Contoh Soal 10 Statistika

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

$\begin{array}{ll}41.& \text{Median dan modus dari data berikut}\\ & 3,6,7,8,4,5,9,6\\ & \text{ adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{7 dan 5}& & & \text{d}.& \text{5 dan 6}\frac{1}{2}\\ \text{b}.& \text{6 dan 6}& \text{c}.& \text{6 dan 7}& \text{e}.& \text{5 dan 6}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui}n=8\\ & \text{Datum diurutkan dari kecil ke besar}\\ & :3,4,5,6,6,7,8,9\\ & \bullet \mathbf{\text{modus}}=M_o=6,\\ & \bullet \mathbf{\text{mean}}=\bar{x}=6\end{array}\end{array}$.

$\begin{array}{ll}42.& \text{Hasil tes matematika di suatu kelas yang}\\ & \text{diikuti tes 49 siswa menghasilkan nilai}\\ & \text{rata-rata 7}.\text{Jika Andi ikut tes susulan}\\ & 7,04.\text{Nilai Andi adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{7,5}& & & \text{d}.& \text{9}\\ \text{b}.& \text{8}& \text{c}.& \text{8,5}& \text{e}.& \text{9,5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}\text{Misalkan}& y=\text{besar nilai Andi}\\ {\bar{x}}_{gabungan}& ={\displaystyle \frac{n.\bar{x}+y}{n+1}}\\ 7,04& ={\displaystyle \frac{49\times 7+y}{49+1}}\\ 7,04& ={\displaystyle \frac{343+y}{50}}\\ 343+y& =50\times (7,04)\\ 343+y& =352\\ y& =352-343\\ & =9\end{array}\end{array}$.

$\begin{array}{ll}43.& \text{Mean dari tabel berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 50-54& 4\\ 55-59& 6\\ 60-64& 10\\ \hline\end{array}\\ & \begin{array}{llllll}\text{a}.& \text{60,5}& & & \text{d}.& \text{58,5}\\ \text{b}.& \text{60}& \text{c}.& \text{59,5}& \text{e}.& \text{57}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \begin{array}{|ccccc|}\hline \text{Ukuran}& x_i& d_i=x_i-x_s& f_i& f_i\times d_i\\ 50-54& 52& -10& 4& -40\\ 55-59& 57& -5& 6& -30\\ 60-64& 62& 0& 10& 0\\ \text{Jumlah}& & & 20& -70\\ \hline\end{array}\\ & \bar{x}=x_s+{\displaystyle \frac{{\displaystyle \sum _{i=1}^n{f}_i\times d_1}}{{\displaystyle \sum _{i=0}^n{f}_i}}=62+{\displaystyle \frac{-70}{20}}}\\ & =62-3,5=58,5\end{array}\end{array}$.

$\begin{array}{ll}44.& \text{Median dari tabel berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 47-49& 1\\ 50-52& 6\\ 53-55& 6\\ 56-58& 7\\ 59-61& 4\\ \hline\end{array}\\ & \begin{array}{llllll}\text{a}.& \text{55,5}& & & \text{d}.& \text{53,5}\\ \text{b}.& \text{55}& \text{c}.& \text{54,5}& \text{e}.& \text{53}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 47-49& 1\\ 50-52& 6\\ 53-55& 6\\ 56-58& 7\\ 59-61& 4\\ & 24\\ \hline\end{array}\\ & \text{Median posisi datumnya}:\\ & \text{datum ke}-\left({\displaystyle \frac{24}{2}}\right)=x_{{.}_{12}}\\ & \text{dan terletak di interval}\\ & 53-55,\text{dengan}{f}_k=1+6=7\\ & \text{serta}c=3\\ & M_e=Q_2=L_i+c\left({\displaystyle \frac{{\displaystyle \frac{n}{2}-f_k}}{f}}\right)\\ & M_e=Q_2=52,2+3\left({\displaystyle \frac{{\displaystyle \frac{24}{2}-7}}{6}}\right)\\ & =52,5+3\left({\displaystyle \frac{12-7}{6}}\right)\\ & =52,5+\left({\displaystyle \frac{5}{2}}\right)\\ & =52,5+2,5\\ & =55\end{array}\end{array}$.

$\begin{array}{ll}45.& \text{Modus dari tabel berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 50-54& 4\\ 55-59& 6\\ 60-64& 8\\ 65-69& 16\\ 70-74& 10\\ 75-79& 4\\ 80-84& 2\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& \text{67,32}& & & \text{d}.& \text{70,12}\\ \text{b}.& \text{67,36}& \text{c}.& \text{67,56}& \text{e}.& \text{70,36}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Perhatikan tabelnya lagi}\\ & \text{dengan penambahan warna berikut ini}\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 50-54& 4\\ 55-59& 6\\ 60-64& 8\\ 65-69& 16\\ 70-74& 10\\ 75-79& 4\\ 80-84& 2\\ \hline\end{array}\\ & \text{Diketahui}n=50\\ & \text{modus terletak pada kelas}\\ & \text{interval dengan frekuensi}\\ & \text{terbesar, yaitu}:16.\text{Kelas intervalnya}\\ & 65-69,\text{dengan}c=5\\ & \text{serta}\{\begin{array}{ll}{\mathrm{△}}_1& =f-f_1=16-8=8\\ {\mathrm{△}}_2& =f-f_2=16-10=6\end{array}\\ & M_o=L+c\left({\displaystyle \frac{{\mathrm{△}}_1}{{\mathrm{△}}_1+{\mathrm{△}}_2}}\right)\\ & =64,5+5\left({\displaystyle \frac{8}{8+6}}\right)\\ & =64,5+{\displaystyle \frac{40}{14}}\\ & =64,5+2,857\\ & =67,36\end{array}\end{array}$.