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2022年1月10日 星期一

111年公務人員初等考試-統計學大意詳解

111年公務人員初等考試

等 別: 初等考試
類 科: 統計
科 目: 統計學大意

解答:$$P(A^c\cap B^c) =1-P(A\cup B)=1-{2\over 3}= {1\over 3},故選\bbox[red,2pt]{(B)}$$
解答:$$\sum f(x)=1 \Rightarrow A({1\over 1}+ {1\over 2}+ {1\over 3}+ {1\over 4}+{1\over 5}) = A\cdot {137\over 60} =1 \Rightarrow A={60\over 137},故選\bbox[red,2pt]{(B)}$$
解答:$$f(x)可能會大於1,例如:X\sim U(a=0.1,b=0.2) \Rightarrow f(x)={1\over 0.2-0.1}=10\ge 1\\只是連續型機率密度函數在區間內才有機率(其值\le 1),故選\bbox[red,2pt]{(B)}$$
解答:$$Var(aX+bY)= a^2Var(X)+ b^2Var(Y)+ 2abCov(X,Y),故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{E(X)=0\cdot {1\over 3}+1\cdot{1\over 2}+ 2\cdot 0+3\cdot {1\over 6}=1\\ E(X^2)= 0^2\cdot {1\over 3}+1^2\cdot{1\over 2}+ 2^2\cdot 0+3^2\cdot {1\over 6}=2} \\ \Rightarrow Var(X)= E(X^2)-(E(X))^2 = 2-1^2=1,故選\bbox[red,2pt]{(B)}$$
解答:$$\begin{array} {|c|c|c|c|}\hline & A & B & C \\\hline 觀察值 & 100  & 427  & 273 \\\hline 期望值& 0.12\times 800=96 & 0.54\times 800=432 & 0.34\times 800=272\\\hline\end{array} \\ \Rightarrow 卡方檢定統計量\chi^2= {(100-96)^2\over 96} + {(427-432)^2\over 432} + {(273-271)^2\over 272} =0.2282 \\查試題附表可得\chi_{0.05}^2(2)= 5.99147 \Rightarrow 0.228 \lt \chi_{0.05}^2(2) \Rightarrow 不拒絕H_0,故選\bbox[red,2pt]{(D)}$$
解答:$$將20位消費者年齡,由小到大排序如下:\\ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20\\ \hline 13 & 14 & 14 & 14 & 15 & 15 & 15 & 16 & 16 & 16 & 17 & 17 & 18 & 18 & 19 & 21 & 23 & 25 & 29 & 30 \\\hline\end{array}\\中位數為第10及第11的平均,即(16+17)\div 2 = 16.5,故選\bbox[red,2pt]{(B)}$$
解答:$$全距=最大值-最小值,可表達群體分散情形,故選\bbox[red,2pt]{(D)}$$
解答:$$A\cap (B\cup C)= (A \cap B)\cup (A\cap C),故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{P(A)=1/2\\ P(B)=1/3\\ P(A\cap B)=1/6} \Rightarrow P(A\cap B)=P(A)P(B) \Rightarrow A、B兩事件獨立,故選\bbox[red,2pt]{(D)}$$
解答:$$E(\bar X)=\mu=30,故選\bbox[red,2pt]{(A)}$$
解答:$$P(4\le \bar X\le 13)= P({4-10\over \sqrt{441/49}} \le Z\le {13-10\over \sqrt{441/49}}) =P(-2\le Z\le 1) =P(Z\le 1)-P(Z\le -2)\\ =0.8413-0.0228 = 0.8185,故選\bbox[red,2pt]{(B)}$$
解答:$$觀察值:\begin{array}{c|ccc|c} & 所得水準高 & 所得水準中 &所得水準低 & 小計\\\hline 信用狀況良好& 18 & 22 & 12 &52 \\  信用狀況不良 &16 & 20 & 14 & 50\\\hline 小計 & 34 & 42 & 26 & 102\end{array}\\ 令p=52/102  \\ \Rightarrow 期望值:\begin{array}{c|ccc|c} & 所得水準高 & 所得水準中 &所得水準低 \\\hline 信用狀況良好& 34p=17.333 & 42p =21.412 & 26p  =13.255 \\  信用狀況不良 &34(1-p)=16.667 & 42(1-p) =20.588 & 26(1-p) =12.745\\\hline\end{array} \\ \Rightarrow 檢定統計量\chi^2 ={ (18-17.333)^2\over 17.333} +{(22-21.412)^2 \over 21.412} +{(12-13.255)^2\over 13.255} \\\qquad +{(16-16.667)^2 \over 16.667} +{(20-20.588)^2 \over 20.588} +{(14-12.745)^2 \over 12.745} =0.3276;\\查試題附表\chi_{0.025}^2(2) =7.37776 \Rightarrow 0.3276 \lt \chi_{0.025}^2(2) \Rightarrow 不具顯著相關性,故選\bbox[red,2pt]{(A)}$$
解答:$$\bar x=(614+665 + 836+622 + 506+568+580+545)\div 8=4936\div 8=617\\ \Rightarrow 信賴區間=\bar x\pm z_{\alpha/2}{\sigma\over \sqrt n} =617\pm 1.96\cdot {100\over \sqrt {8}} =617\pm 69.296= (547.704,686.296)\\,故選\bbox[red,2pt]{(A)}$$
解答:$$\alpha=H_0為值卻拒絕H_0的機率;\beta=H_0為偽卻不拒絕H_0,故選\bbox[red,2pt]{(C)}$$
解答:$$\bar x=(3+7 +5+ 13+16 +12 +17+8 + 11+14)\div 10 =10.6 \\ s^2= ((3-10.6)^2+ (7-10.6)^2 +\cdots +(14-10.6)^2) \div (10-1) =198.4/9= 22.044\\ \Rightarrow 信賴區間=\bar x\pm t_{\alpha/2}\sqrt{\sigma^2\over  n} =10.6\pm 3.25\cdot \sqrt{22.044\over 10} = 10.6\pm 4.825= (5.775,15.425)\\,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{男性贊成廢比例p_1=125/1000 =0.125\\ 女性贊成廢比例p_2=120/1200 =0.1} \\\Rightarrow 比例差之信賴區間= (p_1-p_2) \pm z_{\alpha/2}\cdot \sqrt{{p_1(1-p_1) \over n_1} +{p_2(1-p_2)\over n_2}}\\ =0.025 \pm 1.96\cdot \sqrt{{0.125\cdot 0.875\over 1000} +{0.1\cdot 0.9\over 1200}} \\ =0.025\pm 0.0266 =(-0.0016,0.0516),故選\bbox[red,2pt]{(C)}$$
解答:$$\bar x=(12+10+15+18 +22+19)\div 6=96\div 6=16 \\ \Rightarrow s^2 =((12-16)^2+ (10-16)^2+(15-16)^2 +(18-16)^2 +(22-16)^2 +(19-16)^2)\div 5\\= 102\div 5=21\\ 檢定統計量z={\bar x-\mu\over \sqrt{\sigma^2/n}} ={16-12\over \sqrt{20/6}} = 2.19 \gt 1.645(查試題附表:z_{0.05}=1.645) \Rightarrow 拒絕H_0\\,故選\bbox[red,2pt]{(B)}$$
解答:$$查表可得P(z\gt 2.19)=1-0.9857=0.0143,故選\bbox[red,2pt]{(D)}$$
解答:$$自由度=(2-1)\times (3-1)=2,又\alpha=0.05,故選\bbox[red,2pt]{(C)}$$
解答:$$F={10/2\over 120/12}={5\over 10}=0.5,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{a\div 10=1.9 \\ 30\div b=6 \\ c=10.5\div 1.9} \Rightarrow \cases{a=19 \\b=5\\ c=5.526},故選\bbox[red,2pt]{(B)}$$
解答:$$\mu={ \sum x_{ij} \over 3\times 5} ={315\over 15}=21 \\\Rightarrow \tau_2= ((17-21)+(19-21)+(21-21)+ (23-21)+(25-21))\div 5=0\\,故選\bbox[red,2pt]{(C)}$$
解答:$$\begin{array}{cc|rrr} X & Y & X^2 & Y^2 & XY\\ \hline 5 & 21 & 25 & 441 & 105\\3 & 15 & 9 & 225 & 45\\6 & 23 & 36 & 529 & 138\\2 & 12 & 4 & 144 & 24\\4 & 18 & 16 & 324 & 72\end{array} \Rightarrow \cases{\sum X=20 \\ \sum Y=89\\ \sum X^2=90 \\ \sum Y^2=1663\\ \sum XY=384} \Rightarrow \cases{\bar X=20/5=4\\ \bar Y=89/5=17.8}\\ \Rightarrow \hat b_1= {\sum XY-\sum X\sum Y/n\over \sum X^2-(\sum X)^2/5} ={384-20\cdot 89/5\over 90-20^2/5} =2.8 \Rightarrow y=\hat b_1(x-\bar X)+\bar Y =2.8(x-4)+17.8\\ \Rightarrow y=2.8x+6.6 \Rightarrow \hat b_0=6.6,故選\bbox[red,2pt]{(D)}$$
解答:$$\hat y_i=2.8(x_i-4)+17.8 \Rightarrow \begin{array}{} y_i & \hat y_i &   (y_i-\hat y_i)^2\\\hline 21 & 20.6 & 0.4^2 \\ 15 & 15 & 0^2 \\ 23 & 23.4 & 0.4^2 \\ 12 & 12.2^2 & 0.2^2\\ 18 & 17.8 & 0.2^2 \\\hline \end{array}\\ \Rightarrow SSE= \sum (y_i-\hat y_i)^2 =0.16+0+ 0.16+0.04 +0.04 = 0.4,故選\bbox[red,2pt]{(A)}$$
解答:$$相關係數r=b_1\times {\sigma(Y)\over \sigma(X)} =2.8\times \sqrt{\sum X^2-(\sum X)^2/n\over \sum Y^2-(\sum Y)^2/n} =2.8\times \sqrt{90-20^2/5\over 1663-89^2/5} = 0.99746\\ \Rightarrow R^2=r^2=0.99746^2 = 0.9949,故選\bbox[red,2pt]{(B)}$$
解答:$$SSE=0.4 \Rightarrow MSE=SSE/(n-2)=0.4/3=0.133,故選\bbox[red,2pt]{(C)}$$
解答:$$P(B\mid A) ={P(A\cap B)\over P(A)} \Rightarrow P(A\cap B)=P(A)\times P(B\mid A) \ne P(B)\times P(B\mid A),故選\bbox[red,2pt]{(D)}$$
解答:$$F(分子自由度,分母自由度),故選\bbox[red,2pt]{(C)}$$
解答:$$f(x)=q^{x-1}p \Rightarrow E(X)= \sum_{x=1}^\infty xq^{x-1}p = p\sum_{x=0}^\infty {d\over dx}q^x = p{d\over dx}\sum_{x=0}^\infty q^x = p{d\over dx}{1\over 1-q} =p{1\over (1-q)^2}\\ =p\cdot {1\over p^2}={1\over p},故選\bbox[red,2pt]{(B)}$$
解答:$$信賴區間的上界=p+ z_{\alpha/2}\cdot \sqrt{p(1-p)\over n} =0.8+ 1.96\cdot \sqrt{0.8\cdot 0.2\over 900} =0.826\\ 為了拒絕H_0 \Rightarrow p\gt 0.826 \Rightarrow 訂閱數\gt 900\times 0.826=743.4,故選\bbox[red,2pt]{(C)}$$
解答:$$X\sim Poiss(\lambda) \Rightarrow P(Y=y)=e^{-\lambda}{\lambda^{y}\over y!} \Rightarrow E(X)=Var(X)=\lambda \\ 由題意知: P(Y=0) = e^{-\lambda}=p\Rightarrow \lambda =E(X)=-\ln p,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{E(X)=E(Y)= E(Z)= \mu\\ E(X^2)= E(Y^2)=E(Z^2) = \sigma^2+\mu^2\\ E(XY)=E(X)E(Y)= \mu^2} \\\Rightarrow \cases{E(X+cY)= E(Y+cZ)= (1+c)\mu\\ E(X+cY)(Y+cZ)= E(XY)+cE(XZ)+ cE(Y^2)+c^2E(YZ) \\\qquad = (c^2+c+1)\mu^2+c(\sigma^2+\mu^2) =(c^2+2c+1)\mu^2+ c\sigma^2\\ E(X+cY)^2 =E(X^2)+2cE(XY)+c^2E(Y^2) =\mu^2+\sigma^2+ 2c\mu^2 +c^2(\mu^2 +\sigma^2)\\ \qquad =(c^2+2c+1)\mu^2+(c^2+1)\sigma^2 \\ E(Y+cZ)^2 = E(X+cY)^2= (c+1)^2\mu^2+(c^2+1)\ \sigma^2}\\ 相關係數={E(X+cY)(Y+cZ)-E(X+cY)E(Y+cZ) \over \sqrt{E(X+cY)^2-(E(X+cY))^2}\cdot \sqrt{E(Y+cZ)^2-(E(Y+cZ))^2}} \\ ={(c^2+2c+1)\mu^2+c\sigma^2-(1+c)^2\mu^2 \over (c^2+1)\sigma^2} ={c\sigma^2\over (c^2+1)\sigma^2} ={c\over c^2+1},故選\bbox[red,2pt]{(B)}$$
解答:$$注射停止代表最後一次是不良反應,而且過去也有一次不良反應,再加上5次沒有不良反應;\\也就是共七次注射,前六次是「五良一不良」機率為C^6_1(0.1)(0.9)^5再乘上最後一次不良機率0.1\\,即C^6_1(0.1)^2(0.9)^5, 故選\bbox[red,2pt]{(B)}$$
解答:$$X_1,X_2 期望值為 \mu,變異數為\sigma^2  \Rightarrow \bar X={X_1+X_2\over 2} 的期望值為\mu,變異數為 \sigma^2/2\\利用柴比雪夫不等式: P(|X-E(X)|\ge b) \le {Var(X)\over b^2}  \Rightarrow P(|\bar X-\mu|\ge 2\sigma) \le { \sigma^2/2\over 4\sigma^2} ={1\over 8}\\ \Rightarrow P(|\bar X-\mu|\ge 2\sigma) \le {1\over 8},故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{女性有使用理財機器人比例p_1=200/1000=0.2 \\男性有使用理財機器人比例p_2=300/1200=0.25 } \\ \Rightarrow 檢定統計量z= {p_1- p_2 \over \sqrt{{p_1(1-p_1)\over n_1}+{p_2(1-p_2)\over n_2}} } ={0.2-0.25\over \sqrt{{0.2\cdot 0.8\over 1000} +{0.25\cdot 0.75\over 1200}}} = -2.812,故選\bbox[red,2pt]{(C)}$$
解答:$$區間長度與z_{\alpha/2},\sigma, 1/\sqrt n成正比,故選\bbox[red,2pt]{(A)}$$
解答:$$W與X正負號相反,但仍為線性轉換,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{E(X+3)=E(X)+3=4+3=7 \\ \sigma(X+3)= \sigma(X)=8},故選\bbox[red,2pt]{(B)}$$
解答:$$對稱圖形不一定是單峰,也可能是雙峰,也就是眾數與平均數不相近,故選\bbox[red,2pt]{(D)}$$

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