(一)$$樣本空間S=\{(x,y)\mid 1\le x,y\le 6, x,y\in Z\}$$
(二)$$S_{ X=1 }=\{ (1,2),(2,1),(2,3),(2,4),(2,5),(2,6),(3,2),(4,2),(5,2),(6,2)\} \Rightarrow \#(S_{ X=1 })=10\\ S_{ X=2 }=\{ (2,2)\} \Rightarrow \#(S_{ X=2 })=1\\\Rightarrow \#(S_{ X=0 })=\#(S)-\#(S_{ X=1 })-\#(S_{ X=2 })=36-10-1=25 \\ \Rightarrow \begin{cases} P(X=0)=25/36 \\ P(X=1)=10/36 \\ P(X=2)=1/36 \end{cases} \Rightarrow 動差母函數m_X(t)=E(e^{tX}) =\sum_{X=0}^2{e^{tX}P(X)}\\ = e^0\cdot{25\over 36}+e^t\cdot{10\over 36} + e^{2t}\cdot {1\over 36} = \bbox[red,2pt] {{1\over 36}\left( e^{2t}+10e^t+25 \right)}$$
解:
解:
(一)$$積分區域如上圖,因此\int_0^2{\int_0^y{f(x,y)\,dx}dy}=1 \Rightarrow \int_0^2{\int_0^y{{1\over c}\,dx}dy}=1\\ \Rightarrow \int_0^2{{1\over c}ydy}=1 \Rightarrow {2\over c}=1 \Rightarrow \bbox[red,2pt] {c=2}
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(三)$$f_{Y|X}(Y|X)={f_{X,Y}\over f_X}={1/2\over {2-x\over 2}} =\bbox[red,2pt] {{1\over 2-x},0\le x<2}$$
(四)$$\mu _{ Y|X }=\int _{ x }^{ 2 }{ yf_{ Y|X }\, (y|x)dy } =\int _{ x }^{ 2 }{ y\cdot \frac { 1 }{ 2-x } dy } =\frac { 1 }{ 2-x } \left. \left[ \frac { 1 }{ 2 } { y }^{ 2 } \right] \right| _{ x }^{ 2 }=\frac { 1 }{ 2-x } \cdot \frac { 4-x^{ 2 } }{ 2 } =\frac { 2+x }{ 2 }\\ \Rightarrow \bbox[red, 2pt]{\mu _{ Y|X }=\frac { 2+x }{ 2 },0\le x\le 2}\\E\left( Y^{ 2 }|X=x \right) =\int _{ x }^{ 2 }{ y^{ 2 }f_{ Y|X }\, (y|x)dy } =\int _{ x }^{ 2 }{ y^{ 2 }\cdot \frac { 1 }{ 2-x } dy } =\frac { 1 }{ 2-x } \left. \left[ \frac { 1 }{ 3 } { y }^{ 3 } \right] \right| _{ x }^{ 2 }=\frac { 2^{ 3 }-x^{ 3 } }{ 3(2-x) } \\ =\frac { \left( 2-x \right) \left( 2^{ 2 }+2x+x^{ 2 } \right) }{ 3(2-x) } =\frac { 1 }{ 3 } \left( 2^{ 2 }+2x+x^{ 2 } \right) \Rightarrow \sigma _{ Y|X }^{ 2 }=E\left( Y^{ 2 }|X=x \right) -\mu _{ Y|X }^{ 2 }\\ =\frac { 1 }{ 3 } \left( 2^{ 2 }+2x+x^{ 2 } \right) -{ \left( \frac { 2+x }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 12 } \left( 4-4x+x^{ 2 } \right) =\frac { 1 }{ 12 } { \left( x-2 \right) }^{ 2 }\\ \Rightarrow \bbox[red, 2pt]{\sigma _{ Y|X }^{ 2 }=\frac { 1 }{ 12 } { \left( x-2 \right) }^{ 2 },0\le x\le 2}$$(五)$$由(三)可知為\bbox[red,2pt]{均勻分配},即\bbox[red, 2pt]{Y\mid X=x \sim U(x, 2)}$$
解:
(一)$$先計算50筆資料的樣本平均數\bar{x}={\sum_{i=1}^{50}{x_i}\over 50}={3037 \over 50} = 60.74\\ 樣本標準差s=\sqrt{\sum_{i=1}^{50}{x_i^2}-50\cdot (\bar{x})^2\over 50-1} =\sqrt{ 189785-50\cdot 60.74^2\over 49}= \sqrt{5317.62\over 49} = 10.42\\ \Rightarrow 四組組界\begin{cases} \bar{x}+s\cdot z_{0.25}=60.74-10.42\cdot 0.6745=53.71\\\bar{x}=60.74\\\bar{x}+s\cdot z_{0.75} = 60.74+10.42\cdot 0.6745=67.77\end{cases}\\ 理論次數每組都是50\div 4=12.5,觀察次數將原資料依組界分組,可得以下表格\\ \begin{array}{|c|c|c|c|c|}\hline
組界&最小至53.71& 53.71至60.74 &60.74至67.77&67.77至最大\\\hline
觀察值o_i & 12 & 11 & 14 &13\\\hline
理論值e_i & 12.5 &12.5 &12.5&12.5\\\hline
\end{array}\\ H_0:資料服從常態分配\\ H_1:資料不服務常態分配\\顯著水準\alpha=0.05\\ 卡方檢定值\chi^2=\sum_{i=1}^4{(e_i-o_i)^2\over e_i}= {5\over 12.5}=0.4\\ 查表\chi_{df=1,\alpha=0.05}^2=3.841 \Rightarrow 拒絕區域R=\{\chi^2\mid \chi^2>3.841\}\\ 由於0.4\notin R,因此不能拒絕H_0,即資料為常態分配$$(二)$$已知資料\bar{x}=60.74,s=10.42, n=50\\ H_0: \mu=60\\H_1:\mu \ne 60\\ 顯著水準\alpha=0.1\\檢定值t={\bar{x}-\mu \over s/\sqrt{n}} ={60.74-60 \over 10.42/\sqrt{50}}=0.502\\ 查表t_{n-1,\alpha/2}=t_{49,0.05}=1.677 \Rightarrow 拒絕區域R=\{t \mid |t|>1.677\}\\ 由於t=0.502\notin R \Rightarrow 不能拒絕H_0,即\mu=60$$
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