ข้อสอบคณิต O-NET - 2560

$$\begin{gathered} \mathrm{จากโจทย์} \\ \mathrm{เทียบช่วง} \left(\overline{x}-3S, \overline{x}+3S\right) = \left(1400, 1580\right) \\ \mathrm{จะได้} \overline{x}-3S = 1400 \dots \overline{)1} \\ \overline{x}+3S = 1580 \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \overline{)1}+\overline{)2} : 2\overline{x} = 2980 \\ \overline{x} = 1490 \\ \mathrm{แทน} \overline{x} \mathrm{ใน} \overline{)2} : 1490+3S = 1580 \\ 3S = 90 \\ S = 30 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \overline{x} = 1490 \\ S^2 = {30}^2 = 900 \to \mathrm{ตอบ} 5. \end{gathered}$$