ข้อสอบ PAT 1 - กุมภาพันธ์ 2561

$$\begin{gathered} \mathbf{นิยาม} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อมูลชุดที่} 1 \hspace{0.17em}\hspace{0.17em}{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3, \dots ,{\mathrm{x}}_{\mathrm{i}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อมูลชุดที่} 2 \hspace{0.17em}\hspace{0.17em}{\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3, \dots ,{\mathrm{y}}_{\mathrm{i}} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{โดย} {\mathrm{y}}_{\mathrm{i}}={\mathrm{ax}}_{\mathrm{i}}+\mathrm{b} \\ \mathrm{จะได้}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{{\mathrm{y}}_{\mathrm{i}}}=\mathrm{a}\overline{{\mathrm{x}}_{\mathrm{i}}}+\mathrm{b} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{S}}_{\mathrm{y}}=\left|\mathrm{a}\right|{\mathrm{S}}_{\mathrm{x}} \to {\mathrm{S}}_{\mathrm{y}}^2={\mathrm{a}}^2{\mathrm{S}}_{\mathrm{x}}^2\to \boxed{1} \end{gathered}$$


$$\begin{gathered} \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sum \left({\mathrm{x}}_{\mathrm{i}}-4\right)=40 \hspace{0.17em}\mathrm{และ} \hspace{0.17em}\sum {\left({\mathrm{x}}_{\mathrm{i}}-4\right)}^2=170 \\ \mathrm{กำหนดให้}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{y}}_{\mathrm{i}}={\mathrm{x}}_{\mathrm{i}}-4 \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sum {\mathrm{y}}_{\mathrm{i}}=40 \mathrm{และ} \sum {\mathrm{y}}_{\mathrm{i}}^2=170 \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{x}}=\frac{\sum \mathrm{x}}{\mathrm{N} } \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{y}}=\frac{\sum \mathrm{y}}{\mathrm{N}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{y}}=\frac{40}{10} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{y}}=4 \\ \mathrm{จากสูตร} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{S}}_{\mathrm{x}}=\sqrt{\frac{\sum {\mathrm{x}}^2}{\mathrm{N}}-{\overline{\mathrm{x}}}^2} \end{gathered}$$
$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{S}}_{\mathrm{x}}^2=\frac{\sum {\mathrm{x}}^2}{\mathrm{N}}-{\overline{\mathrm{x}}}^2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{S}}_{\mathrm{y}}^2=\frac{\sum {\mathrm{y}}^2}{\mathrm{N}}-{\overline{\mathrm{y}}}^2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{S}}_{\mathrm{y}}^2=\frac{170}{10}-4^2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{S}}_{\mathrm{y}}^2=1 \end{gathered}$$

$$\begin{gathered} \mathbf{จากโจทย์}\mathbf{\hspace{0.17em}} \\ \mathrm{ความแปรปรวน} 2\left({\mathrm{x}}_1+3\right) , 2\left({\mathrm{x}}_2+3\right) , 2\left({\mathrm{x}}_3+3\right) , \dots , 2\left({\mathrm{x}}_{10}+3\right) \\ \mathrm{พิจารณา} 2\left({\mathrm{x}}_1+3\right) , 2\left({\mathrm{x}}_2+3\right) , 2\left({\mathrm{x}}_3+3\right) , \dots , 2\left({\mathrm{x}}_{10}+3\right) \\ 2{\mathrm{x}}_1+6 , 2{\mathrm{x}}_2+6 , 2{\mathrm{x}}_3+6 , \dots , 2{\mathrm{x}}_{10}+6 \to 2{\mathrm{x}}_{\mathrm{i}}+6 \\ \mathbf{เปรียบเทียบกับ} {\mathrm{y}}_{\mathrm{i}}={\mathrm{x}}_{\mathrm{i}}-4 \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2\left({\mathrm{x}}_{\mathrm{i}}-4\right)+14=2{\mathrm{x}}_{\mathrm{i}}+6 (\mathrm{คูณ} 2 \mathrm{แล้วบวก} 14) \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{a}=2 \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ความแปรปรวนของข้อมูล}={\mathrm{a}}^2{\mathrm{S}}_{\mathrm{y}}^2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=2^2\left(1\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=4 \end{gathered}$$