104年專科學力鑑定-初級統計詳解

教育部104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

解答：$$\sum f(x)=1 \Rightarrow f(1)+f(2)+f(3) +f(4)=1\Rightarrow kx+2kx+3kx+4kx=10kx=1\\ \Rightarrow k={1\over 10} =0.1，故選\bbox[red,2pt]{(A)}$$

解答：$$P(B\mid A)= {P(B\cap A)\over P(A)} ={0.3\over 0.6} ={1\over 2}，故選\bbox[red,2pt]{(B)}$$

解答：$$\begin{array}{rrr|rrr} & X & Y & X^2 & XY & Y^2\\\hline & 1 & 1 & 1 & 1 & 1\\ & 2& 2& 4& 4 & 4\\ & 3& 4& 9 & 12 & 16 \\ & 4 & 3& 16 & 12 & 9\\ & 5 & 6 & 25 & 30 & 36\\\hdashline \sum & 15 & 16 & 55 & 59 & 66\\\hline\end{array}\\ 斜率b_1= {\sum XY - \sum X\sum Y/n\over \sum X^2 -(\sum X)^2/n} ={59-15\times 16/5 \over 55-15^2/5} ={11\over 10}=1.1，故選\bbox[red,2pt]{(D)}$$

解答：$$迴歸直線過點(\bar x,\bar y) \Rightarrow y=1.1(x-15/5)+16/5 \Rightarrow y=1.1x-0.1 \Rightarrow 截距為-0.1，故選\bbox[red,2pt]{(A)}$$

解答：$$相關係數r={\sum XY - \sum X\sum Y/n\over \sqrt{\sum X^2 -(\sum X)^2/n} \cdot \sqrt{\sum Y^2 -(\sum Y)^2/n}} ={59-15\times 16/5 \over \sqrt{55-15^2/5} \cdot \sqrt{66-16^2/5}}\\ ={11\over \sqrt{10} \cdot \sqrt{74/5}} ={11\sqrt{37}\over 74} =0.904，故選\bbox[red,2pt]{(C)}$$

解答：$$判定係數=r^2=0.904^2 =0.817，故選\bbox[red,2pt]{(B)}$$

解答：$$10位同學\Rightarrow 自由度=10-1=9，故選\bbox[red,2pt]{(B)}$$

解答：$$E(X)=\lambda T=2\times 2=4，故選\bbox[red,2pt]{(C)}$$

解答：$$X\sim B(n=3,p=1/2) \Rightarrow P(X=3)=C^3_3p^3 = {1\over 8}，故選\bbox[red,2pt]{(B)}$$

解答：$$顯著水準就是型I錯誤的機率，故選\bbox[red,2pt]{(A)}$$

解答：$$資料依大小排序，中間位置即為中位數，故選\bbox[red,2pt]{(B)}$$

解答：$$單一母體變異數區間估計，且為小樣本，故選\bbox[red,2pt]{(D)}$$

解答：$$\bar x=(6+9+8+5+6)\div 5=34/5=6.8，故選\bbox[red,2pt]{(C)}$$

解答：$$出現頻率最高的數，故選\bbox[red,2pt]{(A)}$$

解答：$$最大值減去最小值=19-3=16，故選\bbox[red,2pt]{(D)}$$

解答：$$\begin{array}{c|cc} & 正面& 反面\\\hline 觀察值& 70 & 30\\ 期望值& 50 & 50\end{array} \Rightarrow \chi^2={(70-50)^2 \over 50}+ {(30-50)^2 \over 50} ={800\over 50}=16，故選\bbox[red,2pt]{(A)}$$

解答：$$平均成績是否等於75，並非大於或小於，故選\bbox[red,2pt]{(C)}$$

解答：$$不偏性:估計式的抽樣分配平均數(期望值)等於母體參數，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{A=\{2,4,6\} \\B=\{1,2\}} \Rightarrow A\cap B=\{2\} \Rightarrow \cases{P(A)=1/2\\ P(B)=1/3\\ P(A\cap B)=1/6} \Rightarrow P(A\mid B) ={P(A\cap B)\over P(B)} ={1\over 2}\\，故選\bbox[red,2pt]{(C)}$$

解答：$${\quad A機器生產的不良品\quad \over A機器生產的不良品+B機器生產的不良品 \qquad } ={0.4\times 0.01 \over 0.4\times 0.01 +0.6\times 0.005} ={4\over 7}\\，故選\bbox[red,2pt]{(C)}$$

解答：$$甲\cap 乙=\varnothing \Rightarrow P(甲\cap 乙)=P(甲)+P(乙)-P(甲\cap 乙)=0.15+0.2-0=0.35，故選\bbox[red,2pt]{(D)}$$

解答：$$\bar x=(10+12+8 +18)/4= 12 \\\Rightarrow 樣本標準差=\sqrt{(10-12)^2+ (12-12)^2+(8-12)^2+ (18-12)^2\over 4-1}=4.32\\，故選\bbox[red,2pt]{(B)}$$

解答：$$E(Y)= E(2X+2) =2E(X)+2=2\times {1\over 2}+2=3，故選\bbox[red,2pt]{(D)}$$

解答：$$X\sim B(n=3,p=1/2) \Rightarrow Var(X)=np(1-p)= 3\times {1\over 2}\times {1\over 2}= 0.75，故選\bbox[red,2pt]{(C)}$$

解答：$$E(P)=E(2X-3Y+4Z) =2E(X)-3E(Y) +4E(Z) =2\times 4-3\times 9+4\times 3=-7\\，故選\bbox[red,2pt]{(A)}$$

解答：$$已知\cases{E(X)=5\\ E(X^2)-(EX)^2=1} \Rightarrow E(X^2)=1+(E(X))^2=1+25=26 \\ \Rightarrow E(Y)= E(X^2+2X+1) = E(X^2)+2E(X) +E(1) =26+10+1=37，故選\bbox[red,2pt]{(B)}$$

解答：$$E(X)=(1+2+3+4+5+6)/6 ={21\over 6} \Rightarrow E(X_1+X_2)= 2E(X) =2\times {21\over 6} =7，故選\bbox[red,2pt]{(B)}$$

解答：$$第1次抽中且第2次沒抽中不良品或第1次沒抽中且第2次抽中不良品\\ ={4\over 16}\times {12\over 15} +{12\over 16}\times {4\over 15} ={96\over 240}={2\over 5}=0.4，故選\bbox[red,2pt]{(C)}$$

解答：$$P(p\lt 10\%)\Rightarrow P\left(Z\lt {10\%-7\%\over \sqrt{10\%\times 90\%\over 100}}\right) =P(Z\lt -1)，故選\bbox[red,2pt]{(A)}$$

解答：$$P(X\lt 60) = P(Z\lt {60-56\over 4}) =P(Z\lt 1) =P(0\le Z\le 1) +0.5 = 0.3413+ 0.5=0.8413\\，故選\bbox[red,2pt]{(C)}$$

解答：$$A、B獨立\Rightarrow P(A\cap B)= P(A)P(B) \Rightarrow P(A\cup B)=P(A)+P(B)-P(A)P(B)\\ \Rightarrow 0.8 = P(A)+0.4-0.4P(A) \Rightarrow P(A)=0.4/0.6=2/3，故選\bbox[red,2pt]{(D)}$$

解答：$$區間長度=2\times t_{\alpha/2}(n-1)\cdot {s\over \sqrt n} =2\times 1.86\times {2.1\over \sqrt 9} = 2.604，故選\bbox[red,2pt]{(B)}$$

解答：$$n=(z_{\alpha/2})^2 \times {s^2\over E^2} =1.645^2 \times {2.1^2 \over 0.2^2} = 298.34，故選\bbox[red,2pt]{(A)}$$

解答：$$P(X\lt 12)= P(Z\lt {12-16\over 4})= P(Z\lt -1)= 0.5-P(0\le Z\le 1) =0.5-0.3413=0.1587\\，故選\bbox[red,2pt]{(B)}$$

解答：$$p+z_{\alpha/2}\cdot \sqrt{p(1-p)\over n} ={1\over 4}+1.96\times \sqrt{0.25\cdot 0.75\over 80} =0.345，故選\bbox[red,2pt]{(B)}$$

解答：$$A=\{(1,1),(2,2),\dots,(6,6)\} \Rightarrow P(A)={6\over 36}={1\over 6}，故選\bbox[red,2pt]{(A)}$$

解答：$$甲、乙、丙三條生產線\Rightarrow df_B=3-1=2；共有60\times 3=180罐飲料\\，因此隨機誤差自由度=(180-1)-df_B=179-2=177，故選\bbox[red,2pt]{(A)}$$

解答：

$$\cases{五家銀行\Rightarrow df_A=5-1=4\\ n=行員人數=6+5+6+7 +6=30} \quad \Rightarrow df_e=(n-1)-df_A= 29-4=25 \\ \Rightarrow MS_e = 76.44/df_e= 76.44/25 = 3.0576 \Rightarrow F=64.12 /MS_e = 64.12/3.0576 = 20.97\\，故選\bbox[red,2pt]{(D)}$$

解答：$$觀察值\begin{array}{c| ccc}  & 部門甲 & 部門乙 & 部門丙 & 小計\\\hline 高中職& 30 & 20 & 50 & 100\\ 大專& 110 & 40 & 50 & 200\\ 研究所& 60 & 40 & 100 & 200 \\\hline 小計& 200 & 100 & 200 & 500\end{array} \\ 期望值\begin{array}{c| rrr}  & 部門甲 & 部門乙 & 部門丙 & 小計\\\hline 高中職& 100\times {2\over 5}=40 & 100\times {1\over 5}=20& 100\times {2\over 5}=40 & 100\\ 大專& 200\times {2\over 5}=80 & 200\times {1\over 5}=40 & 200\times {2\over 5}=80 & 200\\ 研究所& 200\times {2\over 5}=80 & 200\times {1\over 5}=40 & 200\times {2\over 5}=80 & 200 \\\hline 小計& 200 & 100 & 200 & 500\end{array}\\ \Rightarrow \chi^2 ={(30-40)^2\over 40}+ {(20-20)^2\over 20}+{(50-40)^2\over 40}+{(110-80)^2\over 80}+ {(40-40)^2\over 40}+{(50-80)^2\over 80}\\ \qquad \qquad + {(60-80)^2\over 80}+ {(40-40)^2\over 40}+{(100-80)^2\over 80}\\ ={5\over 2}+0+{5\over 2} +{45\over 4}+0 +{45\over 4}+ 5+0+5={75\over 2}=37.5，故選\bbox[red,2pt]{(B)}$$

解答：$$\bar x-z_{\alpha/2}\cdot {s\over \sqrt n} =82-1.96\times {6\over \sqrt{250}} =81.256，故選\bbox[red,2pt]{(B)}$$

解答：$$n=(z_{\alpha/2})^2 \cdot {p(1-p)\over E^2} =1.96^2\cdot {(1/4)\cdot (3/4)\over 0.03^2} =800.33，故選\bbox[red,2pt]{(D)}$$

解答：$$0.02\times 100=2，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{甲銀行:n_1=10, \bar x_1=15, s_1=3\\ 乙銀行:n_2=6, \bar x_2=10, s_2=4} \Rightarrow s_p^2 = {(n_1-1)s_1^2 +(n_2-1)s_2^2 \over n_1+n_2-2} ={9\cdot 3^2+ 5\cdot 4^2\over 10+6-2} =11.5\\ \Rightarrow t={(\bar x_1-\bar x_2)-(\mu_1-\mu_2) \over \sqrt{s_p^2\left( {1\over n_1}+{1\over n_2}\right)}} ={(15-10)-0\over \sqrt{11.5({1\over 10}+{1\over 6})}} ={5\sqrt{690}\over 46}=2.855，故選\bbox[red,2pt]{(B)}$$

解答：$$\chi^2 ={(n-1)s^2 \over \sigma^2} ={(15-1)6^2\over 5^2} =20.16，故選\bbox[red,2pt]{(D)}$$

解答：$${p_1-p_2\over \sqrt{{p_1(1-p_1)\over n_1}+{p_1(1-p_1)\over n_1}}} ={45/50-40/50\over \sqrt{{(45/50)(5/50)\over 50} +{(40/50)(10/50)\over 50}}} =\sqrt 2=1.41，故選\bbox[red,2pt]{(C)}$$

解答：$${(z_{\alpha_1/2})^2 \over n_1}={(z_{\alpha_2/2})^2 \over n_2}  \Rightarrow {z_{0.025}^2 \over 100} ={z_{0.05}^2\over n_2} \Rightarrow {1.96^2 \over 100} ={1.645^2\over n_2} \Rightarrow n_2=70.44，故選\bbox[red,2pt]{(C)}$$

解答：$$n_1E_1^2 = n_2E_2^2 \Rightarrow 50\times 2^2 = n_2\times 1^2 \Rightarrow n_2=200，故選\bbox[red,2pt]{(D)}$$

解答：$$\sigma^2(\hat \mu) =\sigma^2({2x_1+x_2+3x_3+2x_4 \over 8}) ={1\over 64}(4\sigma^2(x_1)+ \sigma^2(x_2)+ 9\sigma^2(x_3)+4 \sigma^2(x_4)) ={18\over 64}\sigma^2(x)\\ \Rightarrow \sigma(\hat \mu)= \sqrt{{18\over 64}\sigma^2(x)} ={3\sqrt 2\over 8}\sigma =0.53\sigma，故選\bbox[red,2pt]{(A)}$$

解答：$$\sigma(Y)= \sigma(100-5X)= |-5|\sigma(X)= 5\times 1=5，故選\bbox[red,2pt]{(A)}$$

解答：$$P(7\le X\le 11) =P({7-9 \over 3/\sqrt 9}\le  Z\le {11-9 \over 3/\sqrt 9}) =P(-2\le Z\le 2) = 2P(0\le Z\le 2)\\ =2\times 0.4772 =0.9544，故選\bbox[red,2pt]{(D)}$$

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解題僅供參考，其他專科學力鑑定試題及詳解