ข้อสอบ PAT 1 - ตุลาคม 2558

$$\begin{gathered} \mathbf{สูตรสถิติ} \left(\mathrm{กลุ่มตัวอย่าง}\right) \\ \overline{\mathrm{x}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}}\to \boxed{1} \\ \mathrm{s}=\sqrt{\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}}=\sqrt{\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\overline{\mathrm{x}}}^2}{\mathrm{n}-1}} \\ {\mathrm{s}}^2=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\overline{\mathrm{x}}}^2}{\mathrm{n}-1}\to \boxed{2} \\ \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \sum _{\mathrm{i}=1}^5{\mathrm{x}}_{\mathrm{i}}^2=214 \mathrm{และ} \sum _{\mathrm{i}=1}^5{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2=34 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} \mathrm{จะได้} {\mathrm{s}}^2=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}=\frac{\sum _{\mathrm{i}=1}^5{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\overline{\mathrm{x}}}^2}{\mathrm{n}-1} \\ \frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\left({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}}\right)}^2}{\mathrm{n}-1}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\overline{\mathrm{x}}}^2}{\mathrm{n}-1} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{n}=5 \\ \frac{34}{5-1}=\frac{214-5\left({\overline{\mathrm{x}}}^2\right)}{5-1} \\ \frac{34}{\cancel{4}}=\frac{214-5\left({\overline{\mathrm{x}}}^2\right)}{\cancel{4}} \end{gathered}$$

$$\begin{gathered} 34=214-5\left({\overline{\mathrm{x}}}^2\right) \\ \left({\overline{\mathrm{x}}}^2\right)=36 \\ \overline{\mathrm{x}}=6,-6 ; \mathrm{จากโจทย์}\hspace{0.17em}\overline{\mathrm{x}}>0 \\ \mathrm{ดังนั้น} \overline{\mathrm{x}}=6 \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathbf{ข้อมูลกลุ่มตัวอย่างใหม่}\mathbf{ }5 \mathrm{จำนวนคือ} \\ {\mathrm{x}}_1+2{\mathrm{x}}_2 , {\mathrm{x}}_2+2{\mathrm{x}}_3 , {\mathrm{x}}_3+2{\mathrm{x}}_4 , {\mathrm{x}}_4+2{\mathrm{x}}_5 , {\mathrm{x}}_5+2{\mathrm{x}}_1 \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{x}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} ; \mathrm{n}=5 \end{gathered}$$

$\overline{\mathrm{x}\text{'}}=\frac{\left({\mathrm{x}}_1+2{\mathrm{x}}_2\right) +\left({\mathrm{x}}_2+2{\mathrm{x}}_3\right) +\left({\mathrm{x}}_3+2{\mathrm{x}}_4\right)+\left( {\mathrm{x}}_4+2{\mathrm{x}}_5\right)+\left({\mathrm{x}}_5+2{\mathrm{x}}_1\right)}{5}$
$$\begin{gathered} =\frac{3\left({\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4+{\mathrm{x}}_5\right)}{5} \\ =3\left(\overline{\mathrm{x}}\right) \\ =3\left(6\right) \\ =18 \\ \mathrm{ดังนั้น}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \overline{\mathrm{x}\text{'}}=18 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ส่วนเบี่ยงเบนมาตรฐานเท่ากับ} 16 \left(\mathrm{s}\text{'}=16\right) \\ \mathrm{จาก} \boxed{2} \mathrm{จะได้} \hspace{0.17em}\hspace{0.17em}{\mathrm{s}}^2=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\overline{\mathrm{x}}}^2}{\mathrm{n}-1} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{s}{\text{'}}^2=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\mathrm{n}{\left(\overline{\mathrm{x}}\text{'}\right)}^2}{\mathrm{n}-1} \end{gathered}$$
$$\begin{gathered} {\left(16\right)}^2=\frac{\left[{\left({\mathrm{x}}_1+2{\mathrm{x}}_2\right)}^2+{\left({\mathrm{x}}_2+2{\mathrm{x}}_3\right)}^2+{\left({\mathrm{x}}_3+2{\mathrm{x}}_4\right)}^2+{\left({\mathrm{x}}_4+2{\mathrm{x}}_5\right)}^2+{\left({\mathrm{x}}_5+2{\mathrm{x}}_1\right)}^2\right]-5{\left(\overline{\mathrm{x}}\text{'}\right)}^2}{5-1} \\ 256=\frac{\left({\mathrm{x}}_1^2+4{\mathrm{x}}_1{\mathrm{x}}_2+4{\mathrm{x}}_2^2\right)+\left({\mathrm{x}}_2^2+4{\mathrm{x}}_2{\mathrm{x}}_3+4{\mathrm{x}}_3^2\right)+\left({\mathrm{x}}_3^2+4{\mathrm{x}}_3{\mathrm{x}}_4+4{\mathrm{x}}_4^2\right)+\left({\mathrm{x}}_4^2+4{\mathrm{x}}_4{\mathrm{x}}_5+4{\mathrm{x}}_5^2\right)+\left({\mathrm{x}}_5^2+4{\mathrm{x}}_5{\mathrm{x}}_1+4{\mathrm{x}}_1^2\right)-5{\left(18\right)}^2}{4} \end{gathered}$$

$$\begin{gathered} 256=\frac{5\left({\mathrm{x}}_1^2+{\mathrm{x}}_2^2+{\mathrm{x}}_3^2+{\mathrm{x}}_4^2+{\mathrm{x}}_5^2\right)+4\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1\right)-1620}{4} \\ 256\times 4=5\left(214\right)+4\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1\right)-1620 \\ 1574=4\left({\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1\right) \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1=393.5 \end{gathered}$$

$$\begin{gathered} \mathbf{หาค่าของ} \mathrm{ค่าเฉลี่ยเลขคณิตของ} {\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1 \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \overline{\mathrm{x}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}}{\mathrm{n}} \\ =\frac{{\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1}{5} \\ =\frac{393.5}{5} \\ =78.7 \\ \mathrm{ค่าเฉลี่ยเลขคณิตของ} {\mathrm{x}}_1{\mathrm{x}}_2+{\mathrm{x}}_2{\mathrm{x}}_3+{\mathrm{x}}_3{\mathrm{x}}_4+{\mathrm{x}}_4{\mathrm{x}}_5+{\mathrm{x}}_5{\mathrm{x}}_1=78.7 \end{gathered}$$