網頁

2021年7月21日 星期三

109年專科學力鑑定-初級統計詳解

教育部109年自學進修專科學校學力鑑定考試

專業科目(一): 初級統計

解答:$$估計目的在於推測母體參數,並非推測樣本參數,故選\bbox[red,2pt]{(A)}$$
解答:$$樣本為母體之部份,藉由樣本來推論母體,,故選\bbox[red,2pt]{(B)}$$
解答:$$5位都沒有超過175公分的機率:({7\over 10})^5;5位中有1位超過175公分的機率C^5_1({7\over 10})^4\times {3\over 10};\\ 因此所求之機率為1-({7\over 10})^5 -C^5_1({7\over 10})^4\times {3\over 10}=0.47178,故選\bbox[red,2pt]{(B)}$$
解答:$$最適合應是圓餅圖,長條圖次之 ,故選\bbox[red,2pt]{(B)}$$
解答:$$樣本平均數與母體分配相同,故選\bbox[red,2pt]{(C)}$$
解答:$$依定義,故選\bbox[red,2pt]{(B)}$$
解答:$$X\sim P(k,T) = {(\lambda T)^k \over k!}e^{-\lambda T} \Rightarrow P(0,T)= e^{-\lambda T} \Rightarrow 欲求之機率為1-P(0,T)=1-e^{-\lambda T}\\,將\cases{\lambda ={1\over 100}\\T=200} 代入,可得1-P(0,T)=1-{1\over e^2} =1-0.135=0.865\\ 當然也可用二項式來計算,也就是1-({99\over 100})^{200}=1-0.134=0.866,故選\bbox[red,2pt]{(D)}$$
解答:$$假設真正會做此題的學生佔比為p,不會做的佔比為1-p;\\真正會做的,答對機率為1,不會做的,答對機率為{1\over 5}\\,因此p\times 1+ (1-p)\times {1\over 5}={80\over 100} \Rightarrow p={3\over 4}=75\%,故選\bbox[red,2pt]{(D)}$$
解答:$$點數和最大值為2+6=8,不可能大於等於9,故選\bbox[red,2pt]{(A)}$$
解答:$$中央極限定理需在大樣本的條件下才成立,故選\bbox[red,2pt]{(B)}$$
解答:$$正面次數大於等於2的情形:正正正、正正負、負正正,正負正,共4種情況\\,全部有2^3=8種,因此機率為{4\over 8}=0.5,故選\bbox[red,2pt]{(D)}$$
解答:$$點數和大於等於9的情形:(6,3-6)\to 4種、(5,4-6)\to 3種、(4,5-6)\to 2 種、(3,6)\to 1種\\共有4+3+2+1=10種,因此機率為{10\over 6\times 6} \approx 0.278,故選\bbox[red,2pt]{(A)}$$
解答:$$擲骰子出現2的機率為{1\over 6},擲5次都沒出現2的機率為({5\over 6})^5\\,至少出現1次2的機率為1-({5\over 6})^5= 0.598,故選\bbox[red,2pt]{(C)}$$
解答:$$相對變異\text{(Coefficient Variance)} = {標準差\over 平均數} \Rightarrow \cases{A:5/71=0.07\\ B:6/89=0.067\\ C:4/56= 0.071} \Rightarrow B班最小,故選\bbox[red,2pt]{(B)}$$
解答:$$在大樣本的條件下,樣本平均數分配近似母體平均數;\\其他標準差、變異數與母體皆不同,故選\bbox[red,2pt]{(A)}$$
解答:$$六個數字由小至排序為25,25,28,32,33,60,\\中位數為第3與第4的平均值,即(28+32)\div 2=30,故選\bbox[red,2pt]{(B)}$$
解答:$$區間估計值:\bar x\pm t_{\alpha/2}{s\over \sqrt n} \Rightarrow 區間長度為2\times t_{\alpha/2}\times {s\over \sqrt n},其中\cases{n=9\\ s=30\\ t分配自由度為n-1=8\\ 1-\alpha=0.95 \Rightarrow \alpha/2=0.025}\\,再由試題所給的t值可知: P(t(8)\gt \color{blue}{2.306})=0.025 \Rightarrow t_{0.025}=2.306\\ \Rightarrow 2\times t_{\alpha/2}\times {s\over \sqrt n}=2\times 2.306\times {30\over \sqrt 9} =46.12,故選\bbox[red,2pt]{(C)}$$
解答:$$n\ge (z_{\alpha/2})^2{\sigma^2 \over E^2} =1.96^2\times {30^2 \over 8^2}= 54.0225,故選\bbox[red,2pt]{(C)}$$
解答:$$\bar x+z_{\alpha/2}\times {\sigma \over \sqrt n} = 70+1.645\times {5\over \sqrt {36}} =70+1.37 =71.37,故選\bbox[red,2pt]{(A)}\\ 註:試題提供P(Z\gt 1.645)=\alpha/2=0.1/2=0.05 \Rightarrow z_{\alpha/2} =1.645$$
解答:$$區間長度=2\times z_{\alpha/2}\times \sqrt{p(1-p)\over n} =2\times 1.96 \times \sqrt{(10/40)(1-10/40)\over 40} =0.268,故選\bbox[red,2pt]{(D)}$$
解答:$$男生:X\sim N(\mu,\sigma^2)=N(170,4^2)\Rightarrow P(X\lt 166)=P({X-170\over 4} \lt {166-170\over 4}) =P(Z\lt -1)\\ =P(Z\gt 1)=0.1587\\ 女生:Y\sim N(\mu,\sigma^2)=N(164,5^2) \Rightarrow P(Y\lt 166) = P({Y-164\over 5} \lt {166-164\over 5}) = P(Z\lt 0.4) \\ = 1-P(Z\gt 0.4)=1-0.3446=0.6554\\ 假設全校人數為n,則男生、女生各有{n\over 2}人,因此{男生低於166 人數\over 全校低於166人數 }\\ ={{n\over 2}\times 0.1587 \over {n\over 2}\times 0.1587+ {n\over 2}\times 0.6554} ={  0.1587 \over   0.1587+   0.6554} ={ 0.1587 \over 0.8141} =0.1949,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{\mu = np = 50\times 4\% = 2 \\ \sigma = \sqrt{np(1-p)} =\sqrt{50\times 4\% \times 96\%}\approx 1.386} \\ 50的5\% =2.5 \Rightarrow P(X\gt 2.5)= P(Z\gt {2.5-2\over 1.386}) = P(Z\gt 0.36)=1-P(Z\lt 0.36)\\ =1-0.6406=0.3594,故選\bbox[red,2pt]{(D)}$$
解答:$$無論出現正或反面,機率皆為1/2,因此機率的多寡取決於共擲次數;\\乙贏的情形:反正、反反反反正、反反反反反反反正、\dots \Rightarrow 共擲3k+2次,k\in \mathbb{Z}\\ 因此機率為\sum_{k=0}^\infty ({1\over 2})^{3k+2} ={1\over 2^2} +{1\over 2^5} +{1\over 2^{8}} +\cdots = {1/2^2\over 1-1/2^3} = {1/4 \over 7/8} ={2\over 7} =0.2857,故選\bbox[red,2pt]{(C)}$$
解答:$$E=z_{\alpha/2}\times \sqrt{p(1-p)\over n} =z_{0.05}\times \sqrt{{20\over 50}(1-{20\over 50}) \over 50} = 1.645\times 0.069 \approx 0.114 =11.4\%,故選\bbox[red,2pt]{(C)}$$
解答:$$顯著水準越大代表拒絕H_0的機率越大,型二錯誤的機率就越小,故選\bbox[red,2pt]{(A)}$$
解答:$$擲骰子240次,每個數字出現的期望值(E_i)都是240\div 6=40\\\begin{array}{c|cccccc} &1 & 2 & 3& 4 & 5 & 6\\\hline O_i & 25 & 46 &51 & 28 & 55 & 35 \\\hline E_i & 40 & 40 & 40 & 40 & 40 & 40\end{array}\\ 因此卡方統計量為\sum_{i=1}^6{(O_i-E_i)^2\over E_i} = (15^2+6^2 +11^2+12^2+15^2+5^2)\div 40 \\=776\div 40=19.4,故選\bbox[red,2pt]{(C)}$$
解答:$$\begin{array}{l|l|l|l } &df & SS & MS& F\\\hline \text{Treatment} & 4-1=3 & SSB &MSB=55 &  \\\hline \text{Error} & 20-4=16 & SSW =55& MSW &  \\\hline \text{total} & 20-1=19 & SST   \end{array}\\ 由題意知\cases{k=4\\ n=5k=20\\ MSB=55 \\ SSW=55} \Rightarrow MSW={SSW \over 16} ={55\over 16} \Rightarrow    F={MSB\over MSW} = 55\times {16\over 55} =16\\,故選\bbox[red,2pt]{(A)}$$
解答:$$\begin{array}{l|l|l|l } &df & SS & MS& F\\\hline \text{Treatment} & 3-1=2 & SSB &MSB=50 &  \\\hline \text{Error} & 90-3=87 & SSW & MSW &  \\\hline \text{total} & 90-1=89 & SST=180  \end{array}\\ 由題意知\cases{k=3\\ n=30k=90\\ MSB=50 \\ SST=180} \Rightarrow SSB=MSB\times 2=50\times 2=100 \\\Rightarrow SSW=SST-SSB=180-100=80 \Rightarrow MSW= SSW/87= 80/87\\ \Rightarrow F=MSB/MSW = 50\times {87\over 80} =54.375,故選\bbox[red,2pt]{(D)}$$
解答:$$\begin{array}{l|l|l|l }  X & Y & X^2 &Y^2 &XY\\\hline 1 & 0 & 1 & 0 & 0 \\\hline 1 & 2 & 1& 4 & 2 \\\hline 2 & 0 & 4 & 0 & 0 \\\hline 3 & 4 & 9 & 16 & 12\\\hline \sum X= 7 & \sum Y=6 & \sum X^2=15 & \sum Y^2=20 & \sum XY=14  \end{array}\\ 相關係數 r={n\sum XY- \sum X \sum Y\over \sqrt{n\sum X^2-(\sum X)^2 }\cdot \sqrt{n\sum Y^2-(\sum Y)^2 }} ={4\times 14-7\times 6 \over \sqrt{4\times 15-7^2}\cdot \sqrt{4\times 20-6^2}}\\ = {14\over \sqrt{11} \cdot \sqrt{44}} ={14\over 22} =0.636,故選\bbox[red,2pt]{(C)}$$
解答:$${檢查有病且真有病\over 檢查有病} ={有病被正常檢出\over 有病被正常檢出+沒病被錯誤檢出}={2\%\times 90\% \over 2\%\times 90\% + 98\% \times 15\%} \\= 0.109,故選\bbox[red,2pt]{(B)}$$
解答:$$\langle x_i\rangle =80,35,28,39,75 \Rightarrow \cases{\sum_{i=1}^5 x_i =257 \\ \sum_{i=1}^5 x_i^2=15555} \\\Rightarrow 樣本變異數={\sum x_i^2-{1\over 5}(\sum x_i)^2 \over 5-1} ={1555-(257^2)/5 \over 4} =586.3,故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{P(A)=0.6\\ P(B)=0.3\\ P(A\cup B)=0.8} \Rightarrow P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.1\\ (A)\times: 1-P(A)=1-0.6=0.4 \ne P(B)=0.3\\ (B) \times: P(A)\times P(B)=0.6\times 0.3=0.18 \ne P(A\cap B)=0.1 \\(C)\bigcirc: 由(A)及(B)可知:A、B非互斥也非獨立\\(D)\times: 可判斷\\,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{P(A)=0.6\\ P(B)=0.3\\ P(A\cap B)=0.2} \Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.7\\ \Rightarrow P(A\cup B')= 1-(P(A\cup B)-P(A)) = 1-(0.7-0.6)=0.9,故選\bbox[red,2pt]{(D)}$$
解答:$$A\cap B= \varnothing \Rightarrow P(A\cap B')= P(A)=0.5,故選\bbox[red,2pt]{(A)}$$
解答:$$A、B獨立\Rightarrow P(A\cap B)=P(A)\times P(B) \Rightarrow P(A\cup B)= P(A)+P(B)-P(A)\times P(B)\\ \Rightarrow 0.8 = 0.5+P(B)-0.5P(B) \Rightarrow 0.3=0.5P(B) \Rightarrow P(B)={3\over 5}=0.6,故選\bbox[red,2pt]{(D)}$$
解答:$$\sigma^2(W)= \sigma^2(3X-2Y)=3^2\sigma^2(X)+2^2\sigma(Y)-2\cdot 3\cdot 2\cdot Cov(X,Y)=9\cdot 2+4\cdot 4-0\\ =34 \Rightarrow \sigma(W)= \sqrt{34}= 5.83,故選\bbox[red,2pt]{(A)}$$
解答:$$至少有一件的機率=1-0件的機率=1-e^{-4\times 0.5} =1-{1\over e^2} =1-0.135 =0.865,故選\bbox[red,2pt]{(B)}$$
解答:$$數據與第28題相同,因此斜率=r\times {\sigma(Y)\over \sigma (X)} =0.636 \times {\sqrt{n\sum Y^2-(\sum Y)^2} \over \sqrt{n\sum X^2-(\sum X)^2}} \\=0.636 \times {\sqrt{4\times 20-6^2} \over \sqrt{4\times 15-7^2}}  =0.636\times 2= 1.272,故選\bbox[red,2pt]{(B)}$$
解答:$$X\sim N(165,4^2) \Rightarrow \bar X\sim N(165,\sqrt{4^2\over 16})=N(165,1) \Rightarrow P(\bar X\lt 163) =P(Z\lt {163-165 \over 1})\\ = P(Z\lt -2) =P(Z\gt 2) =1-P(Z \lt 2)=1-0.9772=0.0228,故選\bbox[red,2pt]{(D)}$$
解答:$$\cases{P(X=0)=0.25\\ P(X=1)=0.5\\ P(X=2)=0.25} \Rightarrow E(X)=0\times 0.25+1\times 0.5 +2\times 0.25=1\\ \Rightarrow E(Y)= E(3X-4)= 3E(X)-4=3-4=-1,故選\bbox[red,2pt]{(C)}$$
解答:$$抽到的都是襄理的機率=C^4_2 / C^6_2 =0.4 \Rightarrow 至少抽到一位經理的機率=1-0.4= 0.6\\,故選\bbox[red,2pt]{(B)}$$
解答:$$P(X\lt 163) = P(Z\lt {163-165\over 4})=P(Z\lt -0.5) =P(Z\gt 0.5)=1-P(Z\lt 0.5) \\=1-0.6915= 0.3085,故選\bbox[red,2pt]{(C)}$$
解答:$$由P(Z\lt 0.25)=0.5987 \approx 60\%可知:只要Z\gt 0.25就代表成績在前40\%\\ Z={X-280\over 20}=0.25 \Rightarrow X=285,故選\bbox[red,2pt]{(C)}$$
解答:$$(A)\times: 點數不是奇數就是偶,因此A、B互斥\\(B) \times:\cases{P(A)\gt 0\\ P(B)  \gt 0\\ P(A\cap B)=0} \Rightarrow P(A)P(B) \ne P(A\cap B) \Rightarrow A、B不獨立\\(C)\bigcirc: \cases{B=\{2,4,6\} \\ C=\{1,2\} \\ B\cap C=\{2\}} \Rightarrow \cases{P(B)=1/2\\ P(C)=1/3 \\P(B\cap C)=1/6} \Rightarrow P(B)P(C)= {1\over 6}=P(B\cap C)\Rightarrow B、C獨立 \\(D)\times: \cases{A=\{1,3,5\} \\C=\{1,2\}\\ A\cap C]\{1\}} \Rightarrow \cases{P(A)=1/2\\ P(B)=1/3\\ P(A\cap C)=1/6} \Rightarrow P(A)P(C)={1\over 6}=P(A\cap C) \Rightarrow A、C獨立\\,故選\bbox[red,2pt]{(C)}$$
解答:$$\cases{P(A)=0.5\\ P(A\cap B')=0.2 \\ P(A\cap B)=P(A)P(B)} \Rightarrow P(A\cap B)=0.5-0.2=0.3 =P(A)P(B)=0.5\cdot P(B) \\ \Rightarrow P(B)=0.6 \Rightarrow P(A\cup B')= P(A)+P(B')-P(A\cap B')= 0.5+(1-0.6)-0.2=0.7\\,故選\bbox[red,2pt]{(D)}$$
解答:$$n\ge (z_{\alpha/2})^2 \times {p(1-p)\over E^2} =(z_{0.025})^2 \times {{10\over 40}\times (1-{10\over 40}) \over 0.05^2} =1.96^2 \times {3\over 16}\times 400=288.12 \\ \Rightarrow n=289,故選\bbox[red,2pt]{(B)}$$
解答:$$\cases{P(X=1)=0.1\\ P(X=2)=0.2\\ P(X=3)=0.3\\ P(X=4)=0.4} \Rightarrow \cases{E(X)=1\times 0.1+2\times 0.2+3\times 0.3 +4\times 0.4=3 \\ E(X^2)=1^2\times 0.1+2^2 \times 0.2+3^2 \times 0.3 +4^2\times 0.4= 10}\\ \Rightarrow \sigma(X)= \sqrt{EX^2 -(EX)^2} =\sqrt{10-3^2}=1 \Rightarrow \sigma(Y)=\sigma(2X-2)=2\sigma(X)=2,故選\bbox[red,2pt]{(A)}$$
解答:$$P(X\lt 20\%) =P(Z \lt {0.2-0.15\over \sqrt{0.2\times 0.8/200}})= P(Z\lt 1.77)=0.962 \\\Rightarrow P-\text{value} =1-0.962=0.038,故選\bbox[red,2pt]{(B)}$$這題..再想想!!
解答:$$ P-\text{value} =0.0158 \Rightarrow 1-0.0158=0.9842 \Rightarrow z=2.15 \Rightarrow {74-72\over 8/\sqrt n}=2.15 \\ \Rightarrow \sqrt n=2.15\times 4=8.6 \Rightarrow n=8.6^2=73.96 ,故選\bbox[red,2pt]{(D)}$$
解答:$$A、B、C獨立\Rightarrow \cases{A、B'獨立\\ A、B'、C獨立} \quad \Rightarrow  P((A\cap B')\cup C) =P(A\cap B')+ P(C)-P(A\cap B'\cap C) \\=P(A)P(B')+P(C)-P(A)P(B')P(C) =0.4(1-0.2) +0.1-0.4(1-0.2)\cdot 0.1 = 0.388\\,故選\bbox[red,2pt]{(C)}$$
=================== END ==========================
 

沒有留言:

張貼留言