Contoh Soal 6 Statistika

$\begin{array}{l}\\ 26.& \text{Jika rata-rata dari}{x}_1,x_2,x_3,x_4,...,x_{10}\\ & \text{adalah}{x}_0,\text{maka rata-rata dari data}\\ & (x_1-1),(x_2+2),(x_3-3),(x_4+4),..\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x_0+5,5\\ \text{b}.& x_0+25\\ \text{c}.& x_0+0,5\\ \text{d}.& x_0-0,5\\ \text{e}.& x_0-2,5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Rata}-& \text{ratanya adalah}:\\ \bar{x}=x_0& ={\displaystyle \frac{x_1+x_2+x_3+...+x_{10}}{10}}\\ \iff 10x_0& =x_1+x_2+x_3+...+x_{10}\\ \text{Selanju}& \text{tnya penghitungan rata-rata yang data baru},\\ {\bar{x}}_{baru}& ={\displaystyle \frac{(x_1-1)+(x_2+2)+(x_3-3)+...+(x_{10}+10)}{10}}\\ & ={\displaystyle \frac{x_1+x_2+...+x_{10}+(2-1+4-3+..+10-9)}{10}}\\ & ={\displaystyle \frac{10x_0+5}{10}}\\ & =x_0+0,5\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Nilai rata-rata dari 20 bilangan adalah}14,2\\ & \text{Jika rata-rata dari 12 bilangan pertama adalah}\\ & \text{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ & \text{adalah 18,2, rata-rata 2 bilangan tersisa}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10,4\\ \text{b}.& 11,8\\ \text{c}.& 12,2\\ \text{d}.& 12,8\\ \text{e}.& 13,8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_{total}& ={\displaystyle \frac{{\bar{x}}_{12}\times (12)+{\bar{x}}_6\times (6)+{\bar{x}}_2\times (2)}{20}}\\ 14,2& ={\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+{\bar{x}}_2\times (2)}{20}}\\ 284& =151,2+109,2+2{\bar{x}}_2\\ 2{\bar{x}}_2& =284-(151,2+109,2)=23,6\\ {\bar{x}}_2& ={\displaystyle \frac{23,6}{2}}\\ & =11,8\end{array}\end{array}$

$\begin{array}{l}\\ 28.& \text{Dari 3 bilangan yang terkecil adalah 39}\\ & \text{dan terbesarnya adalah 75, maka rata-rata}\\ & \text{hitung ketiga bilangan tersebut tidak}\\ & \text{mungkin sama dengan}....\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 52\\ \text{c}.& 53\\ \text{d}.& 59\\ \text{e}.& 60\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_3& ={\displaystyle \frac{39+A+75}{3}}\\ \text{Se}& \text{lanjutnya rentang nilai}A\text{akan berada di}\\ :& 39\le A\le 75\\ \text{Se}& \text{hingga},\\ \text{un}& \text{tuk}A=39,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+39+75}{3}=\frac{153}{3}=51,\text{dan}}\\ \text{un}& \text{tuk}A=75,\text{maka}\\ {\bar{x}}_3& ={\displaystyle \frac{39+75+75}{3}=\frac{189}{3}=63}\end{array}\end{array}$

$\begin{array}{l}\\ 29.& (\mathbf{\text{SPMB 04}})\\ & \text{Nilai rata-rata tes Matematika dari kelompok}\\ & \text{siswa dan kelompok siswi di suatu kelas berturut-}\\ & \text{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ & \text{tersebut adalah 6,2, maka perbandingan banyaknya}\\ & \text{siswa dan siswi adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2:3\\ \text{b}.& 3:4\\ \text{c}.& 2:5\\ \text{d}.& 3:5\\ \text{e}.& 4:5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\bar{x}}_{gabungan}& ={\displaystyle \frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}}\\ 6,2& ={\displaystyle \frac{n_1(5)+n_2(7)}{n_1+n_2}}\\ 6,2(n_1+n_2)& =5n_1+7n_2\\ 6,2n_1-5n_1& =7n_2-6,2n_2\\ 1,2n_1& =0,8n_2\\ {\displaystyle \frac{n_1}{n_2}}& ={\displaystyle \frac{0,8}{1,2}=\frac{2}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 30.& (\mathbf{\text{SPMB 05}})\\ & \text{Nilai rata-rata ulangan kelas A adalah}{\bar{x}}_A\text{dan}\\ & \text{kelas B adalah}{\bar{x}}_B.\text{Setelah kedua kelas digabungkan}\\ & \text{nilai rata-ratanya adalah}\bar{x}.\text{Perbandingan nilai}\\ & \text{kelas A dan B adalah}10:9.\text{Jika perbandingan nilai}\\ & \text{rata-rata kedua kelas dan kelas B adalah}85:81,\\ & \text{maka perbandinganbanyaknya siswa kelas A dan B}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 8:9\\ \text{b}.& 4:5\\ \text{c}.& 3:4\\ \text{d}.& 3:5\\ \text{e}.& 9:10\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\bar{x}}_A:{\bar{x}}_B& =10:9=90:81\\ \bar{x}:{\bar{x}}_B& =85:81,\text{maka}\\ \bar{x}:{\bar{x}}_A:{\bar{x}}_B& =85:90:81\\ (n_A+n_B)\bar{x}& =n_A\times {\bar{x}}_A+n_B\times {\bar{x}}_B\\ {\displaystyle \frac{n_A}{n_B}}& ={\displaystyle \frac{\bar{x}-{\bar{x}}_B}{{\bar{x}}_A-\bar{x}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{85}{81}{\bar{x}}_B-{\bar{x}}_B}}{{\displaystyle \frac{10}{9}{\bar{x}}_B-{\displaystyle \frac{85}{81}{\bar{x}}_B}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{4}{81}{\bar{x}}_B}}{{\displaystyle \frac{5}{81}{\bar{x}}_B}}}\\ & ={\displaystyle \frac{4}{5}}\end{array}\end{array}$