ข้อสอบคณิต 2 - มีนาคม 2561

$$\begin{gathered} \mathrm{จากโจทย์} \\ \mathrm{มัธยฐาน} \mathrm{จะอยู่ตัวที่} \frac{N+1}{2} = \frac{19+1}{2} \\ = 10 \end{gathered}$$

$$\begin{gathered} \mathrm{มัธยฐาน} = x_{10} \\ \mathrm{ค่าเฉลี่ยของข้อมูล} = 30 \\ \frac{x_1+x_2+\dots +x_{19}}{19} = 30 \\ x_1+x_2+\dots +x_{19} = 570 \dots \overline{)1} \end{gathered}$$

$$\begin{gathered} \mathrm{ค่าเฉลี่ย} x_1, x_2, \dots , x_{10} = 24.5 \\ \frac{x_1+x_2+\dots +x_{10}}{10} = 24.5 \\ x_1+x_2+\dots +x_{10} = 245 \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \mathrm{ค่าเฉลี่ย} x_{10}, x_{11}, \dots , x_{19} = 35 \\ \frac{x_{10}+x_{11}+\dots +x_{19}}{10} = 35 \\ x_{10}+x_{11}+\dots +x_{19} = 350 \dots \overline{)3} \end{gathered}$$

$$\begin{gathered} \overline{)2} + \overline{)3} : x_1+x_2+\dots +x_{10}+x_{10}+x_{11}+\dots +x_{19} = 245+350 \\ x_1+x_2+\dots +x_{10} +x_{11}+\dots +x_{19}+x_{10} = 595 \\ x_1+x_2+ \dots +x_{19}+x_{10} = 595 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \overline{)1} 570 + x_{10} = 595 \\ x_{10} = 25 \\ \mathrm{ดังนั้น} \mathrm{มัธยฐาน} = x_{10} = 25 \end{gathered}$$