教育部107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：$$(10+x+20)\div 3=12 \Rightarrow 30+x=36 \Rightarrow x=6，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{P(x\lt 63) =P(Z\lt {63-75\over 4})=P(Z\lt -3 ) \\P( X\gt 87) =P(Z\gt {87-75\over 4}) =P(Z\gt 3) } \Rightarrow P(|Z|\gt 3)\approx 1-0.997=0.003\\\Rightarrow 900\times 0.003=2.7\Rightarrow 挑人數最少的選項，故選\bbox[red,2pt]{(A)}$$
解答：$$Y=4X-6 \Rightarrow X={1\over 4}Y+{6\over 4} \Rightarrow E(X)= E({1\over 4}Y+{6\over 4})={1\over 4}E(Y)+{6\over 4} ={1\over 2}+{3\over 2}=2，故選\bbox[red,2pt]{(A)}$$
解答：$$擲骰子一次的期望值E(X)=(1+2+3+4+5+6)\div 6={7\over 2}\\ \Rightarrow 擲骰子三次的期望值E(3X)= 3E(X)=3\times {7\over 2}={21\over 2}，故選\bbox[red,2pt]{(D)}$$
解答：$$X\sim B(9,{1\over 2}) \Rightarrow 標準差= \sqrt{np(1-p)} =\sqrt{9\times {1\over 2} \times {1\over 2}} ={3\over 2}，故選\bbox[red,2pt]{(B)}$$
解答：$$只有兩骰子都出現1，其平均數才會是1，因此機率為{1\over 6} \times {1\over 6}={1\over 36}，故選\bbox[red,2pt]{(B)}$$
解答：$$\sigma(Y)= \sigma(3X+2) = 3\sigma(X)=3\times 2=6，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{B=\{(6,1),(6,2),(6,3),(6,4),(6,5), (6,6)\} \\A\cap B=\{(6,1), (6,2),(6,3),(6,4), (6,5)\}} \Rightarrow P(A\mid B)= {P(A\cap B)\over P(B)} ={5\over 6}，故選\bbox[red,2pt]{(D)}$$
解答：$$A、B獨立 \Rightarrow P(A\cap B)=P(A)P(B) \Rightarrow P(A\mid B)={P(A\cap B)\over P(B)} ={P(A)P(B)\over P(B)}=P(A)\\，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{前4次都是反面的機率=1/2^4\\ 第5次出現正面的機率=1/2} \Rightarrow {1\over 2^4} \times {1\over 2}={1\over 2^5} ={1\over 32}，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{第1次出現3的機率=1/6\\ 第2次才出現3的機率= (5/6)(1/6)=5/36} \\\Rightarrow P(1\le X\le 2) = P(X=1)+P(X=2)={1\over 6}+ {5\over 36}= {11\over 36}，故選\bbox[red,2pt]{(D)}$$
解答：$$P(X\lt 4)= P(X=1)+P(X=2) +P(X=3) = {1\over 6}+{1\over 6}+{1\over 6} ={1\over 2}，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{取到白球的機率:2/5 \\ 取到紅球的機率:3/5 } \Rightarrow 取出(白,白,紅)的機率為{2\times 2\times 3\over 5^3} ={12\over 125}\\，又(白,白,紅)的排列數為3，因此取到2白1紅的機率為{12\over 125} \times 3={36\over 125}，故選\bbox[red,2pt]{(D)}$$
解答：$$X\sim B(25,p) \Rightarrow EX=np \Rightarrow 10=25p \Rightarrow p=0.4，故選\bbox[red,2pt]{(D)}$$
解答：$$(1-5\%)^2 = (95\%)^2 = 90.25\% =0.9025，故選\bbox[red,2pt]{(D)}$$
解答：$$f(x)={3!\over (3-x)!x!}\times {1\over 8},x=0,1,2,3 \Rightarrow P(X\ge 1) =P(X=1)+P(X=2)+P(x=3)\\ =f(1)+f(2)+f(3) = {1\over 8}({6\over 2} +{6\over 2} +{6\over 6} ) ={1\over 8} \times 7 ={7\over 8}，故選\bbox[red,2pt]{(C)}$$
解答：$$X\sim Pois(\lambda) \Rightarrow P(X=k)={\lambda^k \cdot e^{-\lambda}\over k!} \Rightarrow P(X=0)= e^{-\lambda}= e^{-1} \Rightarrow \lambda =1 \Rightarrow E(X)=\lambda =1\\ \Rightarrow E(Y)=E(3X)=3E(X)= 3，故選\bbox[red,2pt]{(D)}$$
解答：$$X\sim Pois(\lambda) \Rightarrow Var(X)=\lambda =3 \Rightarrow E(Y)=E(2X) = 2E(X)=2\lambda =6，故選\bbox[red,2pt]{(B)}$$
解答：$$X\sim Pois(\lambda) \Rightarrow E(X)=\lambda =2 \Rightarrow P(X=0)=e^{-\lambda}=e^{-2}，故選\bbox[red,2pt]{(C)}$$
解答：$$10件產品中有1件是不良品，其餘9件是良品；\\因此\cases{第1次抽中不良品且第2次抽中良品的機率={1\over 10} \times {9\over 9} ={1\over 10}\\第1次抽中良品且第2次抽中不良品的機率={9\over 10}\times {1\over 9} ={1\over 10}}\\ \Rightarrow 欲求之機率為{1\over 10}+ {1\over 10}= {1\over 5}，故選\bbox[red,2pt]{(B)}$$
解答：$$題目的g(x)是\bbox[blue,2pt]{錯誤}的，正確的應該是g(x)={1\over 1000}e^{-x\over 1000}，\\因此P(X\ge 1000)=\int_{1000}^\infty {1\over 1000}e^{-x\over 1000}\;dx= \left. \left[ -e^{-x\over 1000} \right] \right|_{1000}^\infty =0+e^{-1}，故選\bbox[red,2pt]{(A)}$$
解答：$$E(X)=\int xf(x)\;dx = \int_0^1 {1\over 2}x\;dx =\left. \left[{1\over 4}x^2 \right] \right|_0^1 ={1\over 4}，故選\bbox[red,2pt]{(B)}$$
解答：$$\cases{X\sim N(\mu_x,16) \\ Y\sim N(\mu_y,9) } \Rightarrow Var(X-Y)=16+9=25 \Rightarrow \sigma(X-Y)=\sqrt{25}=5，故選\bbox[red,2pt]{(A)}$$
解答：$$P(Z\gt 1)= 0.1587 \Rightarrow P(Z\lt 1)=1- 0.1587=0.8413 \Rightarrow P(Z\gt -1)=0.8413\\ 而P(Z\gt -1)= P({x-10\over 2}\gt -1) = P(x\gt 8)，故選\bbox[red,2pt]{(B)}$$
解答：$$合格機率=P(19\lt x\lt 20) = P({19-19.5\over 0.5} \lt z\lt {20-19.5\over 0.5})= P(-1\lt z\lt 1)\\ \Rightarrow 不合格比率=P(z\lt -1)+P(z\gt 1)= 2P(z\gt 1) =2 \times 0.1587= 0.3174，故選\bbox[red,2pt]{(C)}$$
解答：$$依有效性之定義，較小之變異估計式較具有效性，故選\bbox[red,2pt]{(B)}$$
解答：$$依信賴區間公式可知，其長度與變異數(標準差)有關，與平均數無關，故選\bbox[red,2pt]{(B)}$$
解答：$$與上題類似，與平均數無關，故選\bbox[red,2pt]{(D)}$$
解答：$$\bar x+z_{\alpha/2} \cdot {s\over \sqrt n} =84+1.96\times {5\over \sqrt{100}} =84+0.98=84.98，故選\bbox[red,2pt]{(A)}$$
解答：$$E=z_{\alpha/2} \cdot {s\over \sqrt n} = 1.645\times {1.22\over \sqrt{64}} = 0.25，故選\bbox[red,2pt]{(C)}$$
解答：$$2\cdot z_{\alpha/2} \cdot \sqrt{p(1-p)\over n} = 2\cdot 1.96\times \sqrt{0.2\times 0.8\over 100} = 0.1568，故選\bbox[red,2pt]{(B)}$$
解答：$$母體平均數\mu 的估計誤差= z_{\alpha/2} \cdot {\sigma \over \sqrt n} \Rightarrow \cases{(A)\times: \sigma 已知則估計誤差較小\\ (B)\times:\alpha 越大\Rightarrow z_{\alpha/2}越大\Rightarrow 誤差越大\\ (C)\bigcirc: n越大\Rightarrow 誤差越小\\ (D) \times: 若母體標準差未知，需以樣本標準差推估}\\，故選\bbox[red,2pt]{(C)}$$
解答：$$E= z_{\alpha/2} \cdot {\sigma\over \sqrt n} \Rightarrow 2.575 \cdot {2\over \sqrt n} \le 0.7 \Rightarrow n\ge 54.127，故選\bbox[red,2pt]{(A)}$$
解答：$$E= z_{\alpha/2} \cdot \sqrt{p(1-p)\over n} \Rightarrow 1.645 \cdot \sqrt{(1/8)(7/8)\over n} \le 0.04 \Rightarrow n\ge 184.98，故選\bbox[red,2pt]{(D)}$$
解答：$$E= z_{\alpha/2} \cdot \sqrt{p(1-p)\over n} \Rightarrow 1.96 \cdot \sqrt{(120/200)(80/200)\over 200} =1.96\times {\sqrt 3\over 50} = 0.0678\approx 6.8\%\\，故選\bbox[red,2pt]{(C)}$$
解答：$$n_1E_1^2 =n_2E_2^2 \Rightarrow 70\times 2.5^2 =n_2\times 1.8^2 \Rightarrow n_2=135.03，故選\bbox[red,2pt]{(B)}$$
解答：$${z_{0.025}\over \sqrt{215}} ={z_{0.005} \over \sqrt n} \Rightarrow {1.96\over \sqrt{215}} ={2.575 \over \sqrt n} \Rightarrow n=371.09，故選\bbox[red,2pt]{(A)}$$
解答：$$E_1\sqrt {n_1} =E_2\sqrt{n_2 } \Rightarrow 0.05\times \sqrt{59} =0.03\times \sqrt{n_2} \Rightarrow n_2=163.88，故選\bbox[red,2pt]{(D)}$$
解答：$$信賴區間=(86.43,93.57) \Rightarrow 誤差E=(93.57-86.43)\div 2=3.57 = z_{\alpha/2}{s \over \sqrt n}=1.645 \times {15\over \sqrt n} \\ \Rightarrow n=\left( 1.645\times 15\times {1\over 3.57}\right)^2 =47.77，故選\bbox[red,2pt]{(A)}$$
解答：$$(A)\times: 顯著水準:H_0成立，拒絕H_0的機率\\(C)\times:型I錯誤=顯著水準:H_0成立，拒絕H_0的機率\\(D)\times:型II錯誤:H_a正確，但接受H_0的機率\\只有(B)正確，故選\bbox[red,2pt]{(B)}$$
解答：$$z={20000-18600\over 3850/\sqrt{30}}=1.99，由P(Z\gt 1.99)=0.023及雙尾檢定，其P值=0.023\times 2=0.046\\，故選\bbox[red,2pt]{(D)}$$
解答：$$p={15\over 50}= 0.3 \Rightarrow z={0.3-0.5\over \sqrt{0.5\times 0.5}/\sqrt {50}} =-2.828 \Rightarrow P(Z\gt -2.828) =1-P(Z \gt 2.83) \\ \Rightarrow P值=P(Z\gt 2.83)=0.0023，故選\bbox[red,2pt]{(B)}$$
解答：$$\cases{P(X\gt 75)=P(Z\gt {75-77\over 6/\sqrt n}) =P(Z\lt {2\over 6/\sqrt n}) =1-0.004=0.996\\ P(Z\gt 2.65) = 0.004 \Rightarrow P(Z\lt 2.65)=0.996} \\ \Rightarrow {2\over 6/\sqrt n} =2.65 \Rightarrow n=63.2，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{n_1=50,p_1=20/50\\ n_2=50,p_2=30/50} \Rightarrow \cases{q_1=1-p_1=30/50\\ q_2=1-p_2=20/50} \Rightarrow s^2={p_1(1-p_1) \over n_1} +{p_2(1-p_2)\over n_2} ={12\over 1250}\\ \Rightarrow s=0.098 \Rightarrow z={p_2-p_1\over s} ={0.2\over 0.098}= 2.04 \Rightarrow P(Z\gt 2.04)\approx 0.0228，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{n_1=10,\bar x_1=8,s_1=3\\ n_2=10,\bar x_2=7,s_2=2\\ n_3=10, \bar x_3=4,s_3=1} \Rightarrow \bar x=(n_1\bar x_1+n_2\bar x_2+ n_3\bar x_3)\div (n_1+n_2+n_3)\\\qquad = (80+70+40)\div 30={19\over 3}\\ \Rightarrow SSB= n_1(\bar x_1-\bar x)^2 + n_2(\bar x_2-\bar x)^2 + n_3(\bar x_3-\bar x)^2 =10({25\over 9}+{4\over 9}+{49\over 9})={260\over 3}\\ \cases{3^2= {\sum_{i=1}^{10}(x_{1i}-\bar x_1)^2\over 10-1} \Rightarrow 81= \sum x_{1i}^2-(\sum x_{1i})^2/10 =\sum x_{1i}^2-80^2/10\\ 2^2={\sum_{i=1}^{10}(x_{2i}-\bar x_2)^2\over 10-1} \Rightarrow 36= \sum x_{2i}^2-(\sum x_{2i})^2/10 =\sum x_{2i}^2-70^2/10 \\ 1^2= {\sum_{i=1}^{10}(x_{2i}-\bar x_2)^2\over 10-1} \Rightarrow 9= \sum x_{3i}^2-(\sum x_{3i})^2/10 =\sum x_{3i}^2-40^2/10} \Rightarrow \cases{\sum x_{1i}^2=721 \\\sum x_{2i}^2=526 \\\sum x_{3i}^2=169 }\\ SST= \sum \sum x_{ij}^2 -(\sum \sum x_{ij})^2/(n_1+n_2+n_3) = (721+526+169)-190^2/30= 638/3\\ \Rightarrow SSW = SST-SSB = (638-260)/3=126 \Rightarrow \cases{MSB = SSB/2=130/3\\ MSW=SSW/27 =14/3} \\ \Rightarrow F=MSB/MSW = 130/14=9.28，故選\bbox[red,2pt]{(C)}$$
解答：$$\begin{array}{l|rrrrrr} & 1 & 2 & 3 & 4 & 5 & 6\\\hline 觀察值O_i & 8 & 9 & 13 & 7 & 14 & 9 \\ 期望值E_i & 10 & 10 & 10 & 10 & 10 & 10\end{array} \\ \Rightarrow \chi^2=\sum_{i=1}^6{(O_i-E_i)^2 \over E_i} ={1\over 10}(2^2+1^2+3^2+3^2+4^2+1^2) =4，故選\bbox[red,2pt]{(D)}$$
解答：$$迴歸直線方程式: y=5x+50；將x=5代入可推估成績為5\times 5+50=75分\\ P(X\gt 77)=P(Z\gt {77-75\over \sqrt{16}})= P(Z\gt 0.5)=0.31，故選\bbox[red,2pt]{(D)}$$
解答：$$迴歸直線斜率b_1 =r\times {\sigma(y)\over \sigma(x)} =0.6\times {10\over 5} =1.2 \Rightarrow 迴歸直線方程式: y=1.2x+60\\ 將x=10代入上式\Rightarrow y=1.2\times 10+60=72，故選\bbox[red,2pt]{(C)}$$
解答：$$\cases{迴歸直線斜率b_1 \\ x平均值\bar x\\ y平均值\bar y} \Rightarrow 迴歸直線方程式\;y-\bar y=b_1(x-\bar x) \Rightarrow y=5(x-10)+90\\ 將x=12 代入方程式\Rightarrow y=5(12-10)+90 =100，故選\bbox[red,2pt]{(A)}$$
解答：$$\cases{迴歸直線斜率b_1 \\ 相關係數r\\ 判定係數R^2} \Rightarrow b_1= r\cdot {\sigma(y)\over \sigma(x)} \Rightarrow 1.4=r\cdot {10\over 5} \Rightarrow r=0.7 =\sqrt{R^2} \Rightarrow R^2=0.7^2=0.49\\，故選\bbox[red,2pt]{(A)}$$
=============== END =====================
沒有留言:
張貼留言