Contoh Soal 6 Statistika

$\begin{array}{ll}\\ 26.&\textrm{Jika rata-rata dari}\: \: x_{1},x_{2},x_{3},x_{4},...,x_{10}\\ &\textrm{adalah}\: \: x_{0},\: \textrm{maka rata-rata dari data}\\ &(x_{1}-1),(x_{2}+2),(x_{3}-3),(x_{4}+4),..\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x_{0}+5,5\\ \textrm{b}.&x_{0}+25\\ \color{red}\textrm{c}.&x_{0}+0,5\\ \textrm{d}.&x_{0}-0,5\\ \textrm{e}.&x_{0}-2,5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Rata}-&\textrm{ratanya adalah}:\\ \overline{x}=x_{0}&=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{10}}{10}\\ \Leftrightarrow 10x_{0}&=x_{1}+x_{2}+x_{3}+...+x_{10}\\ \color{black}\textrm{Selanju}&\color{black}\textrm{tnya penghitungan rata-rata yang data baru},\\ \overline{x}_{baru}&=\displaystyle \frac{(x_{1}-\color{red}1)\color{blue}+(x_{2}+\color{red}2)\color{blue}+(x_{3}-\color{red}3)\color{blue}+...+(x_{10}+\color{red}10)}{10}\\ &=\displaystyle \frac{x_{1}+x_{2}+...+x_{10}+\color{red}(2-1+4-3+..+10-9)}{10}\\ &=\displaystyle \frac{10x_{0}+5}{10}\\ &=\color{red}x_{0}+0,5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Nilai rata-rata dari 20 bilangan adalah}\: \: 14,2\\ &\textrm{Jika rata-rata dari 12 bilangan pertama adalah}\\ &\textrm{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ &\textrm{adalah 18,2, rata-rata 2 bilangan tersisa}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&10,4\\ \color{red}\textrm{b}.&11,8\\ \textrm{c}.&12,2\\ \textrm{d}.&12,8\\ \textrm{e}.&13,8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\overline{x}_{total}&=\displaystyle \frac{\overline{x}_{12}\times (12)+\overline{x}_{6}\times (6)+\overline{x}_{2}\times (2)}{20}\\ 14,2&=\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+\overline{x}_{2}\times (2)}{20}\\ 284&=151,2+109,2+2\overline{x}_{2}\\ 2\overline{x}_{2}&=284-(151,2+109,2)=\color{red}23,6\\ \overline{x}_{2}&=\displaystyle \frac{23,6}{2}\\ &=\color{red}11,8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Dari 3 bilangan yang terkecil adalah 39}\\ &\textrm{dan terbesarnya adalah 75, maka rata-rata}\\ &\textrm{hitung ketiga bilangan tersebut tidak}\\ &\textrm{mungkin sama dengan}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&49\\ \textrm{b}.&52\\ \textrm{c}.&53\\ \textrm{d}.&59\\ \textrm{e}.&60 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{3}&=\displaystyle \frac{39+\color{red}A\color{blue}+75}{3}\\ \textrm{Se}&\textrm{lanjutnya rentang nilai}\: \: A\: \: \textrm{akan berada di}\\ :\: &39\leq \color{red}A\color{blue}\leq 75\\ \textrm{Se}&\textrm{hingga},\\ \textrm{un}&\textrm{tuk}\: \: \color{red}A=39,\: \: \color{blue}\textrm{maka}\\ \overline{x}_{3}&=\displaystyle \frac{39+39+75}{3}=\frac{153}{3}=\color{red}51,\: \: \color{black}\textrm{dan}\\ \textrm{un}&\textrm{tuk}\: \: \color{red}A=75,\: \: \color{blue}\textrm{maka}\\ \overline{x}_{3}&=\displaystyle \frac{39+75+75}{3}=\frac{189}{3}=\color{red}63 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 29.&(\textbf{SPMB 04})\\ &\textrm{Nilai rata-rata tes Matematika dari kelompok}\\ &\textrm{siswa dan kelompok siswi di suatu kelas berturut-}\\ &\textrm{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ &\textrm{tersebut adalah 6,2, maka perbandingan banyaknya}\\ &\textrm{siswa dan siswi adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2:3\\ \textrm{b}.&3:4\\ \textrm{c}.&2:5\\ \textrm{d}.&3:5\\ \textrm{e}.&4:5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\displaystyle \frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}\\ \color{black}6,2&\color{blue}=\displaystyle \frac{n_{1}(5)+n_{2}(7)}{n_{1}+n_{2}}\\ 6,2(n_{1}+n_{2})&=5n_{1}+7n_{2}\\ 6,2n_{1}-5n_{1}&=7n_{2}-6,2n_{2}\\ 1,2n_{1}&=0,8n_{2}\\ \displaystyle \frac{n_{1}}{n_{2}}&=\displaystyle \frac{0,8}{1,2}=\color{red}\frac{2}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 30.&(\textbf{SPMB 05})\\ &\textrm{Nilai rata-rata ulangan kelas A adalah}\: \: \overline{x}_{A}\: \: \textrm{dan}\\ &\textrm{kelas B adalah}\: \: \overline{x}_{B}.\: \textrm{Setelah kedua kelas digabungkan}\\ &\textrm{nilai rata-ratanya adalah}\: \: \overline{x}.\: \textrm{Perbandingan nilai}\\ &\textrm{kelas A dan B adalah}\: \: 10:9.\: \textrm{Jika perbandingan nilai}\\ &\textrm{rata-rata kedua kelas dan kelas B adalah}\: \: 85:81,\\ &\textrm{maka perbandinganbanyaknya siswa kelas A dan B}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&8:9\\ \color{red}\textrm{b}.&4:5\\ \textrm{c}.&3:4\\ \textrm{d}.&3:5\\ \textrm{e}.&9:10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\overline{x}_{A}:\overline{x}_{B}&=10:9=90:81\\ \overline{x}:\overline{x}_{B}&=85:81,\: \textrm{maka}\\ \color{red}\overline{x}:\overline{x}_{A}:\overline{x}_{B}&=\color{red}85:90:81\\ (n_{A}+n_{B})\overline{x}&=n_{A}\times \overline{x}_{A}+n_{B}\times \overline{x}_{B}\\ \displaystyle \frac{n_{A}}{n_{B}}&=\displaystyle \frac{\overline{x}-\overline{x}_{B}}{\overline{x}_{A}-\overline{x}}\\ &=\displaystyle \frac{\displaystyle \frac{85}{81}\overline{x}_{B}-\overline{x}_{B}}{\displaystyle \frac{10}{9}\overline{x}_{B}-\displaystyle \frac{85}{81}\overline{x}_{B}}\\ &=\displaystyle \frac{\displaystyle \frac{4}{81}\overline{x}_{B}}{\displaystyle \frac{5}{81}\overline{x}_{B}}\\ &=\color{red}\displaystyle \frac{4}{5} \end{aligned} \end{array}$