ข้อสอบคณิต 2 - มีนาคม 2562

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{ตำแหน่งของ} P_r \\ \mathrm{ตน}. P_r = \frac{r}{100}·\left(N+1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} P_{12} \mathrm{อยู่ตัวที่} \frac{12}{100}\left(24+1\right) = 3 \\ P_{12} = X_3 = 12 \\ P_{80} \mathrm{อยู่ตัวที่} \frac{80}{100}\left(24+1\right) = 20 \\ P_{80} = X_{20} = 20.5 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} X_3 = X_1+\left(3-1\right)d \\ 12 = X_1+2d \dots \overline{)1} \\ X_{20} = X_1+\left(20-1\right)d \\ 20.5 = X_1+19d \dots \overline{)2} \end{gathered}$$

$$\begin{gathered} \overline{)2} - \overline{)1} ; 20.5-12 = \left(X_1+19d\right)-\left(X_1+2d\right) \\ 8.5 = 17d \\ 0.5 = d \mathrm{แทนใน} \overline{)1} \\ \mathrm{จะได้} 12 = X_1+2\left(0.5\right) \\ X_1 = 11 \end{gathered}$$

$$\begin{gathered} \mathrm{ตน}. Me = \frac{24+1}{2} = 12.5 \\ \mathrm{จะได้} X_{12} = 11+11\left(0.5\right) = 16.5 \\ X_{13} = 16.5+0.5 = 17 \\ \mathrm{ดังนั้น} \mathrm{มัธยฐาน} = \frac{X_{12}+X_{13}}{2} \\ = \frac{16.5+17}{2} = 16.75 \to \mathrm{ตอบ} 4 \end{gathered}$$