Contoh Soal 13 Statistika

$\begin{array}{ll} 56.&\textrm{Simpangan baku dari data berikut}:\\ &6,7,4,5,3\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{2}\quad &\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\quad&\textrm{e}.&\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,4,5,3\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+4+5+3}{5}=\displaystyle \frac{25}{5}=\color{red}5 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((6-5)^{2}+(7-5)^{2}+(4-5)^{2}+(5-5)^{2}+(3-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(1^{2}+2^{2}+1^{2}+0+2^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 1+4+1+4 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(10)}=\sqrt{\displaystyle \frac{10}{5}}=\color{red}\sqrt{2}\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 57.&\textbf{UN 2010}\\ &\textrm{Simpangan baku dari data berikut}:\\ &2,3,4,5,6\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\sqrt{15}&&&\textrm{d}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \sqrt{10}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&\sqrt{2} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,3,4,5,6\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+3+4+5+6}{5}=\displaystyle \frac{20}{5}=\color{red}4 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((2-5)^{2}+(3-5)^{2}+(4-5)^{2}+(5-5)^{2}+(6-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(3^{2}+2^{2}+1^{2}+0+1^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 9+4+1+1 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(15)}=\sqrt{\displaystyle \frac{15}{5}}=\color{red}\sqrt{3}\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 58.&\textrm{Simpangan baku dari data berikut}:\\ &7,9,11,13,15\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&2,4&&&\textrm{d}.&2,8\\ \textrm{b}.&\displaystyle 2,5\quad &\textrm{c}.&\displaystyle 2,7\quad&\textrm{e}.&2,9 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &7,9,11,13,15\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{7+9+11+13+15}{5}=\displaystyle \frac{55}{5}=\color{red}11 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((7-11)^{2}+(9-11)^{2}+(11-11)^{2}+(13-11)^{2}+(15-11)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(4^{2}+2^{2}+0+2^{2}+4^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 16+4+4+16 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(40)}=\sqrt{\displaystyle \frac{40}{5}}=\color{red}\sqrt{8}=2,82..\\ \end{aligned} \end{array}$.

 $\begin{array}{ll} 59.&\textrm{Simpangan baku dari data berikut}:\\ &2,4,4,5,6,6,7,8,9,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&4\sqrt{3}&&&\textrm{d}.&\displaystyle \frac{2}{5}\sqrt{30}\\ \textrm{b}.&2\displaystyle \frac{2}{5}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,4,4,5,6,6,7,8,9,9\quad \\ &\textrm{Simpangan bakunya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+4+4+5+6+6+7+8+9+9}{10}=\displaystyle \frac{60}{10}=\color{red}6 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left((2-6)^{2}+2(4-6)^{2}+(5-6)^{2}+2(6-6)^{2}+(7-6)^{2}+(8-6)^{2}+2(9-6)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left(4^{2}+2.2^{2}+1^{2}+0+1^{2}+2^{2}+2.3^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left( 16+8+1+1+4+18 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}(48)}=\sqrt{\displaystyle \frac{48}{10}}=\sqrt{\displaystyle \frac{120}{25}}=\color{red}\displaystyle \frac{2}{5}\sqrt{30}\\ \end{aligned} \end{array}$.

Contoh Soal 12 Statistika

$\begin{array}{ll} 51.&\textrm{Simpangan kuartil dari data}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \: \textrm{adalah}\: \: 1\displaystyle \frac{1}{2}.\: \textrm{Jika median datanya}\\ & \textrm{adalah}\: \: 5\displaystyle \frac{1}{2},\: \textrm{maka rata-rata data}\: \: \textrm{tersebut adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{4}&&&\textrm{d}.&\textrm{5}\displaystyle \frac{1}{2}\\ \textrm{b}.&\textrm{4}\displaystyle \frac{1}{2}\quad &\textrm{c}.&\textrm{5}\quad&\textrm{e}.&\textrm{6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \Rightarrow \: n=6\\&\color{red}\textrm{karena mediannya}=M_{e}=Q_{2}=5\frac{1}{2}=\frac{11}{2}\\&\color{blue}\textrm{ data menjadi}\: :\: \color{black}a\: \: 3\: \: 5\: \: 6\: \: 7\: \: 8\: \: \color{blue}\textrm{atau}\color{black}\: \: 3\: \: a\: \: 5\: \: 6\: \: 7\: \: 8\\ &\textrm{Simpangan kuartilnya adalah}\: 1\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2}\\ &Q_{d}=\displaystyle \frac{1}{2}\left (Q_{3}-Q_{1}  \right ) =\displaystyle \frac{3}{2}\: \Rightarrow\: Q_{3}-Q_{1}=3\\&\textrm{maka}\\ &3=Q_{3}-Q_{1}=x_{._{\frac{3}{4}n+\frac{1}{2}}}-x_{._{\frac{1}{4}n+\frac{1}{2}}}=x_{._{\frac{3}{4}6+\frac{1}{2}}}-x_{._{\frac{1}{4}6+\frac{1}{2}}} \\ &\quad\: =\left (x_{._{5}}-x_{._{2}}  \right )=7-x_{._{2}}=\color{red}3\: \color{black}\Rightarrow\: x_{._{2}}=4\\ &\textrm{Jadi, rata-ratanya adalah}:\\ &\overline{x}=\frac{3+4+5+6+7+8}{6}=\color{red}5,5   \end{aligned} \end{array}$.

$\begin{array}{ll} 52.&\textrm{Simpangan kuartil dari data}\\ &\textrm{berikut}\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{10}&&&\textrm{d}.&\textrm{6}\\ \textrm{b}.&\textrm{9}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{-}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\color{blue}\textrm{data durutkan}\\ &50,50,50,53,53,53,53\\ &61,61,61,61,61,70,70,70\\ &\textrm{Simpangan kuartil adalah}\: Q_{d},\\ &Q_{d}=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( 61-53 \right )=\displaystyle \frac{1}{2}\times 8=4\\ &\textrm{Jadi},\: Q_{d}=4 \end{aligned} \end{array}$.

$\begin{array}{ll} 53.&\textrm{Simpangan rata-rata dari data berikut}:\\ &6\quad 4\quad 2\quad 8\quad 10\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{2}&&&\textrm{d}.&\textrm{3},0\\ \textrm{b}.&\textrm{2},4\quad &\textrm{c}.&\textrm{2},5\quad&\textrm{e}.&\textrm{3},5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6\quad 4\quad 2\quad 8\quad 10\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+4+2+8+10}{5}=\displaystyle \frac{30}{5}=\color{red}6 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 6-6 \right|+\left| 4-6 \right|+\left| 2-6 \right|+\left| 8-6 \right|+\left| 10-6 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 0 \right|+\left| -2 \right|+\left| -4 \right|+\left| 2 \right|+\left| 4 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( 0+2+4+2+4 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}(12)=\displaystyle \frac{12}{5}=\color{red}2,4\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 54.&\textrm{Simpangan rata-rata dari data berikut}:\\ &10,8,7,10,7,5,8,6,10,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1,0}&&&\textrm{d}.&\textrm{8,0}\\ \textrm{b}.&\textrm{1,4}\quad &\textrm{c}.&\textrm{6,0}\quad&\textrm{e}.&\textrm{14,0} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &10,8,7,10,7,5,8,6,10,9\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{10+8+7+10+7+5+8+6+10+9}{10}\\ &\: \: \: \, =\displaystyle \frac{80}{10}=\color{red}8 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| 5-8 \right|+\left| 6-8 \right|+2\left| 7-8 \right|+2\left| 8-8 \right|+\left| 9-8 \right|+3\left| 10-8 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| -3 \right|+\left| -2 \right|+2\left| -1 \right|+2\left| 0 \right|+\left| 1 \right|+3\left| 2 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( 3+2+2+0+1+6 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}(14)=\displaystyle \frac{14}{10}=\color{red}1,4\\ \end{aligned} \end{array}$.

$\begin{array}{ll} 55.&\textrm{Nilai variansi  dari  data}\\ &6,7,7,8,8,8,8,12\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1}&&&\textrm{d}.&\textrm{8}\\ \textrm{b}.&\displaystyle \frac{26}{8}\quad &\textrm{c}.&\displaystyle \frac{11}{4}\quad&\textrm{e}.&\textrm{22} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{c}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,7,8,8,8,8,12\quad \\ &\textrm{Variannya adalah}: \\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+7+8+8+8+8+12}{8}=\displaystyle \frac{64}{8}=\color{red}8 \\ &\textrm{maka nilai}\\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( (6-8)^{2}+2(7-6)^{2}+4(8-8)^{2}+(12-8)^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 2^{2}+2.1+4.0+4^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 4+2+0+16 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}(22)=\color{red}\displaystyle \frac{11}{4}=\color{black}2,75\\ \end{aligned} \end{array}$.

Contoh Soal 11 Statistika

$\begin{array}{ll} 46.&\textrm{Rata-rata dari data yang disajikan }\\ &\textrm{dengan hitogram berikut adalah}\: ....\\   \end{array}$.

$.\qquad\begin{array}{ll} &\begin{array}{lllllll}\\ \textrm{a}.&41,372&&&\textrm{d}.&43,135\\ \textrm{b}.&42,150\quad&\textrm{c}.&43,125\quad&\textrm{e}.&44,250 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned} &\begin{array}{|c|c|c|c|}\hline \begin{aligned}&\textrm{Berat}\\ &\textrm{Badan} \end{aligned}&\color{red}\textrm{f}_{1}&\begin{aligned}&\textrm{Nilai Tengah}\\ &\qquad(\textrm{x}_{1}) \end{aligned}&\color{red}\textrm{f}_{1}\textrm{x}_{1}\\\hline 30-34&5&32&160\\ 35-39&7&37&259\\ 40-44&12&42&504\\ 45-49&9&47&423\\ 50-54&4&52&208\\ 55-59&3&57&171\\\hline &\sum \textrm{f}_{1}=\color{red}40&&\sum \textrm{f}_{1}\textrm{x}_{1}=\color{red}1725\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: \overline{\textrm{x}}\: \: \textrm{adalah}:\\ &\overline{\textrm{x}}=\displaystyle \frac{\sum \textrm{f}_{1}\textrm{x}_{1}}{\sum \textrm{f}_{1}}=\displaystyle \frac{1725}{40}=\color{red}43,125 \end{aligned}  \end{array}$.

$\begin{array}{ll} 47.&(\textbf{UN Mat IPA 2006})\\ &\textrm{Perhatikan gambar berikut}\end{array}$.

$.\qquad\begin{array}{ll} &\textrm{Berat badan pada suatu kelas tersaji dengan}\\ &\textrm{bentuk histogram seperti pada gambar di atas}\\ &\textrm{Rata-rata berat badan tersebut adalah}\: ....\: \textrm{Kg}\\ &\begin{array}{lllllll}\\ \textrm{a}.&64,5&&&\textrm{d}.&66\\ \textrm{b}.&65\quad&\textrm{c}.&65,5\quad&\textrm{e}.&66,5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned} &\begin{array}{|c|c|c|c|}\hline \begin{aligned}&\textrm{Berat}\\ &\textrm{Badan} \end{aligned}&\color{red}\textrm{f}_{1}&\begin{aligned}&\textrm{Nilai Tengah}\\ &\qquad(\textrm{x}_{1}) \end{aligned}&\color{red}\textrm{f}_{1}\textrm{x}_{1}\\\hline 50-54&4&52&208\\ 55-59&6&57&342\\ 60-64&8&62&496\\ 65-69&10&67&670\\ 70-74&8&72&576\\ 75-79&4&77&308\\\hline &\sum \textrm{f}_{1}=\color{red}40&&\sum \textrm{f}_{1}\textrm{x}_{1}=\color{red}2600\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: \overline{\textrm{x}}\: \: \textrm{adalah}:\\ &\overline{\textrm{x}}=\displaystyle \frac{\sum \textrm{f}_{1}\textrm{x}_{1}}{\sum \textrm{f}_{1}}=\displaystyle \frac{2600}{40}=\color{red}65 \end{aligned}  \end{array}$.

$\begin{array}{ll} 48.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&3\\ 150-154&5\\ 155-159&17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{Kuartil bawah dapat dinyatakan dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&149,5+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{b}.&150+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{c}.&155+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{d}.&154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{e}.&155,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 \end{array}\\\\  &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}  &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{Ditanyakan kuartil bawah, maka hal ini}\\ &=Q_{1}\: \Rightarrow \: x_{._{\frac{1}{4}n}}=x_{._{\frac{1}{4}50}}=x_{._{12,5}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&\color{red}3\\ 150-154&\color{red}5\\ \color{blue}155-159&\color{blue}17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{1}\: \: \textrm{adalahh}:\\ &Q_{1}=L+\left( \displaystyle \frac{\frac{1}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 ​\end{aligned} \end{array}$.

$\begin{array}{ll} 49.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&2\\ 10-14&8\\ 15-19&10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{Median dari tabel di atas adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&15,0&&&\textrm{d}.&16,5\\ \textrm{b}.&15,5\quad&\textrm{c}.&16,0\quad&\textrm{e}.&17,0 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{e}\\ &\begin{aligned}  &\textrm{Diketahui}\: n=\color{blue}30\\ &\textrm{Ditanyakan median, maka formulanya}\\ &=Q_{2}\: \Rightarrow \: x_{._{\frac{2}{4}n}}=x_{._{\frac{1}{2}.30}}=x_{._{15}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&\color{red}2\\ 10-14&\color{red}8\\ \color{blue}15-19&\color{blue}10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{2}\: \: \textrm{adalah}:\\ &Q_{2}=L+\left( \displaystyle \frac{\frac{2}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}14,5+\left( \displaystyle \frac{15-10}{10} \right).5\\ &\quad\: \: =14,5+\frac{25}{10}=14,5+2,5=\color{red}17,0 \end{aligned}  \end{array}$.

$\begin{array}{ll} 50.&\textrm{Jangkauan antarkuartil dari data}\\ &\textrm{berikut}:\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{20}&&&\textrm{d}.&\textrm{5}\\ \textrm{b}.&\textrm{10}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{a}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\color{blue}\textrm{data durutkan}\\ &25\: \: 28\: \: 35\: \: 36\: \: 38\: \: 40\: \: 42\: \: 43\\ &45\: \: 54\: \: 55\: \: 56\: \: 64\: \: 70\: \: 82\\ &\textrm{Jangkauan antarkuartil adalah}\: H,\\ &H= Q_{3}-Q_{1} \\ &\quad\: =x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}}\\ &\quad\: =x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}}\\ &\quad\: =\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =56-36\\ &\textrm{Jadi},\: H=Q_{3}-Q_{1}=\color{red}20 \end{aligned} \end{array}$.

Contoh Soal 10 Statistika

Soal sebelumnya (yaitu Contoh Soal 9 Statistika) klik di sini

$\begin{array}{ll} 41.&\textrm{Median dan modus dari data berikut}\\ &\color{red}3,6,7,8,4,5,9,6\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7 dan 5}&&&\textrm{d}.&\textrm{5 dan 6}\frac{1}{2}\\ \textrm{b}.&\textrm{6 dan 6} &\textrm{c}.&\textrm{6 dan 7}&\textrm{e}.&\textrm{5 dan 6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\textrm{Diketahui}\: \: \color{blue} n=8\\ &\textrm{Datum diurutkan dari kecil ke besar}\\ &:3,4,5,\color{red}6,6\color{black},7,8,9\\ &\bullet \: \: \textbf{modus} =M_{o}=6,\\ &\bullet \: \: \textbf{mean} =\overline{x}=6 \end{aligned} \end{array}$.

$\begin{array}{ll} 42.&\textrm{Hasil tes matematika di suatu kelas yang}\\ &\textrm{diikuti tes 49 siswa menghasilkan nilai}\\ &\textrm{rata-rata 7}.\: \textrm{Jika Andi ikut tes susulan}\\ &7,04.\: \textrm{Nilai Andi adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7,5}&&&\textrm{d}.&\textrm{9}\\ \textrm{b}.&\textrm{8} &\textrm{c}.&\textrm{8,5}&\textrm{e}.&\textrm{9,5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}\textrm{Misalkan}&\: \: \color{red}y\color{black}=\textrm{besar nilai Andi}\\ \overline{x}_{gabungan}&=\displaystyle \frac{n.\overline{x}+\color{red}y}{n+1}\\ 7,04&=\displaystyle \frac{49\times 7+\color{red}y}{49+1}\\ 7,04&=\displaystyle \frac{343+\color{red}y}{50}\\ 343+\color{red}y&=50\times \left ( 7,04 \right )\\ 343+\color{red}y&=352\\ y&=352-343\\ &=\color{red}9 \end{aligned} \end{array}$.

$\begin{array}{ll} 43.&\textrm{Mean dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&10\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{60,5}&&&\textrm{d}.&\textrm{58,5}\\ \textrm{b}.&\textrm{60} &\textrm{c}.&\textrm{59,5}&\textrm{e}.&\textrm{57} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}&\begin{array}{|c|c|c|c|c|}\hline \textrm{Ukuran}&x_{i}&d_{i}=x_{i}-x_{s}&f_{i}&f_{i}\times d_{i}\\\hline 50-54&52&-10&4&-40\\\hline 55-59&57&-5&6&-30\\\hline 60-64&62&0&10&0\\\hline \textrm{Jumlah}&&&20&-70\\\hline \end{array}\\ &\overline{x}=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}\times d_{1}}{\displaystyle \sum_{i=0}^{n}f_{i}}=62+\displaystyle \frac{-70}{20}\\ &\: \: =62-3,5=\color{red}58,5 \end{aligned} \end{array}$.

$\begin{array}{ll} 44.&\textrm{Median dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&1\\ 50-52&6\\ 53-55&6\\ 56-58&7\\ 59-61&4\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{55,5}&&&\textrm{d}.&\textrm{53,5}\\ \textrm{b}.&\textrm{55} &\textrm{c}.&\textrm{54,5}&\textrm{e}.&\textrm{53} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&\color{red}1\\ 50-52&\color{red}6\\ \color{blue}53-55&\color{blue}6\\ 56-58&7\\ 59-61&4\\\hline &24\\\hline \end{array}\\ &\textrm{Median posisi datumnya}:\\ &\textrm{datum ke}-\left ( \displaystyle \frac{24}{2} \right )=x_{._{12}}\\ &\textrm{dan terletak di interval}\\ &53-55,\: \: \textrm{dengan}\: \: f_{k}=1+6=7\\ &\textrm{serta}\: \: c=3\\ &M_{e}=Q_{2}=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\\ &M_{e}=Q_{2}=52,2+3\left ( \displaystyle \frac{\displaystyle \frac{24}{2}-7}{6} \right )\\ &\: \quad=52,5+3\left ( \displaystyle \frac{12-7}{6} \right )\\ &\: \quad=52,5+\left ( \displaystyle \frac{5}{2} \right )\\ &\: \quad=52,5+2,5\\ &\: \quad=\color{red}55 \end{aligned} \end{array}$.

$\begin{array}{ll} 45.&\textrm{Modus dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&8\\ 65-69&16\\ 70-74&10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\textrm{67,32}&&&\textrm{d}.&\textrm{70,12}\\ \textrm{b}.&\textrm{67,36} &\textrm{c}.&\textrm{67,56}&\textrm{e}.&\textrm{70,36} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&\color{red}8\\ \color{blue}65-69&\color{blue}16\\ 70-74&\color{red}10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{modus terletak pada kelas}\\ &\textrm{interval dengan frekuensi}\\ &\textrm{terbesar, yaitu}:16.\: \textrm{Kelas intervalnya}\\ &65-69,\: \: \textrm{dengan}\: \: c=5\\ &\textrm{serta}\: \: \begin{cases} \triangle _{1} & =f-f_{1}=16-8=8 \\ \triangle _{2} & =f-f_{2}=16-10=6 \end{cases}\\ &M_{o}=L+c\left ( \displaystyle \frac{\triangle _{1}}{\triangle _{1}+\triangle _{2}} \right )\\ &\quad\: \: =64,5+5\left ( \displaystyle \frac{8}{8+6} \right )\\ &\quad\: \: =64,5+\displaystyle \frac{40}{14}\\ &\quad\: \: =64,5+2,857\\ &\quad\: \: =\color{red}67,36 \end{aligned} \end{array}$.

Koefisien Keragaman (Koefisien Variansi)

A. Pengertian

Pada bahasan ini untuk membandingkan dua atau lebih distribusi data yang sejenis dapat digunakan koefisien keragaman. Koefisien variansi adalah nilai dari standar deviasi suatu data dibagi dengan rata-ratanya.

B. Formula koefisien Variansi

Jika diketahui  $S$ adalah simpangan baku dan  $\overline{x}$ adalah rataan hitung suatu data, maka koefidien variansinya (V) dirumuskan dengan:

$V=\displaystyle \frac{S}{\overline{x}}\times 100%$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

Contoh 1

Coba perhatikan lagi data pada halaman ini di sini, dengan datanya adalah:

$\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&47-49&50-52&53-55&56-58&59-61\\\hline \textrm{Frek}&2&4&6&5&3\\\hline \end{array}$. 

Dari perhitungan untuk data tersebut didapatkan besar rataan hitungnya adalah 54,45 dan simpangan bakunya adalah 3,58, maka koefisien dari variansi dari data tersebut adalah:

$\begin{aligned}V&=\displaystyle \frac{S}{\overline{x}}\times 100\%\\ &=\displaystyle \frac{3,58}{54,45}\times 100\%\\ &=\color{red}6,57\% \end{aligned}$.

Contoh 2

Diketahui nilai ulangan matematika suatu kelas di suatu waktu memiliki rataan 78 dengan simpangan bakunya adalah 7, sedangkan untuk nilai ulangan kimia dari kelas tersebut mendapatkan rataan 62 dan simpangan bakunya adalah 6. Tentukanlah mata pelajaran mana dari keduanya yang telah diuhikan itu yang memiliki penyebaran data yang lebih kecil

Jawab:

Dari data di atas, jika kita hanya berpatokan pada hasil simpangan baku kedua mapel yang telah diujikan tersebut tentunya mapel kimia akan memiliki persebaran yang lebih kecil dari pada mapel matematika. Akan tetapi adalah perhitungan yang lebih baik tentang permasalahan di atas, yaitu dengan menggunkan rumus koefisien variansi sebagaimana perhitungan berikut ini:

$\begin{array}{|c|c|}\hline \textrm{Mapel Matematika}&\textrm{Mapel Kimia}\\\hline \begin{aligned}V&=\displaystyle \frac{7}{78}\times 100\%\\ &=8,97\% \end{aligned}&\begin{aligned}V&=\displaystyle \frac{6}{62}\times 100\%\\ &=9,68\% \end{aligned}\\\hline \end{array}$

Tampak dari perhitungan koefisien variansi di atas bahwa nilai ulangan mapel matematika memiliki sebaran relatif lebih kecil dari pada hasil ulangan mapel kimia.

C. Angka Baku

Misalkan ada suatu permasalahan seorang siswa saat ulangan matematika mendapatkan nilai 8 di mana rataan kelasnya adalah 6,5 dan simpangan bakunya adalah 2. Sedangkan untuk hasil ulangan kimia ia berhasil mendapatkan nilai 9 yang rataan kelasnya 7,5 dan simpangan bakunya 3. Pertanyaannnya adalah hasil yang didapatkan anak tersebut kedudukannya mana yang lebih baik?

Untuk menjawab pertanyaan di atas kita dapat menggunkan angka baku, yaitu  $z=\displaystyle \frac{x-\overline{x}}{S}$.

Berdasarkan nilai kita bisa tentukan angka baku nilai siswa tersebut, yaitu:

$\begin{aligned}\textrm{matematika}\: :\: z&=\displaystyle \frac{8-6,5}{2}=\color{blue}0,75\\ \textrm{fisika}\qquad\quad\: :\: z&=\displaystyle \frac{9-7,2}{3}=\color{red}0,60 \end{aligned}$.

Dari perhitungan angka bakunya, tampak bahwa nilai ulangan matematika siswa tersebut lebih besar dari angka baku fisikanya. Hal ini menunjukkan nilai matematika siswa tersebut adalah yang lebih baik.

Ukuran Penyebaran Data Berkelompok (Materi Kelas XII Matematika Wajib) (Bagian 2)

 B. 2 Data Berkelompok

$\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Data Dispersi}&\textrm{Keterangan}\\\hline 1.&\textrm{Jangkauan}&\begin{aligned}\textrm{a}.\: \: &\textrm{selisih titik tengah}\\ &\textrm{kelas tertinggi dengan}\\ &\textrm{titik tengah kelas}\\ &\textrm{terendah}\\ \textrm{b}.\: \: &\textrm{selisih tepi atas kelas}\\ &\textrm{kelas tertinggi dengan}\\ &\textrm{tepi bawah kelas}\\ &\textrm{terendah} \end{aligned}\\\hline 2.&H&Q_{3}-Q_{1}\\\hline 3.&Q_{d}&\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 4.&SR&\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}} \\\hline 5.&S^{2}&\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i} \left (x_{i}-\overline{x} \right )^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}\\\hline 6.&S&\sqrt{\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i} \left (x_{i}-\overline{x} \right )^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukanlah nilai simpangan rata-ratanya}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&\colorbox{white}{47-49}&\colorbox{white}{50-52}&\colorbox{white}{53-55}&\colorbox{white}{56-58}&\colorbox{white}{59-61}\\\hline \textrm{Frek}&2&4&6&5&3\\\hline \end{array}\\\\ &\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{array}{ll} &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&f_{i}.x_{i}&\left | x_{i}-\overline{x} \right |&f_{i}.\left | x_{i}-\overline{x} \right |\\\hline 47-49&48&2&96&6,45&12,49\\\hline 50-52&51&4&204&3,45&13,8\\\hline 53-55&\colorbox{yellow}{54}&6&\colorbox{yellow}{324}&\colorbox{yellow}{0,45}&\colorbox{yellow}{2,7}\\\hline 56-58&57&5&285&2,55&12,75\\\hline 59-61&60&3&180&5,55&16,65\\\hline \textrm{Jumlah}&&20&1089&&58,8\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\begin{aligned}\overline{x}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.x_{i}}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=54+\displaystyle \frac{1089}{20}=54+0,45=\color{red}54,45 \end{aligned}\\ &\begin{aligned}SR&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{58,8}{20}\\ &=\color{red}2,94 \end{aligned}\\ &\textrm{Jadi, simpangan rata-ratanya adalah}\: SR=2,94 \end{array} \\\\ &\textbf{Alternatif 2}\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&d_{i}&f_{i}.d_{i}&\left | x_{i}-\overline{x} \right |&f_{i}.\left | x_{i}-\overline{x} \right |\\\hline \colorbox{white}{47-49}&48&2&-6&-12&6,45&12,49\\\hline \colorbox{white}{50-52}&51&4&-3&-12&3,45&13,8\\\hline \colorbox{yellow}{53-55}&\colorbox{yellow}{54}&6&\colorbox{yellow}0&\colorbox{yellow}0&\colorbox{yellow}{0,45}&\colorbox{yellow}{2,7}\\\hline \colorbox{white}{56-58}&57&5&3&15&2,55&12,75\\\hline \colorbox{white}{59-61}&60&3&6&18&5,55&16,65\\\hline \textrm{Jumlah}&&20&&9&&58,8\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\begin{aligned}\overline{x}&=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=54+\displaystyle \frac{9}{20}=54+0,45=\color{red}54,45 \end{aligned}\\ &\begin{aligned}SR&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}.\left | x_{i}-\overline{x} \right |}{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{58,8}{20}\\ &=\color{red}2,94 \end{aligned}\\ &\textrm{Jadi, simpangan rata-ratanya adalah}\: SR=2,94 \end{array}$.

$\begin{array}{ll} 2.&\textrm{Tentukanlah nilai varian/ragamnya}\\ &\textrm{dari data soal no.1 di atas}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|c|c|c|c|c|c|}\hline \textrm{Nilai}&x_{i}&f_{i}&\left | x_{i}-\overline{x} \right |& (x_{i}-\overline{x})^{2}&f_{i}. (x_{i}-\overline{x})^{2} \\\hline 47-49&48&2&6,45&41,6025&83,205\\\hline 50-52&51&4&3,45&11,9025&47,61\\\hline 53-55&54&6&324&0,2025&1,215\\\hline 56-58&57&5&285&6,5025&32,5125\\\hline 59-61&60&3&180&30,8025&92,4075\\\hline \textrm{Jumlah}&&20&&&256,95\\\hline \end{array}\\ &\textrm{ingat}\: \: x_{i}=\textrm{nilai tengah interval kelas}\\ &\textrm{dan}\: \: \overline{x}=\color{red}54,45\: (\textrm{lihat soal no.1})\\ &\textrm{maka}\\ &\begin{aligned}S^{2}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{k}f_{i}. (x_{i}-\overline{x})^{2} }{\displaystyle \sum_{i=1}^{k}f_{i}}\\ &=\displaystyle \frac{256,95}{20}\\ &=\color{red}12,8475 \end{aligned}\\ &\textrm{Jadi, varian/ragamnya adalah}\\ & S^{2}=12,8475 \end{array}$.

$\begin{array}{ll} 3.&\textrm{Tentukanlah nilai simpangan baku dari}\\ &\textrm{dari data soal no.1 di atas}\\\\ &\textbf{Jawab}:\\ &S=\sqrt{S^{2}}=\sqrt{12,8475}\approx \color{red}3,58 \end{array}$.

Ukuran Penyebaran Data Tunggal (Materi Kelas XII Matematika Wajib) (Bagian 1)

A. Pengertian

Ukuran penyebaran data adalah nilai dari ukuran yang memberikan gambaran sejauh mana data menyebar atau menyimpang (dispersi/deviasi) dari ukuran pemusatan data. Dalam hal ini bagian yang akan disinggung dalam materi ini adalah: Jangkauan (Range), Jangkauan antar kuartil, Simpangan kuartil, Simpangan rata-rata, Ragam (Variansi), Simpangan baku (Deviasi Standar), Koefisien variansi.

$\begin{array}{|c|l|c|}\hline \textrm{No}&\: \: \: \: \textrm{Data Dispersi}&\textrm{Simbol}\\\hline 1.&\textrm{Jangkauan}&R\: \: \textrm{atau}\: \: J\\\hline 2.&\textrm{Jangkauan}&H\\ &\textrm{antarkuartil}&\\\hline 3.&\textrm{Simpangan}&Q_{d}\\ &\textrm{kuartil}&\\\hline 4.&\textrm{Langkah}&L\\\hline 5.&\textrm{Pagar dalam}&Q_{1}-L\\\hline 6.&\textrm{Pagar luar}&Q_{3}-L\\\hline 7.&\textrm{Simpangan}&SR\\ &\textrm{rata-rata}&\\\hline 8.&\textrm{Ragam/variansi}&S^{2}\\\hline 9&\textrm{Simpangan baku}&S\\\hline 10.&\textrm{Koefisien variansi}&V\\\hline \end{array}$.

Sebagai catatan bahwa $H$ selain disebut jangkauan antarkuartil sebagaian ada yang menyebut dengan istilah rentang antar kuartil dan terkadang pula dengan sebutan jangkauan interkuartil (Inter Quartile Range) dan juga terkadang menyebutnya dengan hamparan. Untuk $Q_{d}$  selanjutnyanya ada yang buku yang menyebutnya dengan istilah simpangan kuartil terkadang juga rentang semi interkuartil atau jangkauan antarkuartil.

Perhatikan gambar distribusi frekuensi suatu data berikut

B. Ukuran Penyebaran Data

B. 1 Data Tunggal

$\begin{array}{|c|l|c|}\hline \textrm{No}&\quad \textrm{Data}&\textrm{Formula}\\\hline 1.&R\: \: \textrm{atau}\: \: J&x_{max}-x_{min}\\\hline 2.&H&Q_{3}-Q_{1}\\\hline 3.&Q_{d}&\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 4.&L&\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 5.&Q_{1}-L&Q_{1}-L\\\hline 6.&Q_{3}-L&Q_{3}-L\\\hline 7.&SR&\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\\hline 8.&S^{2}&\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2}\\\hline 9&S&\sqrt{S2}\\\hline 10.&V&\displaystyle \frac{S}{\overline{x}}\times 100 \%\\\hline \end{array}$.

Catata: Data ukuran yang kurang dari pagar dalam dan atau lebih besar dari pagar luar dinamakan pencilan.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} 30&32&32&43&50&51\\ 53&53&58&58&58&60\\ 63&64&66&67&68&69\\ 70&72&75&78&80&82\\ 84&85&86&86&83&83 \end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Jangkauan}\\ &\textrm{b}.\quad Q_{1},\, Q_{2},\, \textrm{dan}\: \: Q_{3}\\ &\textrm{c}.\quad \textrm{Jangkauan Antarkuartil}\\ &\textrm{d}.\quad \textrm{Simpangan Kuartil}\\ &\textrm{e}.\quad \textrm{Pagar Dalam}\\ &\textrm{f}.\quad \textrm{Pagar Luar}\\ &\textrm{g}.\quad \textrm{Pencilan}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan sajian data dalam bentuk}\\ &\textrm{diagram}\: \textbf{batang daun}\: \textrm{berikut}\\ &\begin{array}{|c|l|}\hline \textbf{Batang}&\: \quad\quad\textbf{Daun}\\\hline \color{red}3&0\: \: 2\: \: 2\\ \color{red}4&3\\ \color{red}5&0\: \: 1\: \: 3\: \: 3\: \: 8\: \: 8\: \: 8\\ \color{red}6&0\: \: 3\: \: 4\: \: 6\: \: 7\: \: 8\: \: 9\\ \color{red}7&0\: \: 2\: \: 5\: \: 8\\ \color{red}8&0\: \: 2\: \: 3\: \: 3\: \: 4\: \: 5\: \: 6\: \: 6\\\hline \end{array}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\: \: \color{blue}n=30\\ \textrm{a}.\quad \: J&=x_{max}-x_{min}=86-30=\color{red}56\\ \textrm{b}.\: \: \: Q_{1}&=\left ( x_{._{\frac{1}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{1}{4}.30+\frac{1}{2}}} \right )=x_{.8}=53\\ Q_{2}&=\left ( x_{._{\frac{2}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{2}{4}.30+\frac{1}{2}}} \right )\\ &=\displaystyle \frac{x_{.15}+x_{.16}}{2}=\displaystyle \frac{66+67}{2}=66,7\\ Q_{3}&=\left ( x_{._{\frac{3}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{3}{4}.30+\frac{1}{2}}} \right )=x_{.23}=80\\ \textrm{c}.\: \: \: H&=Q_{3}-Q_{1}\\ &=x_{._{23}}-x_{._{8}}=80-53=27\\ \end{aligned} \end{array}$

$.\qquad\begin{aligned}\textrm{d}.\: \: \: Q_{d}&=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &=\displaystyle \frac{1}{2}(H)=\displaystyle \frac{1}{2}\left ( 27 \right )=\color{red}13,5\\ \textrm{e}.\: \quad L&=\displaystyle \frac{3}{2}(H)=\displaystyle \frac{3}{2}(27)=\color{red}40,5\\ \textrm{P}&\textrm{agar dalam}:\\ &=Q_{1}-L=53-40,5=\color{red}12,5\\ \textrm{P}&\textrm{agar luar}:\\ &=Q_{1}-L=80+40,5=\color{red}120,5\\ \textrm{g}.\: \quad \textrm{D}&\textrm{ari fakta yang ada data ukuran}\\ &\textrm{yang besarnya kurang dari}\\ &\textrm{pagar dalam dan lebih besar dari}\\ &\textrm{pagar luar tidak ada, maka} \\ &\textrm{tidak ada}\: \color{red}\textbf{data pencilan} \end{aligned}$.

$\begin{array}{ll} 2.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} 73&74&66&65&68&65\\ 60&64&78&79&81&61\\ 72&74&71&68&75&76\\ 96&56&64&80&84&43\end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Jangkauan}\\ &\textrm{b}.\quad Q_{1},\, Q_{2},\, \textrm{dan}\: \: Q_{3}\\ &\textrm{c}.\quad \textrm{Jangkauan Antarkuartil}\\ &\textrm{d}.\quad \textrm{Simpangan Kuartil}\\ &\textrm{e}.\quad \textrm{Pagar Dalam}\\ &\textrm{f}.\quad \textrm{Pagar Luar}\\ &\textrm{g}.\quad \textrm{Pencilan}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan sajian data dalam bentuk}\\ &\textrm{diagram}\: \textbf{batang daun}\: \textrm{berikut}\\ &\begin{array}{|c|l|}\hline \textbf{Batang}&\: \quad\quad\textbf{Daun}\\\hline \color{red}4&3\\ \color{red}5&6\\ \color{red}6&0\: \: 1\: \: 4\: \: 4\: \: 5\: \: 5\: \: 6\: \: 8\: \: 8\\ \color{red}7&1\: \: 2\: \: 3\: \: 4\: \: 4\: \: 5\: \: 6\: \: 8\: \: 9\\ \color{red}8&0\: \: 1\: \: 3\\ \color{red}9&6\\\hline \end{array}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\: \: \color{blue}n=24\\ \textrm{a}.\quad \: J&=x_{max}-x_{min}=96-43=\color{red}53\\ \textrm{b}.\: \: \: Q_{1}&=\left ( x_{._{\frac{1}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{1}{4}.24+\frac{1}{2}}} \right )=x_{_{6,5}}\\ &=\displaystyle \frac{1}{2}\left ( x_{._{6}}+x_{._{7}} \right )=\displaystyle \frac{64+65}{2}=\color{red}64,5\\ Q_{2}&=\left ( x_{._{\frac{2}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{2}{4}.24+\frac{1}{2}}} \right )=x_{_{12,5}}\\ &=\displaystyle \frac{x_{.12}+x_{.13}}{2}=\displaystyle \frac{71+72}{2}=\color{red}71,5\\ Q_{3}&=\left ( x_{._{\frac{3}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{3}{4}.24+\frac{1}{2}}} \right )=x_{_{18,5}}\\ &=\displaystyle \frac{x_{_{18}}+x_{_{19}}}{2}=\displaystyle \frac{76+78}{2}=\color{red}77\\ \textrm{c}.\: \: \: H&=Q_{3}-Q_{1}\\ &=77-64,5=12,5\\ \end{aligned} \end{array}$.

$.\qquad\begin{aligned}\textrm{d}.\: \: \: Q_{d}&=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &=\displaystyle \frac{1}{2}(H)=\displaystyle \frac{1}{2}\left ( 12,5 \right )=\color{red}6,26\\ \textrm{e}.\: \quad L&=\displaystyle \frac{3}{2}(H)=\displaystyle \frac{3}{2}(12,5)=\color{red}18,75\\ \textrm{P}&\textrm{agar dalam}:\\ &=Q_{1}-L=64,5-18,75=\color{red}45,75\\ \textrm{P}&\textrm{agar luar}:\\ &=Q_{1}-L=77+18,75=\color{red}95,75\\ \textrm{g}.\: \quad \textrm{D}&\textrm{ari fakta di atas terdapat}\: \textbf{pencilan}\\ &\textrm{yaitu}:\: \color{red}43 \: \color{black}\textrm{dan}\: \: \color{red}96 \end{aligned}$.

$\begin{array}{ll} 3.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} \color{black}\textrm{a}.&3&4&5&6&7\\ \color{black}\textrm{b}.&1&2&5&8&9\end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Simpangan rata-rata}\\ &\textrm{b}.\quad \textrm{Ragam}\\ &\textrm{c}.\quad \textrm{Simpangan baku}\\\\ &\textbf{Jawab}:\\ &\textrm{Untuk data}:3,4,5,6,7\\ &\begin{aligned}\textrm{Diketahu}&\textrm{i}\: \: \color{blue}n=5\\ \textrm{a}.\quad \: \overline{x}=&\displaystyle \frac{3+4+5+6+7}{5}=\frac{25}{5}=\color{red}5\\ \textrm{sel}&\textrm{anjutnya}\\ \textrm{SR}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right | \\ &=\displaystyle \frac{1}{5}\left (\left | 3-5 \right | +\left | 4-5 \right |+\left | 5-5 \right |+\left |6-5 \right |+\left | 7-5 \right | \right )\\ &=\displaystyle \frac{1}{5}\left ( \left | -2 \right |+\left | -1 \right |+\left | 0 \right |+\left | 1 \right |+\left | 2 \right | \right )\\ &=\displaystyle \frac{1}{5}(2+1+0+1+2)\\ &=\displaystyle \frac{6}{5}=\color{red}1,2 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \: \textrm{S}^{2}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2} \\ &=\displaystyle \frac{1}{5}\left ((3-5)^{2} +(4-5)^{2}+(5-5)^{2}+(6-5)^{2}+(7-5)^{2} \right )\\ &=\displaystyle \frac{1}{5}\left ( 4+1+0+1+4 \right )\\ &=\displaystyle \frac{1}{5}(8)\\ &=\displaystyle \frac{8}{5}=\color{red}1,6\\ \textrm{c}.\quad \: \: \: S&=\sqrt{S^{2}}\\ &=\sqrt{1,6}\approx \color{red}1,26 \end{aligned}\\\\ &\textrm{Dan untuk data}:1,2,5,8,9\\ &\begin{aligned}\textrm{Diketahu}&\textrm{i}\: \: \color{blue}n=5\\ \textrm{a}.\quad \: \overline{x}=&\displaystyle \frac{1+2+5+8+9}{5}=\frac{25}{5}=\color{red}5\\ \textrm{sel}&\textrm{anjutnya}\\ \textrm{SR}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right | \\ &=\displaystyle \frac{1}{5}\left (\left | 1-5 \right | +\left | 2-5 \right |+\left | 5-5 \right |+\left |8-5 \right |+\left | 9-5 \right | \right )\\ &=\displaystyle \frac{1}{5}\left ( \left | -4 \right |+\left | -3 \right |+\left | 0 \right |+\left | 3 \right |+\left | 4 \right | \right )\\ &=\displaystyle \frac{1}{5}(4+3+0+3+4)\\ &=\displaystyle \frac{14}{5}=\color{red}2,8 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \: \textrm{S}^{2}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2} \\ &=\displaystyle \frac{1}{5}\left ((1-5)^{2} +(2-5)^{2}+(5-5)^{2}+(8-5)^{2}+(9-5)^{2} \right )\\ &=\displaystyle \frac{1}{5}\left ( 16+9+0+9+16 \right )\\ &=\displaystyle \frac{1}{5}(50)\\ &=\displaystyle \frac{50}{5}=\color{red}10\\ \textrm{c}.\quad \: \: \: S&=\sqrt{S^{2}}\\ &=\sqrt{10}\approx \color{red}3,16 \end{aligned} \end{array}$.

$\LARGE\colorbox{aqua}{LATIHAN SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukan nilai Jangkauan},Q_{1},Q_{2},Q_{3}\\ &hamparan,\: \textrm{simpangan kuartil, langkah}\\ &\textrm{pagar dalam, pagar luar, dan pencilan}\\ &\textrm{dari data berikut}\\ &\color{red}\begin{array}{llll} \color{black}\textrm{a}.&3,5,7,9,1,2,8,2,3,4,3,5,7\\ \color{black}\textrm{b}.&10,11,12,13,8,9,4,5,7,5\end{array} \end{array}$.

$\begin{array}{ll} 2.&\textrm{Tentukan simpangan rata-rata}\\ &\textrm{ragam, dan simpangan baku}\\ &\textrm{dari data berikut}\\ &\color{red}\begin{array}{llll} \color{black}\textrm{a}.&3,5,7,9,1\\ \color{black}\textrm{b}.&10,11,12,13,8,9,4,15,7,5\end{array} \end{array}$.

$\begin{array}{ll} 3.&\textrm{Empat buah bilangan memiliki mean,}\\ &\textrm{tentukanlah keempat bilangan tersebut}\\ \end{array}$.

$\begin{array}{ll} 4.&\textrm{Diketahui datum-datum}\\ &:x-4,x-2,x+1,x+2,x+4,x+5\\ &\textrm{tentukanlah}\\ &\textrm{a}.\quad\textrm{nilai simpangan baku(nyatakan dalam)\: }x\\ &\textrm{b}.\quad\textrm{nilai}\: \: x\: \: \textrm{dan simpangan baku jika mean}\\ & \: \: \: \: \quad\textrm{dari data di atas adalah 9} \end{array}$.

$\begin{array}{ll} 5.&\textrm{Diketahui simpangan baku}\\ &:2,4,7,11,9-n,9+n\: \: \textrm{adalah}\: \: \sqrt{11}\\ &\textrm{tentukanlah}\\ &\textrm{a}.\quad\textrm{mean}\\ &\textrm{b}.\quad\textrm{nilai}\: \: n\: \: \textrm{yang mungkin} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Alam Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: Srikandi Empat Widya Utama.
3. Sharma, S.N., dkk. 2017. Jelajah Matematika 3 SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.

Interpolasi Linear

Sumber ada di sini

materi pendukung untuk desil klik di sini dan

materi persentil klik di sini.

Interpolasi linear adalah sebuah metode yang digunakan untuk penentuan titik di antara dua buah titik yang sudah diketahui dan segaris.

Perhatikanlah ilustrasi gambar berikut

dengan proses seperti menentukan persamaan garis lurus diperoleh rumus:

$\begin{aligned}\displaystyle \frac{y-y_{0}}{y_{1}-y_{0}}&=\frac{x-x_{0}}{x_{1}-x_{0}}\\ y-y_{0}&=\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right ) \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukan taksiran nilai dari}\\ &\textrm{a}.\quad \sqrt{5}\qquad\qquad \textrm{c}.\quad \sqrt{12}\\ &\textrm{b}.\quad \sqrt{7}\qquad\qquad \textrm{d}.\quad \sqrt{22}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \textrm{Di}&\textrm{ketahui bahwa}\\ &\begin{cases} \sqrt{4} & =2 \\ \sqrt{5} & =\: \: ? \\ \sqrt{9} & =3 \end{cases}\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ &\approx 2+\displaystyle \frac{5-4}{9-4}(3-2)\\ &\approx2+\displaystyle \frac{1}{5}\\ &\approx2+0,2\\ &\approx\color{red}2,2 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \textrm{Di}&\textrm{ketahui bahwa}\\ &\begin{cases} \sqrt{4} & =2 \\ \sqrt{7} & =\: \: ? \\ \sqrt{9} & =3 \end{cases}\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ &\approx2+\displaystyle \frac{7-4}{9-4}(3-2)\\ &\approx2+\displaystyle \frac{3}{5}\\ &\approx2+0,6\\ &\approx\color{red}2,6 \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{Di}&\textrm{ketahui bahwa}\\ &\begin{cases} \sqrt{9} & =3 \\ \sqrt{12} & =\: \: ? \\ \sqrt{16} & =4 \end{cases}\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ &\approx3+\displaystyle \frac{12-9}{16-9}(4-3)\\ &\approx3+\displaystyle \frac{3}{7}\\ &\approx3+0,43\\ &\approx\color{red}3,43 \end{aligned} \\ &\begin{aligned}\textrm{d}.\quad \textrm{Di}&\textrm{ketahui bahwa}\\ &\begin{cases} \sqrt{16} & =4 \\ \sqrt{22} & =\: \: ? \\ \sqrt{25} & =5 \end{cases}\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ &\approx4+\displaystyle \frac{22-16}{25-16}(5-4)\\ &\approx4+\displaystyle \frac{6}{9}\\ &\approx4+0,67\\ &\approx\color{red}4,67 \end{aligned} \end{array}$.

$\begin{array}{ll} 2.&\textrm{Diberikan data berikut berkaitan dengan}\\ &\textrm{penduduk di suatu daerah A}\\ &\begin{array}{|c|l|l|}\hline \textrm{Tahun}&2015&2020\\\hline \begin{aligned}&\textrm{Jumlah jiwa}\\ &\textrm{daerah A} \end{aligned}&340.000&600.000\\\hline \end{array}\\ &\textrm{Tentukan perkiraan jumlah penduduk}\\ &\textrm{daerah A saat tahun 2018}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Di}&\textrm{ketahui bahwa}\\ &\begin{cases} \sqrt{2015} & =340.000 \\ \sqrt{2018} & =\qquad ? \\ \sqrt{2020} & =600.000 \end{cases}\\ y&=y_{0}+\frac{x-x_{0}}{x_{1}-x_{0}}\left ( y_{1}-y_{0} \right )\\ &\approx 340.000+\displaystyle \frac{(2018-2015)}{(2020-2015)}(600.000-340.000)\\ &\approx 340.000+\displaystyle \frac{3}{5}\left ( 260.000 \right )\\ &\approx 340.000+156.000\\ &\approx\color{red}496.000 \end{aligned} \end{array}$.

Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 3

 D. Persentil

$\begin{aligned}&\textrm{Dilambangkan dengan}\: \: P_{i}\\ &\textrm{dibaca: persentil ke}-i\\ &\color{red}\textrm{Rumus data tunggal}:\\ &\textbf{Dengan rumus pendekatan interpolasi linear}\\ &P_{i}=\textrm{datum ke}-\displaystyle \frac{i}{100}(n+1)\\ &\textrm{jika}\: \: \displaystyle \frac{i}{100}(n+1)\: \: \textrm{tidak bulat, gunakan rumus}\\ &\textrm{interpolasi linear, yaitu}:\\ &\qquad\quad P_{i}=x_{k}+d\left ( x_{k+1}-x_{k} \right )\\ &\textrm{dengan}\: \: d\: \: \textrm{adalah nilai desimalnya}\\ &\textbf{Dengan tanpa rumus interpolasi linear}\\ &\begin{array}{|c|c|}\hline n\: \: \textrm{ganjil}&n\: \: \textrm{genap}\\\hline P_{1}=x_{._{\frac{1}{100}(n+1)}}&P_{1}=\displaystyle \frac{1}{2}\left (x_{._{\frac{1}{100}n}}+x_{._{\frac{1}{100}n+1}} \right )\\ P_{2}=x_{._{\frac{2}{100}(n+1)}}&P_{2}=\displaystyle \frac{1}{2}\left (x_{._{\frac{2}{100}n}}+x_{._{\frac{2}{100}n+1}} \right )\\ \vdots &\vdots \\ P_{99}=x_{._{\frac{99}{100}(n+1)}}&P_{99}=\displaystyle \frac{1}{2}\left (x_{._{\frac{99}{100}n}}+x_{._{\frac{99}{100}n+1}} \right )\\\hline \end{array}\\ &\textrm{Catatan: sesuaikan dengan kondisi soal}\\ &\color{red}\textrm{Rumus data berkelompok/berfrekuensi}:\\ &P_{i}=L_{i}+\left ( \displaystyle \frac{\displaystyle \frac{i}{100}n-f_{k}}{f} \right )\times c\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}D_{i}&=\textrm{persentil ke}-i\\ i&=1,2,3\\ L_{i}&=\textrm{tepi bawah kelas persentil ke}-i\\ f_{k}&=\textrm{frekuensi kumulatif sebelum}\\ &\quad\: \, \textrm{sebelum kelas persentil ke}-i\\ f&=\textrm{frekuensi kelas persentil ke}-i\\ c&=\textrm{panjang kelas interval}\\ n&=\textrm{banyak data/kelas interval} \end{aligned} \end{aligned}$.

Catatan :untuk bahasan interpolasi linear ada di sini

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukan persentil ke-29 dan ke-75 dari data berikut}\\ &4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\\\ &\textrm{Jawab}:\\ &\textrm{Banyak datum}=15\\ &\textbf{dengan rumus pendekatan interpolasi linear}\\ &\textrm{Data mula-mula}\\ &\quad :4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ &\textrm{Data setelah diurutkan}\\ &\quad :2,3,4,4,4,5,5,6,6,6,7,7,8,8,9\\ &\begin{aligned}P_{29}&=\displaystyle \frac{29}{100}\left ( 15+1 \right )\\ &=\displaystyle \frac{464}{100}=4,64\\ &=x_{4}+0,64(x_{5}-x_{4})\\ &=4+0,64(5-5)=4+0=\color{red}4\\ P_{75}&=\displaystyle \frac{75}{100}\left ( 15+1 \right )\\ &=\displaystyle \frac{1200}{100}=12\\ &=x_{12}\\ &=\color{red}7\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Persentil ke-32}\: \: \left ( P_{32} \right )\: \: \textrm{dari data berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Nilai}&f\\\hline 41-45&7\\ 46-50&12\\ 51-55&9\\ 56-60&8\\ 61-65&4\\\hline \end{array}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&46\\ \textrm{b}.&47\\ \textrm{c}.&48\\ \textrm{d}.&51\\ \textrm{e}.&52 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\: \textrm{persentil ke}-32=\color{black}P_{32},\\ &\textrm{dengan}\: \: n=\sum f=40\\ P_{i}&=\color{black}L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{i\times n}{100}-f_{k}}{f} \right )\\ P_{32}&=\color{black} \textrm{datum ke}-\left ( \displaystyle \frac{32n}{100} \right )\\ &=x_{\frac{32\times 40}{100}}=\color{red}x_{12,8}\\ \textrm{Dan}\: \: \color{red}x_{12}\: \: &\textrm{terletak di kelas interval}\: :\: \color{red}46-50 \\ P_{32}&=\color{black}545,5+5\left ( \displaystyle \frac{12,8-7}{12} \right )\\ &=\color{black}45,5+0,48333...\\ &=\color{red}45.9833... \approx 46 \end{aligned} \end{array}$

Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 2

 C. Desil

$\begin{aligned}&\textrm{Dilambangkan dengan}\: \: D_{i}\\ &\textrm{dibaca: desil ke}-i\\ &\color{red}\textrm{Rumus data tunggal}:\\ &\textbf{Dengan rumus pendekatan interpolasi linear}\\ &D_{i}=\textrm{datum ke}-\displaystyle \frac{i}{10}(n+1)\\ &\textrm{jika}\: \: \displaystyle \frac{i}{10}(n+1)\: \: \textrm{tidak bulat, gunakan rumus}\\ &\textrm{interpolasi linear, yaitu}:\\ &\qquad\quad D_{i}=x_{k}+d\left ( x_{k+1}-x_{k} \right )\\ &\textrm{dengan}\: \: d\: \: \textrm{adalah nilai desimalnya}\\ &\textbf{Dengan tanpa rumus interpolasi linear}\\ &\begin{array}{|c|c|}\hline n\: \: \textrm{ganjil}&n\: \: \textrm{genap}\\\hline D_{1}=x_{._{\frac{1}{10}(n+1)}}&D_{1}=\displaystyle \frac{1}{2}\left (x_{._{\frac{1}{10}n}}+x_{._{\frac{1}{10}n+1}} \right )\\ D_{2}=x_{._{\frac{2}{10}(n+1)}}&D_{2}=\displaystyle \frac{1}{2}\left (x_{._{\frac{2}{10}n}}+x_{._{\frac{2}{10}n+1}} \right )\\ \vdots &\vdots \\ D_{9}=x_{._{\frac{9}{10}(n+1)}}&D_{9}=\displaystyle \frac{1}{2}\left (x_{._{\frac{9}{10}n}}+x_{._{\frac{9}{10}n+1}} \right )\\\hline \end{array} \\ &\textrm{Catatan: sesuaikan dengan kondisi soal}\\ &\color{red}\textrm{Rumus data berkelompok/berfrekuensi}:\\ &D_{i}=L_{i}+\left ( \displaystyle \frac{\displaystyle \frac{i}{10}n-f_{k}}{f} \right )\times c\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}D_{i}&=\textrm{desil ke}-i\\ i&=1,2,3\\ L_{i}&=\textrm{tepi bawah kelas desil ke}-i\\ f_{k}&=\textrm{frekuensi kumulatif sebelum}\\ &\quad\: \, \textrm{sebelum kelas desil ke}-i\\ f&=\textrm{frekuensi kelas desil ke}-i\\ c&=\textrm{panjang kelas interval}\\ n&=\textrm{banyak data/kelas interval} \end{aligned} \end{aligned}$.

Catatan :untuk bahasan interpolasi linear ada di sini

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukanlah}\: \: \: D_{1},D_{2},D_{3},D_{4},D_{5},D_{6}\\ &D_{7},D_{8},D_{9}\: \: \: \textrm{dari data berikut}\\ & 2,3,8,9,2,4,5,8,9\\\\ &\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Total datum}=9\\ &\textrm{Data mula-mula}:\: 2,3,8,9,2,4,5,8,9\\ &\textrm{Setelah data diurutkan menjadi}\\ &\quad : 2,2,3,4,5,8,8,9,9\\ &D_{i}=\displaystyle \frac{i}{10}(n+1)\: .\: \textrm{Jika hasilnay tidak bulat}\\ &\textrm{maka dihitung dengan}\: \: D_{i}=x_{k}+d.(x_{k+1}-x_{k})\\ &\color{red}\textrm{Sehingga}\\ &\begin{aligned}D_{1}&=\displaystyle \frac{1}{10}(9+1)=\frac{10}{10}=1\\ &=x_{1}=2\\ D_{2}&=\displaystyle \frac{2}{10}(9+1)=\frac{20}{10}=2\\ &=x_{2}=2\\ D_{3}&=\displaystyle \frac{3}{10}(9+1)=\frac{30}{10}=3\\ &=x_{3}=3\\ D_{4}&=\displaystyle \frac{4}{10}(9+1)=\frac{40}{10}=4\\ &=x_{4}=4\\ D_{5}&=\displaystyle \frac{5}{10}(9+1)=\frac{50}{10}=5\\ &=x_{5}=5\\ D_{6}&=\displaystyle \frac{6}{10}(9+1)=\frac{60}{10}=6\\ &=x_{6}=8\\ D_{7}&=\displaystyle \frac{7}{10}(9+1)=\frac{70}{10}=7\\ &=x_{7}=8\\ D_{8}&=\displaystyle \frac{8}{10}(9+1)=\frac{80}{10}=8\\ &=x_{8}=9\\ D_{9}&=\displaystyle \frac{9}{10}(9+1)=\frac{90}{10}=9\\ &=x_{9}=9\\ \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll} 2.&\textrm{Tentukan desil ke-4 dan ke-6 dari data berikut}\\ &4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\\\ &\textrm{Jawab}:\\ &\textrm{Banyak datum}=15\\ &\textbf{dengan rumus pendekatan interpolasi linear}\\ &\textrm{Data mula-mula}\\ &\quad :4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ &\textrm{Data setelah diurutkan}\\ &\quad :2,3,4,4,4,5,5,6,6,6,7,7,8,8,9\\ &\begin{aligned}D_{4}&=\displaystyle \frac{4}{10}\left ( 15+1 \right )\\ &=\displaystyle \frac{64}{10}=6,4\\ &=x_{6}+0,4(x_{7}-x_{6})\\ &=5+0,4(5-5)=\color{red}5\\ D_{6}&=\displaystyle \frac{6}{10}\left ( 15+1 \right )\\ &=\displaystyle \frac{96}{10}=9,6\\ &=x_{9}+0,6(x_{10}-x_{9})\\ &=6+0,6(6-6)=\color{red}6\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Desil ke-8}\: \: \left ( D_{8} \right )\: \: \textrm{dari data berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Nilai}&f\\\hline 41-45&7\\ 46-50&12\\ 51-55&9\\ 56-60&8\\ 61-65&4\\\hline \end{array}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&58\\ \textrm{b}.&57,5\\ \textrm{c}.&57\\ \textrm{d}.&56,75\\ \textrm{e}.&56,25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\: \textrm{desil ke}-8=\color{black}D_{8},\: \: \textrm{dengan}\: \: n=\sum f=40\\ D_{i}&=\color{black}L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{i\times n}{10}-f_{k}}{f} \right )\\ D_{8}&=\color{black} \textrm{datum ke}-\left ( \displaystyle \frac{8n}{10} \right )=x_{\frac{8\times 40}{10}}=\color{red}x_{32}\\ \textrm{Dan}\: \: \color{red}x_{32}\: \: &\textrm{terletak di kelas interval}\: :\: \color{red}56-60 \\ D_{8}&=\color{black}55,5+5\left ( \displaystyle \frac{32-28}{8} \right )\\ &=\color{black}55,5+2,5\\ &=\color{red}58 \end{aligned} \end{array}$

Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 1

A. Pendahuluan

Sebelumnya telah dipelajari tentang salah satu bentuk ukuran pemusatan data yaitu median yang membagi sebuah data menjadi dua bagian yang sama. Selain median ada juga istilah lain yang dapat membagi sebuah data menjadi beberapa bagian yang sama pula, yaitu kuartl yang membagi sebuah data menjadi 4 bagian yang sama. Kemudian selain kuartil, ada juga desil yang memabgi sebuah data menjadi 10 bagian yang sama serta persentil yang membagi sebuah data menjadi 100 bagian yang sama pula.

B. Kuartil

$\begin{aligned}&\textrm{Dilambangkan dengan}\: \: Q_{i}\\ &\textrm{dibaca: kuartil ke}-i\\ &\color{red}\textrm{Rumus data tunggal}:\\ &\textbf{Dengan rumus pendekatan interpolasi linear}\\ &Q_{i}=\textrm{datum ke}-\displaystyle \frac{i}{4}(n+1)\\ &\textrm{jika}\: \: \displaystyle \frac{i}{4}(n+1)\: \: \textrm{tidak bulat, gunakan rumus}\\ &\textrm{interpolasi linear, yaitu}:\\ &\qquad\quad Q_{i}=x_{k}+d\left ( x_{k+1}-x_{k} \right )\\ &\textrm{dengan}\: \: d\: \: \textrm{adalah nilai desimalnya}\\ &\textbf{Dengan tanpa rumus interpolasi linear}\\ &\begin{array}{|c|c|}\hline n\: \: \textrm{ganjil}&n\: \: \textrm{genap}\\\hline Q_{1}=x_{._{\frac{1}{4}(n+1)}}&Q_{1}=x_{._{\frac{1}{4}n+\frac{1}{2}}}\\ Q_{2}=x_{._{\frac{2}{4}(n+1)}}&Q_{2}=x_{._{\frac{2}{4}n+\frac{1}{2}}}\\ Q_{3}=x_{._{\frac{3}{4}(n+1)}}&Q_{3}=x_{._{\frac{3}{4}n+\frac{1}{2}}}\\\hline \end{array}\\ &\textrm{Catatan: sesuaikan dengan kondisi soal}\\ &\color{red}\textrm{Rumus data berkelompok/berfrekuensi}:\\ &Q_{i}=L_{i}+\left ( \displaystyle \frac{\displaystyle \frac{i}{4}n-f_{k}}{f} \right )\times c\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}Q_{i}&=\textrm{kuartil ke}-i\\ i&=1,2,3\\ L_{i}&=\textrm{tepi bawah kelas kuartil ke}-i\\ f_{k}&=\textrm{frekuensi kumulatif sebelum}\\ &\quad\: \, \textrm{sebelum kelas kuartil ke}-i\\ f&=\textrm{frekuensi kelas kuartil ke}-i\\ c&=\textrm{panjang kelas interval}\\ n&=\textrm{banyak data/kelas interval} \end{aligned} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukanlah}\: \: Q_{1},Q_{2},Q_{3}\: \: \textrm{dari data}\\ &\textrm{berikut}\\ &\textrm{a}.\quad 3,5,7,1,2,4,9,7\\ &\textrm{b}.\quad 2,3,8,9,2,4,5,8,9\\\\ &\textrm{Jawab}:\\ &\textbf{Dengan rumus pendekatan interpolasi linear}\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Total datum}=8\\ &\textrm{Data mula-mula}:\: 3,5,7,1,2,4,9,7\\ &\textrm{Setelah data diurutkan menjadi}\\ &\quad : 1,2,3,4,5,7,7,9\\ &Q_{i}=\displaystyle \frac{i}{4}(n+1)\begin{cases} Q_{1} &=\displaystyle \frac{1}{4}(8+1)=2\displaystyle \frac{1}{4} \\\\ Q_{2} &=\displaystyle \frac{2}{4}(8+1)=4\displaystyle \frac{1}{2} \\\\ Q_{3} &=\displaystyle \frac{3}{4}(8+1)=6\displaystyle \frac{3}{4} \end{cases}\\ &Q_{1}=x_{2}+\displaystyle \frac{1}{4}(x_{3}-x_{2})=2+\displaystyle \frac{1}{4}=2\displaystyle \frac{1}{4}\\ &Q_{2}=x_{4}+\displaystyle \frac{1}{2}(x_{5}-x_{4})=4+\displaystyle \frac{1}{2}=4\displaystyle \frac{1}{2}\\ &Q_{3}=x_{6}+\displaystyle \frac{3}{4}(x_{7}-x_{6})=7+0=7 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\textrm{Total datum}=9\\ &\textrm{Data mula-mula}:\: 2,3,8,9,2,4,5,8,9\\ &\textrm{Setelah data diurutkan menjadi}\\ &\quad : 2,2,3,4,5,8,8,9,9\\ &Q_{i}=\displaystyle \frac{i}{4}(n+1)\begin{cases} Q_{1} &=\displaystyle \frac{1}{4}(9+1)=2\displaystyle \frac{1}{2} \\\\ Q_{2} &=\displaystyle \frac{2}{4}(9+1)=5 \\\\ Q_{3} &=\displaystyle \frac{3}{4}(9+1)=7\displaystyle \frac{1}{2} \end{cases}\\ &Q_{1}=x_{2}+\displaystyle \frac{1}{2}(x_{3}-x_{2})=2+\displaystyle \frac{1}{1}=2\displaystyle \frac{1}{2}\\ &Q_{2}=x_{5}=5\\ &Q_{3}=x_{7}+\displaystyle \frac{1}{2}(x_{8}-x_{7})=8+\displaystyle \frac{1}{2}=8\displaystyle \frac{1}{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Data penjualan suatu barang setiap bulan}\\ &\textrm{di sebuah toko pada tahun 2019 adalah}:\\ &20,3,9,11,4,12,1,9,9,12,8,10.\\ &\textrm{Median, kuartil bawah, dan kuartil atasnya}\\ &\textrm{berturut-turut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&6\displaystyle \frac{1}{2},3\frac{1}{2},\: \textrm{dan}\: \: 9\frac{1}{2}\\ \color{red}\textrm{b}.&9,6,\: \textrm{dan}\: \: 11\displaystyle \frac{1}{2}\\ \textrm{c}.&6\displaystyle \frac{1}{2},9,\: \textrm{dan}\: \: 12\\ \textrm{d}.&9,4,\: \textrm{dan}\: \: 12\\ \textrm{e}.&9,3\displaystyle \frac{1}{2},\: \textrm{dan}\: \: 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\textbf{Dengan tanpa rumus interpolasi linear}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &20,3,9,11,4,12,1,9,9,12,8,10\\ \textrm{Sete}&\textrm{lah data diurutkan}\\ :\: &1,3,4,8,9,9,9,10,11,12,12,20\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=12\: \: \color{black}\textbf{genap}\\ Q_{1}&=x_{\frac{1}{4}n+\frac{1}{2}}=x_{\frac{1}{4}.12+\frac{1}{2}}=x_{3,5}=\color{red}6\\ Q_{2}&=x_{\frac{2}{4}n+\frac{1}{2}}=x_{\frac{2}{4}.12+\frac{1}{2}}=x_{6,5}=\color{red}9=\color{black}M_{e}\\ Q_{3}&=x_{\frac{3}{4}n+\frac{1}{2}}=x_{\frac{3}{4}.12+\frac{1}{2}}=x_{9,5}=\color{red}11\displaystyle \frac{1}{2}\\ \textrm{Sela}&\textrm{njutnya data dapat dituliskan}\\ &1,3,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{1} \end{matrix}}{\underbrace{\color{red}4,8}},9,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{2}=M_{e} \end{matrix}}{\underbrace{\color{red}9,9}},10,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{3} \end{matrix}}{\underbrace{\color{red}11,12}},12,20\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&(\textbf{UN IPA 2014})\\ &\textrm{Kuartil atas dari data pada tabel berikut}\\ &\textrm{adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Data}&f\\\hline 20-25&4\\\hline 26-31&6\\\hline 32-37&6\\\hline 38-43&10\\\hline 44-49&12\\\hline 50-55&8\\\hline 56-61&4\\\hline \end{array}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&49,25\\ \textrm{b}.&48,75\\ \textrm{c}.&48,25\\ \textrm{d}.&47,75\\ \textrm{e}.&47,25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Kuartil atas}=\color{black}Q_{3},\: \textrm{dengan}\: \: n=\sum f=50\\ &\textrm{Kita sertakan lagi tabel di atas berikut}\\ &\begin{array}{|c|c|}\hline \color{black}\textrm{Data}&\color{red}f\\\hline 20-25&4\\\hline 26-31&6\\\hline 32-37&6\\\hline 38-43&10\\\hline \colorbox{white}{44-49}&\colorbox{white}{12}\\\hline 50-55&8\\\hline 56-61&4\\\hline \end{array}\\ &Q_{3}=\textrm{Datum ke}-\left ( \displaystyle \frac{3n}{4} \right )=x_{\frac{3.50}{4}}=x_{37,5}\\ &\textrm{dan}\: \: x_{37,5}\: \: \textrm{terletak di kelas interval}\: \: 44-49\\ &Q_{3}=\color{purple}L_{3}+c\left ( \displaystyle \frac{\displaystyle \frac{3n}{4}-f_{k}}{f} \right )\\ &\: \: \: \, =\color{black}43,5+6\left ( \displaystyle \frac{37,5-26}{12} \right )\\ &\: \: \: \, =\color{black}43,5+\displaystyle \frac{11,5}{2}\\ &\: \: \: \, =\color{black}49,5+5,75\\ &\: \: \: \, =\color{red}49,25 \end{aligned} \end{array}$

Ukuran Pemusatan Data (Materi Kelas XII Matematika Wajib)

 A. Pendahuluan

Dalam statistik deskriptif ada nilai yang yang dapat mewakili pengukuran baik dalam pengukuran secara tunggal ataupun berkelompok dan nilai ini selanjutnya dinamakan sebagai ukuran pemusatan data dikarenakan akan memiliki kecenderungan nilai yang sama. Bahasan yang dimasukkan dalam kelompok ukuran pemusatan data ini ada 3 macam, yaitu: mean, median, dan modus.

B. Ukuran Pemusatan Data

Jika diketahui terdapat data dengan datum-datum: $x_{1},x_{2},x_{3},x_{4},...,x_{n}$, maka

1. Mean(rata-rata)

$\begin{aligned}&\textrm{Dilambangkan dengan}\: \: \overline{x}\\ &\textrm{dibaca: x bar}.\\ &\color{red}\textrm{Rumus data tunggal}:\\ &\overline{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}+...+x_{n}}{n}\\ &\textrm{atau}\\ &\overline{x}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}x_{i}}{n}\\ &\color{red}\textrm{Rumus data berkelompok/berfrekuensi}:\\ &\overline{x}=\displaystyle \frac{f_{1}.x_{1}+f_{2}.x_{2}+f_{3}.x_{3}+...+f_{_{n}}.x_{n}}{f_{1}+f_{2}+f_{3}+...+f_{n}}\\ &\textrm{atau}\\ &\overline{x}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.x_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}&=\textrm{mean/rataan hitung}\\ f_{i}&=\textrm{frekuensi kelas interval ke}-i\\ x_{i}&=\textrm{nilai tengah kelas interval ke}-i\\ n&=\textrm{banyak data/kelas interval} \end{aligned} \end{aligned}$.

$\begin{aligned}&\color{red}\textrm{Rumus data berkelompok}:\\ &\color{red}\textrm{dengan menggunakan rataan sementara}\\ &\overline{x}=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}&=\textrm{mean/rataan hitung}\\ x_{s}&=\textrm{rataan sementara dipilih}\\ x_{i}&=\textrm{nilai tengah kelas interval ke}-i\\ f_{i}&=\textrm{frekuensi kelas interval ke}-i\\ d_{i}&=\textrm{simpangan/deviasi}=x_{i}-x_{s}\\ \end{aligned}\\\\ &\color{red}\textrm{Rumus gabungan}:\\ &\overline{x}_{g}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.\overline{x}_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}_{g}&=\textrm{rataan gabungan}\\ f_{i}&=\textrm{frekuensi kelompok ke}-i\\ \overline{x}_{i}&=\textrm{rata-rata kelompok ke}-i\\ \end{aligned} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah rata-rata dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{purple}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ \overline{x}&=\displaystyle \frac{5.1+6.4+7.6+8.4+9.1}{16}\\ &=\color{black}\frac{112}{16}=\color{red}7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 2-6&2\\\hline 7-11&3\\\hline 12-16&3\\\hline 17-21&6\\\hline 22-26&6\\\hline \end{array}\\ &\textrm{Tentukan rataannya}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\begin{array}{|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&\color{red}f_{i}&\color{black}x_{i}.f_{i}\\\hline 2-6&4&2&8\\\hline 7-11&9&3&27\\\hline 12-16&14&3&42\\\hline 17-21&19&6&114\\\hline 22-26&24&6&144\\\hline &\color{black}\sum_{i=1}^{5}&\color{red}20&\color{black}335\\\hline \end{array}\\ &\color{black}\textrm{maka rata-ratanya}\\ &\overline{x}=\displaystyle \frac{\sum x_{i}.f_{i}}{\sum f_{i}}=\frac{335}{20}=\color{red}16,75 \end{aligned} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Dengan rumus rataan sementara},\: \textrm{yaitu}:\\ &\begin{array}{|c|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&d_{i}=x_{i}-x_{s}&\color{red}f_{i}&\color{black}f_{i}.d_{i}\\\hline 2-6&4&-15&2&-30\\\hline 7-11&9&-10&3&-30\\\hline 12-16&14&-5&3&-15\\\hline 17-21&\color{red}19&\color{blue}0&6&0\\\hline 22-26&24&5&6&30\\\hline &&&\color{red}20&-45\\\hline \end{array}\\ &\begin{aligned}\overline{x}&=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &=19+\displaystyle \frac{-45}{20}\\ &=19-2,25\\ &=\color{red}16,75 \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 3.&\textrm{Nilai rata-rata dari 20 bilangan adalah}\: \: 14,2\\ &\textrm{Jika rata-rata dari 12 bilangan pertama adalah}\\ &\textrm{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ &\textrm{adalah 18,2, tentukan rata-rata 2 bilangan}\\ &\textrm{tersisa}\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\color{black}\displaystyle \frac{\overline{x}_{12}\times (12)+\overline{x}_{6}\times (6)+\overline{x}_{2}\times (2)}{20}\\ 14,2&=\color{black}\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+\overline{x}_{2}\times (2)}{20}\\ 284&=\color{black}151,2+109,2+2\overline{x}_{2}\\ 2\overline{x}_{2}&=\color{black}284-(151,2+109,2)=\color{red}23,6\\ \overline{x}_{2}&=\displaystyle \frac{23,6}{2}\\ &=\color{red}11,8 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&(\textbf{SPMB 04})\\ &\textrm{Nilai rata-rata tes Matematika dari kelompok}\\ &\textrm{siswa dan kelompok siswi di suatu kelas berturut-}\\ &\textrm{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ &\textrm{tersebut adalah 6,2, maka perbandingan banyaknya}\\ &\textrm{siswa dan siswi adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2:3\\ \textrm{b}.&3:4\\ \textrm{c}.&2:5\\ \textrm{d}.&3:5\\ \textrm{e}.&4:5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\displaystyle \frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}\\ \color{black}6,2&\color{blue}=\displaystyle \frac{n_{1}(5)+n_{2}(7)}{n_{1}+n_{2}}\\ 6,2(n_{1}+n_{2})&=5n_{1}+7n_{2}\\ 6,2n_{1}-5n_{1}&=7n_{2}-6,2n_{2}\\ 1,2n_{1}&=0,8n_{2}\\ \displaystyle \frac{n_{1}}{n_{2}}&=\displaystyle \frac{0,8}{1,2}=\color{red}\frac{2}{3} \end{aligned} \end{array}$.

2. Median 

$\begin{aligned}&\textrm{Dilambangkan dengan}:\: Me\\ &\color{red}\textrm{Rumus data tunggal}\\ &\bullet \: \: \textrm{data ganjil},\: Me=x_{._{\frac{n+1}{2}}}\\ &\bullet \: \: \textrm{data genap},\: Me=\displaystyle \frac{1}{2}\left ( x_{._{\frac{n}{2}}}+x_{._{\frac{n}{2}+1}} \right )\\ &\color{red}\textrm{Rumus data berkelompok}\\ &Me=L+\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\times c \\ &\color{blue}\textrm{Penjelasan}\\ &\begin{aligned}L&=\textrm{tepi bawah kelas median}\\ f_{k}&=\textrm{frekuensi kumulatif sebelum}\\ &\: \, \quad \textrm{kelas median}\\ c&=\textrm{lebar kelas interval}\\ n&=\textrm{total datum} \end{aligned} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah median dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{purple}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ Me&=\displaystyle \frac{1}{2}\left ( x_{._{\frac{n}{2}}}+x_{._{\frac{n}{2}+1}} \right )\\ &=\displaystyle \frac{1}{2}\left ( x_{._{\frac{16}{2}}}+x_{._{\frac{16}{2}+1}} \right )=\displaystyle \frac{1}{2}\left ( x_{8}+x_{9} \right )\\ &=\displaystyle \frac{7+7}{2}\\ &=\color{black}\frac{14}{2}=\color{red}7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Dari data berikut yang memiliki}\\ &\textbf{mean}\: \: 7\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \textbf{median}\: 7\: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2,5,6,9,7,8,5,14,8,11\\ \textrm{b}.&6,3,7,8,6,4,11,8,9,8\\ \textrm{c}.&3,7,10,7,9,5,10,2,14,11\\ \textrm{d}.&4,1,6,12,8,11,4,5,8,2\\ \color{red}\textrm{e}.&2,3,4,3, 10,8,12,6,15,12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{array}{|l|l|}\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \textrm{Median}&2,5,6,9,7,8,5,14,8,11\\\hline \quad \textrm{a}&\color{blue}2,5,5,6,\color{red}7,8,\color{blue}8,9,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{70}{10}=7\\\hline \textrm{Median}&6,3,7,8,6,4,11,8,9,8\\\hline \quad \textrm{b}&\color{blue}3,4,6,6,\color{red}7,8,\color{blue}8,8,9,11\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{78}{10}=7,8\\\hline \textrm{Median}&3,7,10,7,9,5,10,2,14,11\\\hline \quad \textrm{c}&\color{blue}2,3,5,7,\color{red}7,9,\color{blue}10,10,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{61}{10}=6,1\\\hline \textrm{Median}&4,1,6,12,8,11,4,5,8,2\\\hline \quad \textrm{d}&\color{blue}1,2,4,4,\color{red}5,6,\color{blue}8,8,11,12\\\hline \color{blue}\textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \color{blue}\textrm{Median}&2,3,4,3, 10,8,12,6,15,12\\\hline \quad \color{red}\textrm{e}&\color{blue}2,3,3,4,\color{red}6,8,\color{blue}10,12,12,15\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 51-60&5\\\hline 61-70&7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline \end{array}\\ &\textrm{Tentukanlah mediannya}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\textrm{Diketahui},\: n=\sum f=40,\: \: \textrm{Perhatikan tabel}\\ &\textrm{berikut ini}\\ &\begin{array}{|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}f_{i}\\\hline 51-60&\color{red}5\\\hline 61-70&\color{red}7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline &\color{red}40\\\hline \end{array}\\ &\color{red}Median=\textrm{Datum ke}-\left ( \displaystyle \frac{40}{2} \right )=x_{._{20}}\\ &x_{._{20}}\: \: \textrm{terletak pada kelas interval}:\: \: 71-80\\ &\textrm{dengan}\: \: f=14,\: \: f_{k}\: \: \textrm{sebelum}\: \: Me=12,\\ &L=70,5,\: \: \textrm{serta}\: \: c=10\\ &\textrm{maka mediannya}\\ &Q_{2}=\color{purple}L+c\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\\ &=\color{black}70,5+10\left ( \displaystyle \frac{20-12}{14} \right )\\ &=\color{black}70,5+5,714=\color{red}76,21 \end{aligned} \end{array}$.

3. Modus

Modus dikatakan ada jika sekelompok data memiliki datum yang paling banyak muncul. Jika frekuensi munculnya datum terjadi kesamaan, maka dikatakan tidak ada modus.

$\begin{aligned}&\textrm{Dilambangkan dengan}:\: Mo\\ &\color{red}\textrm{Rumus data tunggal}\\ &\textrm{hanya untuk datum yang sering muncul}\\ &\color{red}\textrm{Rumus data berkelompok}\\ &Mo=L+\left ( \displaystyle \frac{\triangle _{1}}{\triangle _{1}+\triangle _{2}} \right )\times c \\ &\color{blue}\textrm{Penjelasan}\\ &\begin{aligned}L&=\textrm{tepi bawah kelas modus}\\ \triangle _{1}&=\textrm{frekuensi kelas modus dengan}\\ &\: \, \quad \textrm{frekuensi kelas sebelumnya}\\ \triangle _{2}&=\textrm{frekuensi kelas modus dengan}\\ &\: \, \quad \textrm{frekuensi kelas setelahnya}\\ c&=\textrm{lebar kelas interval} \end{aligned} \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah modus dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{red}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ Mo&=\color{black}\textrm{berupa datum yang sering muncul}\\ &=\color{red}7 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Skor}&f\\\hline 40-49&8\\\hline 50-59&9\\\hline 60-69&22\\\hline 70-79&15\\\hline 80-89&6\\\hline \end{array}\\ &\textrm{Modusnya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&65,50\\ \color{red}\textrm{b}.&66,00\\ \textrm{c}.&66,50\\ \textrm{d}.&67,00\\ \textrm{e}.&85,50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: n=\sum f=60,\: \: \color{red}modusnya\\ &\textrm{terdapat pada kelas dengan frekuensi terbanyak}\\ & \textrm{yaitu}:\: 60-69,\: \: \textrm{dengan}\: \: p=10\\ &\begin{cases} \triangle_{1} & =f-f_{1}=22-9=13 \\ \triangle_{2} & =f-f_{2}=22-15=7 \end{cases}\\ &\color{black}\textrm{Sehingga}\\ &M_{0}=\color{purple}L+c\left ( \displaystyle \frac{\triangle_{1}}{\triangle_{1}+\triangle_{2}} \right )\\ &M_{0}=\color{black}59,5+10\left ( \displaystyle \frac{22-9}{(22-9)+(22-15)} \right )\\ &\: \: \: =\color{black}59,5+\displaystyle \frac{10.13}{13+7}\\ &\: \: \: =\color{black}59,5+6,5=\color{red}66,0 \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Alam KBK. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: Srikandi Empat Widya Utama.
3. Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA

Trik Menyelesaikan Soal Persamaan yang Melibatkan Bentuk Gabungan Eksponen dan Logaritma

Terkadang beberapa soal pada akhir semester gasal dimunculkan soal yang melibatkan bentuk ekponen dan logaritma sekaligus dalam sebuah persamaan. Bentuk soal yang dihadapi para siswa pada suatu waktu tidak hanya fokus pada satu pokok bahasan saja, terkadang tersaji soal yang menuntut siswa untuk mengkombinasikan konsep-konsep yang telah disampaikan dan diajarkan oleh para guru dan pembimbing. Berawal dari sana, di bagian ini dipaparkan beberapa soal yang yang dimaksudkan dengan harapan siswa lebih terbiasa dalam menghadapi tipe soal yang tersaji demikian.

Soal pertama saya pilihkan ada di blog ini, berikut tautannya klik di sini

$\begin{array}{ll}\\ 1.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \color{purple}\textrm{atau}\: \: \color{blue}^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \color{purple}\textrm{atau}\: \: \color{blue}^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \color{purple}\textrm{atau}\: \: \color{blue}x_{2}=3^{3}\\ &\qquad \color{black}\textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\color{red}\frac{1}{9} \end{aligned} \end{array}$.

Soal kedua juga saya pilihkan ada di blog ini, tautannya klik di sini

$\begin{array}{ll}\\ 2.&\textrm{Persamaan}\\ & 10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ & \textrm{mempunyai dua akar yaitu}\: \: x_{1}\: \: \textrm{dan}\: \: x_{2}\\ &\textrm{Nilai}\: \: x_{1}\times x_{2}=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-2\\ \textrm{b}.&-5\\ \color{red}\textrm{c}.&2\\ \textrm{d}.&5\\ \textrm{e}.&10 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&10^{\,^{\, ^{2}}\log x^{2} }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &10^{\, 2^{\, ^{2}}\log x }-7\left ( 10^{\,^{\, ^{2}}\log x } \right )+10=0\\ &\color{black}\textrm{adalah persamaan kuadrat dalam}\: \: \color{red}10^{\,^{\, ^{2}}\log x }\\ &\color{black}\textrm{Misalkan}\: \: \color{red}p=10^{\,^{\, ^{2}}\log x },\: \: \color{black}\textrm{maka persamaan}\\ &\color{black}\textrm{menjadi}\: \: \color{purple}p^{2}-7p+10=0\: \begin{cases} a & =1 \\ b & =-7 \\ c & =10 \end{cases}\\ & \color{red}\textrm{Karena nilai}\: \: \color{black}p_{1}\times p_{2}=\displaystyle \frac{c}{a}\: \: \textrm{maka}\\ &10^{\,^{\, ^{2}}\log x_{1} }\times 10^{\,^{\, ^{2}}\log x_{2} }=\displaystyle \frac{10}{1}=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10\\ &10^{\,^{\, ^{2}}\log x_{1}\: +\: ^{2}\log x_{2} }=10^{1}\\ &\Leftrightarrow \: ^{2}\log x_{1}\: +\: ^{2}\log x_{2}=1\\ &\Leftrightarrow \: ^{2}\log x_{1}\times x_{2}=1\\ &\Leftrightarrow \: \color{red}x_{1}\times x_{2}=2^{1}=2\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Hasil kali semua akar real persamaan }\\ &\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\log \left ( x^{2}-x+4 \right )}}=\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ & \textrm{adalah}\: ....\\ &\qquad\qquad\qquad\textrm{UM UGM 2016 Mat IPA}\\ &\begin{array}{llll}\\ \textrm{a}.&-18\\ \color{red}\textrm{b}.&-6\\ \textrm{c}.&1\\ \textrm{d}.&6\\ \textrm{e}.&18 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ &\Leftrightarrow \color{black}\log \color{red}\sqrt{10}\left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\color{black}\log \color{red}\left ( x^{2}-x+4 \right )^{\frac{3}{2}}\\ &\Leftrightarrow \log \sqrt{10}+\log \left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\Leftrightarrow \log 10^{.^{\frac{1}{2}}}+\log \left ( x^{2}-x+4 \right )^{.^{\color{black}\log \color{blue}\left ( x^{2}-x+4 \right )}}=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\Leftrightarrow \displaystyle \frac{1}{2}+ \log \left ( x^{2}-x+4 \right )\times \log \left ( x^{2}-x+4 \right )=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\displaystyle \frac{1}{2}+ \log^{2} \left ( x^{2}-x+4 \right )=\displaystyle \frac{3}{2}\log \left ( x^{2}-x+4 \right )\\ &\color{red}\textrm{misalkan}\: \: \color{black}\log \left ( x^{2}-x+4 \right )=p,\: \: \textrm{maka}\\ &\displaystyle \frac{1}{2}+p^{2}=\displaystyle \frac{3}{2}p\\ &\Leftrightarrow 2p^{2}-3p+1=0\\ &\Leftrightarrow \left ( p-1 \right )\left ( 2p-1 \right )=0\\ &\Leftrightarrow p=1\: \: \textrm{atau}\: \: p=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \log \left ( x^{2}-x+4 \right )=1\: \: \textrm{atau}\: \: \log \left ( x^{2}-x+4 \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \left ( x^{2}-x+4 \right )=10^{1}\: \: \textrm{atau}\: \: \left ( x^{2}-x+4 \right )=10^{.^{\frac{1}{2}}}\\ &\Leftrightarrow x^{2}-x+4=10\: \: \textrm{atau}\: \: x^{2}-x+4=\sqrt{10}\\ &\Leftrightarrow x^{2}-x-6=0\: \: \textrm{atau}\: \: x^{2}-x+4-\sqrt{10}=0\\ &\textrm{Jelas bahwa yang ada akar real adalah}\\ &\textrm{persamaan}\: :\: \color{red}x^{2}-x-6=0\\ &\textrm{dan hasil kali semua akarnya adalah}:\: x_{1}.x_{2}.\\ &\Rightarrow x_{1}.x_{2}=\displaystyle \frac{c}{a}=\frac{-6}{1}=\color{red}-6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &2x^{.^{^{6}}\log x}+72x^{.^{^{.^{\frac{1}{6}}}}\log x}=24\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&18\: \: \textrm{dan}\: \: \displaystyle \frac{1}{36}\\ \textrm{b}.&24\: \: \textrm{dan}\: \: \displaystyle 2\\ \color{red}\textrm{c}.&6\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\ \textrm{d}.&36\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\ \textrm{e}.&\displaystyle \frac{1}{6}\: \: \textrm{dan}\: \: \displaystyle \frac{1}{18} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&2x^{.^{^{6}}\log x}+72x^{.^{^{.^{\frac{1}{6}}}}\log x}=24\\ &\Leftrightarrow 2x^{.^{^{6}}\log x}+72x^{-1.\left (^{^{6}}\log x \right )}=24\\ &\Leftrightarrow 2x^{.^{^{6}}\log x}+\displaystyle \frac{72}{x^{.^{^{6}}\log x}}-24=0\\ &\color{red}\textrm{misalkan}\: \: \color{black}p=x^{.^{^{6}}\log x},\: \: \textrm{maka}\\ &2p+\displaystyle \frac{72}{p}-24=0\Leftrightarrow p^{2}-12p+36=0\\ &\Leftrightarrow (p-6)^{2}=0\Leftrightarrow p=\pm 6\\ &\Leftrightarrow x^{.^{^{6}}\log x}=6\: \: \textrm{saja},\\ &\textbf{disebabkan} \: \textrm{syarat numerus}\: \: \color{blue}x^{.^{^{6}}\log x}\color{black}\neq -6.\\ & \textrm{Selanjutnya}\\ &\textrm{untuk}\: \: x^{.^{^{6}}\log x}=6,\: \: \textrm{nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{ada} \: \: 2,\: \: \textrm{yaitu} :\: \: 6\: \: \textrm{dan}\: \: \displaystyle \frac{1}{6}\\\\ &\textrm{Karena}\begin{cases} x^{.^{^{6}}\log x} & \Rightarrow 6^{.^{^{6}}\log 6}=6 \\\\ x^{.^{^{6}}\log x} & \Rightarrow \left (\frac{1}{6} \right )^{.^{^{6}}\log \left ( \frac{1}{6} \right )}=6^{-1.^{^{6}}\log 6^{-1}}=6^{.^{^{6}}\log 6^{1}}=6 \end{cases} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &^{2x-3}\log ^{-1}2.^{x}\log 2-\, ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&8\\ \textrm{b}.&9\\ \textrm{c}.&-1\\ \color{red}\textrm{d}.&6\\ \textrm{e}.&5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&^{2x-3}\log ^{-1}2.^{x}\log 2-\, ^{x+6}\log ^{-1}x+\displaystyle \frac{1}{^{x+2}\log x}=1\\ &\Leftrightarrow \: ^{2}\log (2x-3).^{x}\log 2-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log 2.\, ^{2}\log (2x-3)-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log (2x-3)-\, ^{x}\log (x+6)+\, ^{x}\log (x+2)=1\\ &\Leftrightarrow \: ^{x}\log \displaystyle \frac{(2x-3)(x+2)}{(x+6)}=1\\ &\begin{array}{|c|c|}\hline \textrm{Basis}&\textrm{Numerus}\\\hline \begin{aligned}&(1).\quad x>\displaystyle \frac{3}{2}\\ &(2).\quad x>0\\ &(3).\quad x>-2 &\\ &\color{red}\textrm{dipilih}\\ &\: \: \qquad x>\frac{3}{2} \end{aligned}&\begin{aligned}&\quad x>\displaystyle 0\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\\ &\Leftrightarrow \displaystyle \frac{(2x-3)(x+2)}{(x+6)}=x^{1}\\ &\Leftrightarrow (2x-3)(x+2)=x(x+6)\\ &\Leftrightarrow 2x^{2}+x-6=x^{2}+6x\\ &\Leftrightarrow x^{2}-5x-6=0\\ &\Leftrightarrow (x-6)(x+1)=0\\ &\Leftrightarrow x=6\: \: \textrm{atau}\: \: x=-1\\ &\textrm{Jadi, nilai}\: \: x=\color{red}6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Himpunan penyelesaian untuk}\: \: x,y,z\\ &\textrm{yang memenuhi persamaan berikut}\\ &\begin{cases} (2x+3y)^{\log (x-y+2z)}=1 & \\ 3^{2x+y+z}\times 27^{3z+2y+x}=81 & \\ 5x+3y+8z=2 & \end{cases}\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&\left \{ \displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \}\\ \textrm{b}.&\left \{ -\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6} \right \}\\ \textrm{c}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6} \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6} \right \}\\ \color{red}\textrm{e}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\begin{cases} (2x+3y)^{\log (x-y+2z)}=1 & \\ 3^{2x+y+z}\times 27^{3z+2y+x}=81 & \\ 5x+3y+8z=2 & \end{cases}\\ &\textrm{Perhatikan bahwa soal dapat dituliskan ulang}\\ &\textrm{menjadi}\\ &\begin{cases} \log (x-y+2z)=0 & \\ 3^{2x+y+z+3(3z+2y+x)}=3^{4} & \\ 5x+3y+8z=2 & \end{cases}\\ &\textrm{berubah lagi menjadi}\\ &\begin{cases} x-y+2z=1 &----(1) \\ 5x+7y+10z=4 &----(2) \\ 5x+3y+8z=2 &----(3) \end{cases}\\ &\textrm{Dengan eliminasi berikut akan didapatkan}\\ &\begin{array}{|c|c|}\hline \textbf{Persamaan 1&2}&\textbf{Persamaan 2&3}\\\hline \begin{array}{ll} 5x-5y+10z&=5\\ 5x+7y+10z&=4\quad -\\\hline \: \: \quad-12y&=1\\ \qquad\qquad\qquad y&=-\displaystyle \frac{1}{12}\\ &\\ &---(4) \end{array}&\begin{array}{ll} 5x+7y+10z&=4\\ 5x+3y+8z&=2\quad -\\\hline \: \: \qquad 4y+2z&=2\\ &---(5)\\ &\\ & \end{array} \\\hline \end{array}\\ &\textrm{Dari persamaan}\: 4\&5\: \: \textrm{didapatkan}\\ &\textrm{nilai}\: \: z=\displaystyle \frac{7}{6},\: \textrm{selanjutnya juga}\\ &\textrm{akan didapatkan nilai}\: \: x=-\displaystyle \frac{17}{12}.\\ &\textrm{Jadi, HP}=\color{red}\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk SMA/MA Kelas XKelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Cet. II. Jakarta: GRASINDO