ข้อสอบคณิต O-NET - 2554

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{สถิติ} \left(\mathrm{ข้อมูลดิบ}\right) \\ \overline{x} = \frac{\sum x_i}{N} \left(\mathrm{ค่าเฉลี่ยเลขคณิต}\right) \\ S = \sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{N}} \left(\mathrm{ส่วนเบี่ยงเบนมาตรฐาน}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{x}=4 , S = \frac{4}{\sqrt{7}} \\ \mathrm{จะได้} \frac{2+3+3+x+4+y+7}{7} = 4 \\ y = 9-x \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \sqrt{\frac{{\left(2-4\right)}^2+{\left(3-4\right)}^2+{\left(3-4\right)}^2+{\left(x-4\right)}^2+{\left(4-4\right)}^2+{\left(y-4\right)}^2+{\left(7-4\right)}^2}{7}} = \frac{4}{\sqrt{7}} \\ \sqrt{\frac{4+1+1+{\left(x-4\right)}^2+0+{\left(y-4\right)}^2+9}{7}} = \frac{4}{\sqrt{7}} \\ \sqrt{\frac{{\left(x-4\right)}^2+{\left(y-4\right)}^2+15}{7}} = \frac{4}{\sqrt{7}} \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{เอา} \overline{)1} \mathrm{แทน} \overline{)2} \mathrm{จะได้} x=4 , 5 \\ \mathrm{เนื่องจาก} x \mathrm{อยู่ระหว่าง} 3 \mathrm{กับ} 4 \\ \mathrm{จะได้} x=4 \mathrm{แทนใน} \overline{)1} y=5 \\ \mathrm{ดังนั้น} y-x = 5-4 = 1 \end{gathered}$$