朱式幸福: 113年專科學力鑑定-初級統計詳解

教育部113年自學進修專科學校學力鑑定考試試題本

專業科目(一)：初級統計

解答:

$「超過」500次，故選\bbox[red, 2pt]{(B)}$

解答:

$標準常態分配的平均數為0不是1，故選\bbox[red, 2pt]{(D)}$

解答:

$樣本是母體的子集合，故選\bbox[red, 2pt]{(C)}$

解答:

$由小到大排序:50,60,90,120,130,140,150,180，其中第4與第5的平均數=(120+130)\div 2 =125\\，故選\bbox[red, 2pt]{(C)}$

解答:

$只有兩種結果(如:正面反面)，故選\bbox[red, 2pt]{(A)}$

解答:

$常態分配圖為左右對稱，故選\bbox[red, 2pt]{(D)}$

解答:

$盒鬚圖表達四分位數的相對位置，不適合描述婚姻，故選\bbox[red, 2pt]{(A)}$

解答:

$變異係數=標準差/平均數，標準差與平均數同除1000，變異係數值不會改變，故選\bbox[red, 2pt]{(D)}$

解答:

$雨量可以加減乘除，故選\bbox[red, 2pt]{(C)}$

解答:

$連續型之隨機變數，其某點機率皆為0，只能計算區間之機率，故選\bbox[red, 2pt]{(C)}$

解答:

$莖葉圖可計算，非名目尺度；故選\bbox[red, 2pt]{(C)}$

解答:

$P(Z\lt 1.05) =0.8531 \Rightarrow P(Z\gt 1.05) =1-0.8531=0.1469 =P(Z\lt -1.05) \\ \Rightarrow P(-1.05\lt Z\lt 1.05) =P(Z\lt 1.05)-P(Z\lt -1.05) =0.8531-0.1469= 0.7062，故選\bbox[red, 2pt]{(B)}$

解答:

$等紅綠燈時間服從圴勻分配，P(X\gt 20)=1-{20\over 60}={2\over 3}=0.67，故選\bbox[red, 2pt]{(C)}$

解答:

$不偏僅是抽樣估計平均數等於母體參數值，故選\bbox[red, 2pt]{(B)}$

解答:

$P(X\gt 17.6) =P(Z\gt {17.6-12\over 2}) =P(Z\gt 2.8) =1-P(Z\lt 2.8) =1-0.9974 =0.0026\\，故選\bbox[red, 2pt]{(A)}$

解答:

$平均數4.3\gt 中位數3.6\gt 眾數3.3 \Rightarrow 右偏，故選\bbox[red, 2pt]{(A)}$

解答:

$由小到大排序:a_1=5,a_2=6,a_3=12,a_4=13,a_5=15,a_6=18,a_7=22,a_8=50 \\ \Rightarrow \cases{Q_1=(a_2+a_3 )/2=9\\ Q_3= (a_6+a_7)/2= 20} \Rightarrow IQR=Q_3-Q_1= 20-9= 11，故選\bbox[red, 2pt]{(B)}$

解答:

$例:1,1,1,2,2,2,3,4 \Rightarrow 1與2都是眾數，故選\bbox[red, 2pt]{(A)}$

解答:

$P(44000\lt X\lt 5600) =P({44000-50000\over 3000} \lt Z\lt {56000-50000\over 3000}) =P(-2\lt Z\lt 2)，故選\bbox[red, 2pt]{(D)}$

解答:

$0.04\lt 0.05 \Rightarrow 拒絕H_0，故選\bbox[red, 2pt]{(A)}$

解答:

$前後測皆為相同受試樣本，故選\bbox[red, 2pt]{(B)}$

解答:

$適合度檢定\Rightarrow 卡方檢定，故選\bbox[red, 2pt]{(C)}$

解答:

$\cases{X=1:(5,1-4),(5,6),(1-4,5),(6,5) 共10種\\ X=2:(5,5)只有一種} \Rightarrow E(X)={10\over 36}+2\cdot {1\over 36} ={1\over 3}=0.33，故選\bbox[red, 2pt]{(B)}$

解答:

$P(A\cap B^c)=P(A) =0.3，故選\bbox[red, 2pt]{(B)}$

解答:

$(A) \bigcirc:{\sqrt{25} \over 60} =0.083\\ (B)\times: {\sqrt{36} \over 80} =0.075 \ne 0.45\\，故選\bbox[red, 2pt]{(A)}$

解答:

$500\times 0.9\times 0.8=360，故選\bbox[red, 2pt]{(B)}$

解答:

$E(\bar X)={1\over 4}(20+20+20+20)=20，故選\bbox[red, 2pt]{(A)}$

解答:

$\sum_{x=1}^4 p(x)=1 \Rightarrow k+2k+3k+4k=10k=1 \Rightarrow k={1\over 10} \\ \Rightarrow E(X^2) = \sum_{x=1}^4 x^2p(x) =1^2\cdot {1\over 10}+ 2^2\cdot {2\over 10}+ 3^2 \cdot {3\over 10}+ 4^2 \cdot {4\over 10} ={100\over 10}=10，故選\bbox[red, 2pt]{(C)}$

解答:

$P(x;\lambda) ={\lambda^x e^{-\lambda} \over x!} \Rightarrow P(x=0; \lambda=30/10=3) =e^{-3}，故選\bbox[red, 2pt]{(B)}$

解答:

$\cases{P(A)=0.6\\ P(B)=0.4\\ P(A\cap B)=0.2} \Rightarrow P(A\cup B) =P(A)+P(B)-P(A\cap B)=0.6+0.4-0.2=0.8\\ \Rightarrow 1-P(A\cup B) =1-0.8=0.2，故選\bbox[red, 2pt]{(A)}$

解答:

$抽中紅球的機率p={4\over 4+2+4}=0.4 \Rightarrow 前3次都沒抽出紅球，第4次抽中紅球的機率 \\=0.6^3\cdot 0.4=0.0864，故選\bbox[red, 2pt]{(C)}$

解答:

${10個良品抽出3個, 5個不良品抽出1個\over 15個抽4個} ={C^{10}_3C^5_1\over C^{15}_4} ={120\cdot 5\over 1365}={40\over 91}=0.43956，故選\bbox[red, 2pt]{(D)}$

解答:

$P(X\lt 1.4) =P\left(Z\lt {1.4-1.5\over \sqrt{0.5625\over 36}}\right) =P(Z\lt -0.8)=P(Z\gt 0.8) \\=1-P(Z\lt 0.8)=1-0.7881 =0.2119，故選\bbox[red, 2pt]{(D)}$

解答:

$P(X\gt 70)=P(Z\gt {70-60\over 8}) =P(Z\gt 1.25) =1-P(Z\lt 1.25) =1-0.8944=0.1056\\，故選\bbox[red, 2pt]{(A)}$

解答:

$二項隨機變數B(n=5,p=3/5=0.6) \Rightarrow C^5_3p^3(1-p)^2 =10\cdot 0.6^3 \cdot 0.4^2=0.3456，故選\bbox[red, 2pt]{(C)}$

解答:

$信賴區間: \bar x\mp z_{\alpha/2}\cdot {s\over \sqrt n} =1.2\mp 1.96\cdot {0.5\over \sqrt{49}} =1.2\pm 0.14 =(1.06,1.34)，故選\bbox[red, 2pt]{(B)}$

解答:

$n\ge z_{\alpha/2}^2 \cdot {p(1-p)\over E^2} =1.96^2\cdot {0.5\cdot 0.5\over 0.03^2} =1067.11 \Rightarrow n=1068，故選\bbox[red, 2pt]{(B)}$

解答:

$樣本平均數\bar x={1\over 6}(62+61+ 63+58+57+59) =60\\ \Rightarrow 樣本變異數s^2={1\over 6-1}(2^2+1^2+3^2+2^2+3^2+1^2) ={28\over 5}=5.6 \\ \Rightarrow 變異數信賴區間 {(n-1)s^2 \over \chi_{n-1,\alpha/2}^2} \le \sigma^2 \le {(n-1)s^2 \over \chi_{n-1,1-\alpha/2}} \Rightarrow {5\cdot 5.6\over 12.83} \le \sigma^2 \le {5\cdot 5.6\over 0.83} \\ \Rightarrow 2.182\le \sigma^2\le 33.735，故選\bbox[red, 2pt]{(A)}$

解答:

$顯然為單一母體變異數檢定，故選\bbox[red, 2pt]{(B)}$

解答:

$由於去年學生人數未知，無法用兩獨立母體平均數檢定，故選\bbox[red, 2pt]{(B)}$

解答:

$n=16 為小樣本 \Rightarrow t={54.3-50\over 13.2/\sqrt{16}} =1.303，故選\bbox[red, 2pt]{(B)}$

解答:

$計算兩獨立母體平均數差異，故選\bbox[red, 2pt]{(A)}$

解答:

$樣本數\cases{n_1= 10\\ n_2=15}為小樣一，且母體變異數未知，因此檢定統計量為t值\\ t={43900-40300\over \sqrt{{3300^2\over 10}+{2700^2\over 15}}} =2.869，故選\bbox[red, 2pt]{(D)}$

解答:

$ANOVA計算平均時間，並非變異數；而且是計算全體一致性，並非兩兩計算，故選\bbox[red, 2pt]{(B)}$

解答:

$(A)\bigcirc:(1)=696.70-380.50=316.2 \\(B)\bigcirc: (2)=4-1=3\\ (C)\times:4\times 6-4=20 \ne 5\\ (D)\bigcirc: F={(4) \over (5)} ={380.5/3\over 316.2/20} =8.02\\，故選\bbox[red, 2pt]{(C)}$

解答:

$\begin{array} {}& X & Y& X^2 & Y^2 &XY \\\hline & 30& 80& 900&6400& 2400 \\ &33& 81& 1089& 6561& 2673 \\ & 25&67& 625& 4489 & 1675\\ & 42& 90& 1764& 8100& 3780\\ & 48& 100 & 2304& 10000& 4800\\\hdashline \sum& 178 &418&6682& 35550& 15328\end{array} \\ \Rightarrow 相關系數r= {n \sum XY-\sum X \sum Y\over \sqrt{n\sum X^2-(\sum X)^2} \cdot \sqrt{n\sum Y^2-(\sum Y)^2}}\\ ={5\cdot 15328-178\cdot 418\over \sqrt{5\cdot 6682-178^2} \cdot \sqrt{5\cdot 35550-418^2}} =0.978，故選\bbox[red, 2pt]{(D)}$