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2019年9月29日 星期日

107年公務人員初等考試-統計學大意詳解


107年公務人員初等考試試題
等別:初等考試
類科 :統計
科目:統計學大意


$$P(A\mid B)=\frac{P(A\cap B)}{P(B)} \Rightarrow 0.4={P(A\cap B)\over 0.6} \Rightarrow P(A\cap B)=0.24\\ \Rightarrow P(A)=P(A\cap B)+ P(A\cap B^c) =0.24+0.2=0.44 \\\Rightarrow P(A\cup B)= P(A)+P(B)-P(A\cap B)=0.44+0.6-0.24 =0.8,故選\bbox[red,2pt]{(C)}$$


:$$假設\begin{cases}A 硬幣:兩面皆10元\\ B硬幣:兩面皆國父\\ C硬幣:一面10元,另一面國父 \end{cases}\\\Rightarrow \begin{cases}取到A的機率為1/3:,A連擲3次皆10元的機率為1\\ 取到B的機率為1/3:,B連擲3次皆10元的機率為0\\ 取到C的機率為1/3:,C連擲3次皆10元的機率為1/8 \end{cases}\\
\Rightarrow 擲3次皆10元的機率={1\over 3}\times 1+{1\over 3}\times {1\over 8}= {3\over 8} \Rightarrow P(取到A\mid 擲3次皆10元)= {1/3\over 3/8}={8\over 9}\\,故選\bbox[red,2pt]{(C)}$$


:$$P(30\le X\le 60)= P\left({30-50\over 10}\le {X-50\over 10}\le {60-50\over 10}\right) = P(-2\le Z\le 1) =z_{1}-z_{-2}\\
題目所附之Z表為P(0< Z< z)=\alpha \\\Rightarrow P(-2\le Z\le 1) =2P(0< Z< 1)+P(0< Z< 2)-P(0< Z< 1)\\ =2\times 0.3413+(0.4772-0.3413)=0.8185 \Rightarrow 筆數=200\times 0.8185=163.7,故選\bbox[red,2pt]{(B)} $$


:$$P\left( |X-2|<2\right) = P(0< X< 4) = P(X=1)+P(X=2)+P(X=3)\\= {n\choose 1}p(1-p)^{n-1} +{n\choose 2}p^2(1-p)^{n-2} +{n\choose 3}p^3(1-p)^{n-3} \\ = {10\choose 1}0.2(0.8)^{9} +{10\choose 2}(0.2)^2(0.8)^{8} +{10\choose 3}(0.2)^3(0.8)^{7} =0.7717,故選\bbox[red,2pt]{(A)}$$



:$$\int f(x)\;dx=1\Rightarrow \int_0^2 (1-Kx)dx =1 \Rightarrow \left. \left[x-{K\over 2}x^2 \right]\right|_0^2 = 2-2K=1 \Rightarrow K={1\over 2},故選\bbox[red,2pt]{(B)}$$


:$$L(\lambda)=f(x_1;\lambda)\times f(x_2;\lambda)\times \cdots \times f(x_n;\lambda) ={\lambda^{x_1}e^{-\lambda} \over  x_1!} \cdot {\lambda^{x_2}e^{-\lambda} \over x_2!}\cdots {\lambda^{x_n}e^{-\lambda} \over x_n!}\\ ={e^{-n\lambda}\lambda^{\sum_{i=1}^n x_i} / \Pi_{i=1}^n x_i!} \Rightarrow \ln{L(\lambda)} =-n\lambda+ \sum_{i=1}^n x_i\ln \lambda -\ln{\left( \Pi_{i=1}^n x_i!\right)} \\ 令{d\over d\lambda}\ln{L(\lambda)}=0 \Rightarrow -n+ (\sum_{i=1}^n x_i)/\lambda =0 \Rightarrow \lambda = (\sum_{i=1}^n x_i)/n =\bar x \\ \Rightarrow 標準差= \sqrt{\lambda} =\sqrt{\bar x},故選\bbox[red,2pt]{(D)}$$




:$$P(|Z|< z)=0.98, 經查表可知z=2.33 \Rightarrow 2.33 \times \sqrt{0.5^2 \over n} \le 3\% \Rightarrow {0.25\over n}\le ({0.03 \over 2.33})^2 \\ \Rightarrow n\ge {0.25 \over (0.03/2.33)^2}=1508.03,故選\bbox[red,2pt]{(B)}$$



樣本數增加,錯誤機率會減少,故選\(\bbox[red,2pt]{(B)}\)。



$$\begin{cases} X_1,X_2,\dots,X_{n1}\sim N(\mu_1,\sigma_1) \Rightarrow \begin{cases}樣本平均數\bar X=\sum X_i/n_1 \\ 樣本變異數s_x^2= \sum (X_i-\bar X)^2/(n_1-1) \end{cases}\\ Y_1,Y_2,\dots,Y_{n2}\sim N(\mu_2,\sigma_2) \Rightarrow \begin{cases}樣本平均數\bar Y=\sum Y_i/n_2 \\ 樣本變異數s_y^2= \sum (Y_i-\bar Y)^2/(n_2-1) \end{cases}\end{cases}\\
\Rightarrow F={s_1^2/\sigma_1^2\over s_2^2/\sigma_2^2} \sim F(n_1-1,n_2-1),故選\bbox[red,2pt]{(D)} $$


:$$\begin{array}{}\hline
變源& SS & df & MS & F\\\hline
組間& SSB & 3-1=2 & MSB & \\
組內& SSE & 14-2=12 & MSE=5 &\\\hline
總和& SST=100 & 15-1=14 & &\\\hline
\end{array}\\
\Rightarrow  SSE=12\times 5=60 \Rightarrow SSB=SST-SSE=100-60=40 \\\Rightarrow MSB=SSB/2=40/2=20 \Rightarrow F=MSB/MSE = 20/5=4,故選\bbox[red,2pt]{(C)}$$


:$$依判定係數R^2定義,故選\bbox[red,2pt]{(A)}$$


:$$R^2=b_1^2{s_x^2 \over s_y^2} =2^2{1.2^2 \over 3^2} =0.64 ,故選\bbox[red,2pt]{(C)}$$


:$$誤差期望值為0是迴歸的求解的依據,無需檢視 ,故選\bbox[red,2pt]{(A)}$$


:$$ {x_1+2x_2+2x_3+2x_4 +x_5\over 8}= {21+38+46+42+29 \over 8}=22 ,故選\bbox[red,2pt]{(B)}$$


:$$P(\bar X>136)=P\left( {\bar X-\mu \over \sigma/\sqrt{n}> }> {136-\mu \over \sigma/\sqrt{n}}\right) = P\left(Z>{136-130 \over 21/\sqrt{49}} \right) =P(Z>2)\\ 查表可得 P(Z>2)=0.5-0.4772 = 0.0228,故選\bbox[red,2pt]{(B)}$$


:$$發燒比率= A型發燒+B型發燒=30\%\times 70\% + 70\%\times 35\% =21\% +24.5\% =45.5\%\\ \Rightarrow \begin{cases} A型發燒\div 發燒 =21/45.5 = 0.46 \\ B型發燒\div 發燒= 24.5/45.5=0.54\end{cases} \Rightarrow 發燒的病人中屬B型流感的機率大於屬A型流感\\,故選\bbox[red,2pt]{(B)}$$



「沒有證據拒絕」與「一定不拒絕」不完全相等,故選\(\bbox[red,2pt]{(C)}\)



:$$\begin{cases} X\sim Po(\lambda) \Rightarrow P(X=k)={\lambda^ke^{-\lambda} \over k!}\\ e^\lambda = 1+\lambda+ {\lambda^2 \over 2!}+ {\lambda^3 \over 3!}+ \cdots\end{cases}\\ \Rightarrow P(X>1)=P(X=2)+P(X=3)+\cdots = e^{-\lambda}\left({\lambda^2 \over 2!} + {\lambda^3 \over 3!} + \cdots\right) = e^{-\lambda}\left(e^\lambda-1-\lambda\right)\\ = 1-e^{-\lambda}-\lambda e^{-\lambda} \\
由題意知: P(X>1)>0.8就要換系統 \Rightarrow 1-e^{-\lambda}-\lambda e^{-\lambda}>0.8 \Rightarrow e^{-\lambda}(1+\lambda)<0.2\\ \begin{cases}\lambda=4\Rightarrow e^{-\lambda}(1+\lambda)=0.091 \\\lambda=3\Rightarrow e^{-\lambda}(1+\lambda)=0.199 \\ \lambda=2 \Rightarrow e^{-\lambda}(1+\lambda)=0.406\end{cases} \Rightarrow 當\lambda \ge 3時, P(X>1)>0.8 ,故選\bbox[red, 2pt]{(D)}$$


:兩母體變異需相等,才能用t檢定,故選\(\bbox[red,2pt]{(B)}\)


:$$\bar x=(n_1\bar x_1+n_2 \bar x_2+ n_3\bar x_3) \div (n_1 + n_2 +n_3) = 6(33+35+34)\div 18=34\\
SSB = n_1(\bar x_1-\bar x)^2+ n_2(\bar x_2-\bar x)^2 +n_3(\bar x_3-\bar x)^2 \\= 6((33-34)^2 +(35-34)^2 +(34-34)^2) =12\\
SSW = (n_1-1)s_1^2 + (n_2-1)s_2^2 +(n_3-1)s_3^2 = 5(5+6+4)=75\\ \Rightarrow \begin{cases} SSB=12 \\ SSW=75\end{cases}
,故選\bbox[red,2pt]{(D)}$$




$$\begin{array}{c|cc}
i& O_i & E_i\\\hline
1 & 275 & 300\\
2 & 275 & 300\\
3 & 350 & 300
\end{array} \Rightarrow \chi^2=\sum_{i=1}^3(O_i-E_i)^2/E_i ={1\over 300}\left((275-300)^2 +(275-300)^2 +(350-300)^2\right)\\ =(625+625+2500)\div 300 = 12.5,故選\bbox[red,2pt]{(C)}$$


:$$X,Y的相關係數與X/7, Y的相關係數是一樣的,故選\bbox[red,2pt]{(C)} $$


:$$X\sim N(\mu=4,\sigma^2=9) \Rightarrow P(X-4 \le 6)=P\left({x-4\over 3}\le {6\over 3} \right) =P(Z\le 2)\\ 查表可得 z_2=0.5+ 0.4772=0.9772 \Rightarrow P(X-4\le 6)=0.9772,故選\bbox[red,2pt]{(D)} $$


:$$信賴區間的寬度比=\sqrt{n_2\over n_1}=\sqrt{10000\over 100} =\sqrt{100}=10,故選\bbox[red,2pt]{(D)}$$


:$$\begin{cases} P(X=0)=P(X=3) = {25\choose 3}/{50\choose 3}\\ P(X=1)=P(X=2) = {25\choose 2}{25\choose 1}/{50\choose 3}\end{cases} \Rightarrow {P(X=1)\over P(X=0)}= {{25\choose 2}{25\choose 1}\over {25\choose 3}}={75\over 23},故選\bbox[red,2pt]{(D)}$$


:$$\begin{cases} (A)=n-1=20-1=19\\ (B)=40\div 1=40 \\(C)=n-2= 20-2=18\\ (B)\div (D)= 50/3 \Rightarrow (D)=40\times 3\div 50 =2.4\end{cases},故選\bbox[red,2pt]{(D)}$$


:$$\begin{array}{|c|c|c|}\hline
 &贊成 &反對\\\hline
男& O_1=30/E_1=100\times {70\over 200}=35 &O_2=70/E_2=100\times {130\over 200}=65\\\hline
女& O_3=40/E_3=100\times {70\over 200}=35 &O_4=60/E_4=100\times {130\over 200}=65\\\hline
\end{array}\\ \Rightarrow \chi^2={(30-35)^2\over 35}+ {(70-65)^2\over 65}+ {(40-35)^2\over 35}+ {(60-65)^2\over 65} = {10\over 7}+ {10\over 13}={ 200\over 91},故選\bbox[red,2pt]{(C)}$$



$$ \begin{cases} P((\bar P-p)\le E)=0.95 \\ P((\bar P-p)\le E^*)=0.9 \end{cases} \Rightarrow E>E^*,故選\bbox[red,2pt]{(C)}$$


:$$\begin{cases} R_A=[0.25-3\times 0.025,0.25+3\times 0.025] =[0.175,0.325] \\ R_B=[0.3-3\times 0.05,0.3+3\times 0.05] =[0.15,0.45]\end{cases} \Rightarrow \begin{cases} 0.35 \notin R_A \\ 0.4 \in R_B\end{cases}\\ \Rightarrow 聯盟A的推舉球屬於離群點,故選\bbox[red,2pt]{(C)}$$


:$$(B)\lambda={1\over 15} \Rightarrow f(t)=\lambda e^{-\lambda t} = {1\over 15}e^{- {t\over 15}}\\
(C)\int_{20}^{\infty} {1\over 15}e^{- {t\over 15}}\;dt =\left. \left[-e^{- {t\over 15}} \right] \right|_{20}^{\infty} = e^{-4/3}\\ (D)\int_{10}^{15} {1\over 15}e^{- {t\over 15}}\;dt =\left. \left[-e^{- {t\over 15}} \right] \right|_{10}^{15} = e^{-2/3}-e^{-1}\\ \Rightarrow 故選\bbox[red,2pt]{(A)}$$



X是離散型,所以(B)與(C)皆錯誤;又\(0\le f(x)\le 1\),故選\(\bbox[red,2pt]{(D)}\)


:$$\begin{cases}P(M\mid N)= {P(M\cap N) \over P(N)} ={90\over 90+30} ={90\over 120}\\ P(M)= {90+810\over 90+810+30+270} ={900\over 1200 } \end{cases} \Rightarrow P(M\mid N)=P(M),故選\bbox[red,2pt]{(C)}$$



只有(D)正確,故選\(\bbox[red,2pt]{(D)}\)



由第6題可知\(\bar x\)為\(\lambda\)的估計量(也是不偏),因此\(\bar x^2\)為\(\lambda^2\)的估計量,故選\(\bbox[red,2pt]{(B)}\)


:$$\begin{cases}y_i=\beta x_i+\varepsilon_i \\ \hat y_i=\hat \beta x_i \end{cases} \Rightarrow SSE= \sum(y_i-\hat y_i)^2 = \sum(y_i-\hat \beta x_i)^2 =\sum \left(y_i^2-2\hat \beta\sum x_iy_i +\hat \beta^2 \sum x_i^2\right)\\
令{d\over d \hat \beta}SSE=0 \Rightarrow -2\sum x_iy_i+ 2\hat \beta \sum x_i^2=0 \Rightarrow \hat \beta= {\sum x_iy_i \over \sum x_i^2} = {3\over 10}=0.3,故選\bbox[red, 2pt]{(B)}$$註:本題的線性迴歸模式跟一般的不一樣(沒有\(\alpha\))!!



直方圖一般用來表達連續型數據,故選\(\bbox[red,2pt]{(B)}\);離散型一般以長條圖來表達。


:$$\begin{cases} \bar x=165 \\ \bar y= 165\end{cases} 代入 \bar y=b_0+b_1 \bar x \Rightarrow 165=b_0+{4\over 3}\times 165 \Rightarrow b_0=165-220= -55 \\ \Rightarrow \hat y=-55+{4\over 3}\times x =-55+{4\over 3}\times 156 = -55+208 =153,故選\bbox[red, 2pt]{(C)}$$


:$$\bar p= {n_1\hat p_1+n_2 \hat p_2\over n_1+n_2} = {0.6+0.5\over 2}=0.55 \Rightarrow s_{\hat p_1-\hat p_2} = \sqrt{\bar p(1-\bar p)\over 1/n_1+1/n_2} =\sqrt{0.55\times 0.45 \over 1/400+1/400} \approx 0.035\\ \Rightarrow 檢定統計量Z={(\hat p_1-\hat p_2)-(p_1-p_2) \over s_{\hat p_1-\hat p_2}}= {(0.6-0.5)-0 \over 0.035}= {0.1\over 0.035} \approx 2.84,故選\bbox[red,2pt]{(C)}$$


:$$X\sim U(0,\theta) \Rightarrow \begin{cases} \mu_k'=E(X^k)\\ m_k'={1\over n}\sum_{i=1}^n X_i^k\end{cases} \Rightarrow \begin{cases} \mu_1'=E(X)=\theta/2 \\ m_1'={1\over n}\sum_{i=1}^n X_i =\bar X\end{cases} \Rightarrow \theta/2 = \bar X \Rightarrow \theta = 2\bar X\\,故選\bbox[red,2pt]{(D)}$$



ANOVA檢定為顯著性,但不能指出哪一對相互影響,需事後檢定,而Tukey檢定的目的就在此,故選\(\bbox[red,2pt]{(B)}\)




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