bbox[red,2pt]{0.31}\\(二){c+e+f+g\over a+b+c+d+e+f+ g} ={0.1\over 0.68} =\bbox[red,2pt]{0.147} \\(三){g\over b+d+e+g } ={0.01 \over 0.3} = \bbox[red,2pt]{0.0333}$$

解答：

(一)$$假設A_3之處理量為N\Rightarrow \cases{A_1之處理量為1.5N\\ A_2之處理量為1.3N} \Rightarrow {A_3辨識錯誤量\over 全部辨識錯誤量}\\={ N\times 6\% \over 1.5N\times 5\%+ 1.3N\times 4\%+ N\times 6\%} ={0.06\over 0.192} = \bbox[red,2pt]{0.3125}$$(二)$$假設全公司處理量為M \Rightarrow \cases{A_1,A_2,A_3處理量0.6M \Rightarrow \cases{A_1處理量={1.5N\over 1.5N+1.2N +N} \cdot 0.6M ={9\over 37}M \\ A_2處理量={1.2N\over 1.5N+1.2N +N} \cdot 0.6M  ={36\over 185}M \\ A_3處理量={1 N\over 1.5N+1.2N +N} \cdot 0.6M  ={6\over 37}M }\\B_1,B_2,B_3處理量0.4M}\\ 辨識錯誤比率=({9\over 37}+0.4)\times 5\%+ {36\over 185}\times 4\%+ {6\over 37}\times 6\% = \bbox[red,2pt]{0.0497}$$(三)$${0.4\times 5\% \over 0.0497} = \bbox[red,2pt]{0.4026}$$

解答：

(一)$$X\sim N(100,12^2) \Rightarrow \bar X={X_1+X_2+X_3+X_4\over 4} \sim N(100,({12\over 2})^2)\\ P(\bar X\gt 112)= P(Z\gt {112-100\over 6}) = P(Z\gt 2)= 1-0.9772(查表)=\bbox[red,2pt]{0.0228}$$(二)$$\bbox[red,2pt]{\cases{H_0:今年事故次數與去年相同\\ H_1:今年事故次數與去年不同}}$$(三)$$\bar x\pm z_{\alpha/2}{\sigma \over \sqrt n} ={3672\over 36}\pm 2.576\cdot {12\over \sqrt {36}} =102\pm 5.152 = \bbox[red,2pt]{[96.848, 107.152]}$$各子題間關聯敍述似乎不夠清楚...

解答：

(一)$$\begin{array}{c|rrrr|rrrr} &X_1(品牌1)&  X_2(品牌2) &X_3(品牌3)& X_4(品牌4) &X_1^2 &X_2^2 &X_3^2 &X_4^2 \\\hline & 5 & 4 & 5 & 6 & 25 & 16 & 25 & 36 \\ & 7 & 3 & 8 & 5 & 49 & 9 & 64 & 25 \\ & 4 & 2& 7 & 4 &16 & 4 & 49 & 16 & \\ & 4 & 5 & 5 & 6 & 16 & 25 & 25 & 36 \\ & 5 & 6 & 6 & 4 & 25 & 36 & 36 & 16 \\ & 4 & 4 & 7 & 7 & 16 & 16 & 49 & 49 \\ & 6 & &7 & 5& 36 & & 49 & 25 \\ & & & 5 & 4 & & & 25 & 16 \\ & & & 4 & 3 & & & 16 & 9 \\ & & & & 6 & & & & 36\\\hdashline \sum &35 & 24 & 54 & 50 & 183 & 106 & 338 & 264 \end{array}\\ \Rightarrow \cases{n_1=7\\ n_2=6\\ n_3=9\\ n_4=10 \\ n=n_1+n_2 +n_3 + n_4=32 } \Rightarrow \cases{\bar X_1=35/7=5\\ \bar X_2= 24/6=4\\ \bar X_3=54/9=6\\ \bar X_4=50/10=5\\ \bar X=(35+ 24+54+50)\div 32 =5.094}\\ \text{ANOVA table }\\ \begin{array}{|c| c| c| c|c|} \hline 來源& 自由度(df) & 平方和(SS) & 均方(MS) & F\\\hline 組間 & (1) & (4) & (7) & (9)\\\hline 組內& (2) & (5) & (8) & \\\hline 合計& (3) & (6) & &\\\hline\end{array} \\ (1)=品牌數-1=4-1=3\\(2)=(3)-(1)=31-3=28 \\(3)=總樣本數-1=32-1=31\\ (4)=n_1(\bar X_1-\bar X)^2 +n_2 (\bar X_2-\bar X)^2 +n_3(\bar X_3-\bar X)^2 +n_4(\bar X_4-\bar X)^2\\\quad =7(5-5.094)^2 +6(4-5.094)^2 +9(6-5.094)^2 +10(5-5.094)^2=14.7188\\(5)=(6)-(4)= 60.7188-14.7188=46\\(6)= \sum X^2-(\sum X)^2/n =(183+106+338+264)-(35+24+54+50)^2/32=60.7188 \\(7)=(4)\div (1)=14.7188/3=4.9063\\ (8)=(5)\div (2)=46/28=1.6429\\ (9)=(7)\div (8)=4.9063/1.6429 = 2.9864\\ \Rightarrow 變異數分析表:\\ \bbox[red,2pt]{\begin{array}{|c| c| c| c|c|} \hline 來源& 自由度(df) & 平方和(SS) & 均方(MS) & F\\\hline 組間 & 3 & 14.7188 & 4.9063 & 2.9864\\\hline 組內& 28 & 46 & 1.6429 & \\\hline 合計& 31 & 60.7188 & &\\\hline\end{array}}$$(二)$$檢定統計量F=2.9864 \gt F_{0.05}(3,28)=2.95(查試題附表)，\bbox[red,2pt]{已達顯著差異}$$(三)$$小樣本且母體變異數未知，\mu_3-\mu_2的信賴區間為 (\bar x_3-\bar x_2)\pm t_{df,\alpha/2}\cdot S_{\bar x_3-\bar x_2}\\ 其中\cases{\bar x_3=6\\ \bar x_2=4}，\cases{s_2^2= (\sum x_2^2-(\sum x_2)^2/n_2)/(n_2-1) =(106-24^2/6)/5=2 \\ s_3^2= (\sum x_3^2-(\sum x_3)^2/n_3)/(n_3-1) =(338-54^2/9)/8=1.75}\\ S_{\bar x_3-\bar x_2}=\sqrt{{s_2^2\over n_2}+ {s_3^2\over n_3}} =\sqrt{{2\over 6}+ {1.75\over 9}} =0.7265,df={({s_2^2\over n_2}+ {s_3^2\over n_3})^2\over {(s_2^2/n_2)^2\over n_2-1} +{(s_3^2/n_3)^2\over n_3-1}} ={0.2785\over 0.0269}=10(取整數)\\ 因此信賴區間為(6-4)\pm t(10,0.025)\cdot 0.7265 =2\pm 2.2281\cdot 0.7265 =\bbox[red,2pt]{[0.3813,3.6187]}$$

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