ข้อสอบ PAT 1 - ตุลาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ความแปรปรวนเท่ากับ} 25 \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\sigma }^2=25 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sigma =5 \left(\mathrm{ส่วนเบี่ยงเบนมาตรฐาน}=5\right) \\ \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{สัมประสิทธิ์ของการแปรผันเท่ากับ} 62.5\% \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \frac{\sigma }{\mu }=\frac{62.5}{100} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \frac{5}{\mu }=\frac{62.5}{100} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \mu =8 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ก}\right) \mathrm{สูตรส่วนเบี่ยงเบนมาตรฐาน} \\ \sigma =\sqrt{\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{N}}}{\mathrm{x}}_{\mathrm{i}}^2}{\mathrm{N}}-{\mu }^2} \\ 5=\sqrt{\frac{{\displaystyle \sum _{\mathrm{i}=1}^{10}}{\mathrm{x}}_{\mathrm{i}}^2}{10}-{\left(8\right)}^2} \\ 5^2=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{10}}{\mathrm{x}}_{\mathrm{i}}^2}{10}-64 \\ \frac{{\displaystyle \sum _{\mathrm{i}=1}^{10}}{\mathrm{x}}_{\mathrm{i}}^2}{10}=89\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{ค่าเฉลี่ยเลขคณิตของ} x_1^2,{\mathrm{x}}_2^2,...,{\mathrm{x}}_{10}^2=\frac{{\mathrm{x}}_1^2+{\mathrm{x}}_2^2+...+{\mathrm{x}}_{10}^2}{10} \\ \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}\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}\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}\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} =\frac{{\displaystyle \sum _{\mathrm{i}=1}^{10}}{\mathrm{x}}_{\mathrm{i}}^2}{10} ; \mathrm{จาก} \boxed{1} \\ \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}\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}\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}\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}\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} =89 \\ \mathrm{ดังนั้น} \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}\mathrm{ค่าเฉลี่ยเลขคณิตของ} {\mathrm{x}}_1^2,{\mathrm{x}}_2^2,...,{\mathrm{x}}_{10}^2=89 \\ \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}\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}\mathrm{ข้อความ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ข}\right) \mathrm{สมบัติของ} \mathrm{S}.\mathrm{D}. \\ \mathrm{ถ้า} {\mathrm{y}}_{\mathrm{i}}={\mathrm{Ax}}_{\mathrm{i}}+\mathrm{B} \mathrm{จะได้} {\mathrm{\sigma }}_{\mathrm{y}}=\left|\mathrm{A}\right|·\mathrm{\sigma } \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}_1,{\mathrm{x}}_2,...,{\mathrm{x}}_{10} \mathrm{มี} \mathrm{\sigma }=5 \\ \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}-{\mathrm{x}}_1,-{\mathrm{x}}_2,...,-{\mathrm{x}}_{10} \mathrm{มี} {\mathrm{\sigma }}_{\mathrm{ใหม่}}=\left|-1\right|·\left(5\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{\sigma }}_{\mathrm{ใหม่}}=5 \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ส่วนเบี่ยงเบนมาตรฐานของ} -{\mathrm{x}}_1,-{\mathrm{x}}_2,...,-{\mathrm{x}}_{10}=5 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อความ} \left(\mathrm{ข}\right) \mathrm{ถูก} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ค}\right) \mathrm{สมบัติของ} \overline{\mathrm{x}} \to \sum _{\mathrm{i}=1}^{\mathrm{N}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2 \mathrm{มีค่าน้อยที่สุด} \\ \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}_1,{\mathrm{x}}_2,...,{\mathrm{x}}_{10} \hspace{0.17em}\mathrm{มี}\hspace{0.17em} \overline{\mathrm{x}}=8 \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\underset{\mathrm{i}=1}{\overset{\mathrm{N}}{\hspace{0.17em}\sum }}{\left({\mathrm{x}}_{\mathrm{i}}-8\right)}^2 \mathrm{มีค่าน้อยที่สุด} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ข้อความ} \left(\mathrm{ค}\right) \mathrm{ผิด} \end{gathered}$$