教育部106年專科學校畢業程度自學進修學力鑑定考試
專業科目(一)： 初級統計
$$\cases{三家銀行:k=3\\ n_1=4\\ n_2=6\\ n_3=5} \Rightarrow \cases{df_B=k-1=2\\ df_W=n_1+n_2+n_3-1-df_B=12} \\\Rightarrow MS_W=60/df_w=60/12 =5 \Rightarrow F=38/MS_W=38/5= 7.6，故選\bbox[red,2pt]{(B)}$$
解答：$$\begin{array}{l|cc} & 正面 & 反面\\\hline 觀察值 & 19 & 11\\ 期望值 & 15 & 15 \end{array} \Rightarrow \chi^2 ={(19-15)^2\over 15} +{(11-15)^2\over 15} =2.133，故選\bbox[red,2pt]{(C)}$$
解答：$$\begin{array}{cc|ccc } X & Y & X^2 & XY & Y^2\\\hline 1 & 2 & 1 & 2 & 4\\ 2 & 2 & 4 & 4 & 4\\ 4 & 5 & 16 & 20 & 25\\\hdashline \sum X=7 & \sum Y=9 & \sum X^2=21 & \sum XY=26 & \sum Y^2=33 \end{array}\\ \Rightarrow 斜率b_1={\sum XY -\sum X \sum Y/n \over \sum X^2-(\sum X)^2/n} ={26-63/3 \over 21-49/3} = {15\over 14}=1.07，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{\bar x= \sum X/n=7/3\\ \bar y=\sum Y/n= 9/3=3} \Rightarrow 迴歸直線: y=b_1(x-\bar x)+\bar y =1.07(x-7/3)+3\\ \Rightarrow y=1.07x+0.5 \Rightarrow y截距=0.5，故選\bbox[red,2pt]{(D)}$$
解答：$$x=3代入迴歸直線: y=1.07\times 3+0.5=3.71，故選\bbox[red,2pt]{(B)}$$
解答：$$相關係數r=b_1\times \sqrt{\sum X^2-(\sum X)^2/n\over \sum Y^2-(\sum Y)^2/n} =1.07 \times \sqrt{21-49/3\over 33-81/3} =0.944，故選\bbox[red,2pt]{(B)}$$
解答：$$判定係數\;R^2 =r^2 = 0.945^2 = 0.89，故選\bbox[red,2pt]{(D)}$$
解答：$$P(X=k)={e^{-k}\;\lambda^k\over k!} ={e^{-3}\;3^x\over x!} \Rightarrow \lambda=3 \Rightarrow EX=\lambda =3，故選\bbox[red,2pt]{(A)}$$
解答：$$a+b=3 \Rightarrow (a,b)=(1,2)或(2,1)，共2種情況\Rightarrow 機率={2\over 6\times 6} ={1\over 18}，故選\bbox[red,2pt]{(B)}$$
解答：$$常態分配圖形為左右對稱，故選\bbox[red,2pt]{(A)}$$
解答：$$EX=\lambda T=2\times 20=40，故選\bbox[red,2pt]{(D)}$$
解答：$$增加樣本數可以減少誤差，型二錯誤跟著下降，故選\bbox[red,2pt]{(B)}$$
解答：$$EX= \sum xf(x) = 0\times {1\over 2}+1\times {1\over 2}={1\over 2}，故選\bbox[red,2pt]{(C)}$$
解答：$$P(X\le 2)= P(X=1)+P(X=2) = {1\over 10} +{2\over 10}={3\over 10}，故選\bbox[red,2pt]{(C)}$$
解答：$$\int_0^1 g(x)\;dx=1 \Rightarrow \int_0^1 kx\;dx= {1\over 2}k= 1 \Rightarrow k=2，故選\bbox[red,2pt]{(B)}$$
解答：$$信賴區間的上限值=\bar x + z_{\alpha/2} \cdot {\sigma \over \sqrt n} =68+ 1.645\times {4\over \sqrt {64}} = 68.82，故選\bbox[red,2pt]{(D)}$$
解答：$$E(Y)= E(2X+3)= 2E(X)+3=2\times 5+3=13，故選\bbox[red,2pt]{(D)}$$
解答：$$X\sim B(n=3,p=不良率)\Rightarrow P(n=3,X=1)=C^3_1p(1-p)^2 =3\times {1\over 10} \times {81\over 100} =0.243\\，故選\bbox[red,2pt]{(C)}$$
解答：$$X\sim B(n=3,p=1/2)\Rightarrow Var(X)= np(1-p) = 3\times {1\over 2}\times {1\over 2}={3\over 4} ，故選\bbox[red,2pt]{(D)}$$
解答：$$E(T)= E(2X+3Y)= 2E(X)+ 3E(Y)= 16+9=25，故選\bbox[red,2pt]{(C)}$$
解答：$$ P(A\mid B)= {P(A\cap B)\over P(B)} ={0.25\over 0.35} ={5\over 7}，故選\bbox[red,2pt]{(D)}$$
解答：$$1個標準差約68\%，2個標準差約95\%，3個標準差約99.7\%，故選\bbox[red,2pt]{(D)}$$
解答：$$C^5_2/C^{10}_2 =10/45={2\over 9}，故選\bbox[red,2pt]{(A)}$$
解答：$$估計主要是推估母體的統計參數，同時計算其準確性，故選\bbox[red,2pt]{(B)}$$
解答：$$依一致性之定義，故選\bbox[red,2pt]{(D)}$$
解答：$$\cases{X\sim B(n,p) \\ P(X=1)=P(X=2)} \Rightarrow C^5_1p(1-p)^4 = C^5_2p^2(1-p)^3 \Rightarrow 5p(1-p)^4 = 10p^2(1-p)^3\\ \Rightarrow 1-p=2p \Rightarrow p={1\over 3} \Rightarrow E(X)=np = 5\times {1\over 3}={5\over 3}，故選\bbox[red,2pt]{(C)}$$
解答：$$X\sim Poi \Rightarrow P(X=k)={e^{-\lambda}\; \lambda ^k\over k!} \Rightarrow P(X=1)= P(X=2) \Rightarrow e^{-\lambda}\;\lambda = {e^{-\lambda}\;\lambda ^2\over 2} \\ \Rightarrow \lambda =2 \Rightarrow P(X=0)=e^{-\lambda}= e^{-2}，故選\bbox[red,2pt]{(A)}$$
解答：$$n_1E_1^2 =n_2E_2^2 \Rightarrow 60\times 0.03^2 =n_2\times 0.02^2 \Rightarrow n_2=135，故選\bbox[red,2pt]{(B)}$$
解答：$$n_1E_1^2 =n_2E_2^2 \Rightarrow 50\times 2^2 =n_2\times 1.2^2 \Rightarrow n_2=138.89，故選\bbox[red,2pt]{(A)}$$
解答：$$P(X\gt 75) = P(Z\gt {75-77\over 10/\sqrt n})= P(Z\lt {2\over 10/\sqrt n})=1- 0.081 \Rightarrow {2\over 10/\sqrt n}=1.4 \Rightarrow n=49\\，故選\bbox[red,2pt]{(D)}$$
解答：$${\sqrt {n_1}\over z_{\alpha_1/2}} = {\sqrt {n_2}\over z_{\alpha_2/2}} \Rightarrow {\sqrt{50} \over 1.96} ={\sqrt{n_2}\over 2.575} \Rightarrow \sqrt {n_2}=9.29 \Rightarrow n_2=86.3，故選\bbox[red,2pt]{(C)}$$
解答：$$E= z_{\alpha/2} \times \sqrt{p(1-p) \over n} =1.64\times \sqrt{0.45\times 0.55 \over 100} =0.0816，故選\bbox[red,2pt]{(A)}$$
解答：$$\sqrt n= z_{\alpha/2} \times {\sqrt{p(1-p)} \over E} =1.96\times {\sqrt{{38\over 50}\times {12\over 50} } \over 0.05} =16.74 \Rightarrow n=280.28，故選\bbox[red,2pt]{(A)}$$
解答：$$2\times t_{\alpha/2} \times {s \over \sqrt n} =2\times 1.833 \times { 2 \over \sqrt {10}} = 2.318，故選\bbox[red,2pt]{(B)}$$
解答：$$V({2x_1+x_2 \over 3}) ={1\over 9}(4V(x_1)+V(x_2)) ={1\over 9}(4\sigma^2 +\sigma^2)={5\over 9}\sigma^2 \Rightarrow \sigma({2x_1+x_2 \over 3}) =\sqrt{{5\over 9}\sigma^2}={\sqrt 5\over 3}\sigma \\ \Rightarrow c={\sqrt 5\over 3} =0.745，故選\bbox[red,2pt]{(B)}$$
解答：$$ X\sim B(n,p) \Rightarrow \cases{期望值=4=np\\ 變異數=2 =np(1-p)} \Rightarrow {4\over 2}={np\over np(1-p)} \Rightarrow 2={1\over 1-p} \Rightarrow p={1\over 2} \Rightarrow n=8\\ \Rightarrow P(X=0)= p^n=({1\over 2})^8 ={1\over 256}，故選\bbox[red,2pt]{(A)}$$
解答：$$P(X\le 20) = \int_0^{20}{1\over 20}e^{-x/20}dx = \left. \left[ -e^{-x/20}\right] \right|_0^{20} =1-e^{-1}，故選\bbox[red,2pt]{(C)}$$
解答：$$第1次及第2次都不可以出現3且第3次出現3的機率={5\over 6} \times {5\over 6} \times {1\over 6}={25\over 216}，故選\bbox[red,2pt]{(A)}$$
解答：$$P(X\lt 0.2) = P(Z\lt {0.2-0.12\over \sqrt{0.2\times 0.8/100}}) =P(Z\lt 2) \\ \Rightarrow P-值=1-P(Z\lt 2)= P(Z\gt 2)=0.02，故選\bbox[red,2pt]{(A)}$$
解答：$$本題為小樣本(n\lt 30)，採用無母數卡方分配來求母體標準差區間\\\bar x= (3.01+3.01+2.96 +3.02)\div 4 = 3 \Rightarrow s^2= (0.01^2+0.01^2+0.04^2+0.02^2) = 0.0022\\
\Rightarrow 標準差的下界=\sqrt{s^2 \over \chi^2_{\alpha/2}\;(n-1)}= \sqrt{0.0022\over \chi^2_{0.025}\;\;(3)}= \sqrt{0.0022\over 9.35} =0.0153，故選\bbox[red,2pt]{(C)}$$
\Rightarrow 標準差的下界=\sqrt{s^2 \over \chi^2_{\alpha/2}\;(n-1)}= \sqrt{0.0022\over \chi^2_{0.025}\;\;(3)}= \sqrt{0.0022\over 9.35} =0.0153，故選\bbox[red,2pt]{(C)}$$
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解題僅供參考，其他專科學力鑑定試題及詳解
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