Contoh Soal 11 Statistika

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

$.\begin{array}{ll}& \begin{array}{l}\\ \text{a}.& 41,372& & & \text{d}.& 43,135\\ \text{b}.& 42,150& \text{c}.& 43,125& \text{e}.& 44,250\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \begin{array}{rl}& \text{Berat}\\ & \text{Badan}\end{array}& {\text{f}}_1& \begin{array}{rl}& \text{Nilai Tengah}\\ & ({\text{x}}_1)\end{array}& {\text{f}}_1{\text{x}}_1\\ 30-34& 5& 32& 160\\ 35-39& 7& 37& 259\\ 40-44& 12& 42& 504\\ 45-49& 9& 47& 423\\ 50-54& 4& 52& 208\\ 55-59& 3& 57& 171\\ & \sum {\text{f}}_1=40& & \sum {\text{f}}_1{\text{x}}_1=1725\\ \hline\end{array}\\ & \text{maka nilai dari}\bar{\text{x}}\text{adalah}:\\ & \bar{\text{x}}={\displaystyle \frac{\sum {\text{f}}_1{\text{x}}_1}{\sum {\text{f}}_1}={\displaystyle \frac{1725}{40}=43,125}}\end{array}\end{array}$.

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

$.\begin{array}{ll}& \text{Berat badan pada suatu kelas tersaji dengan}\\ & \text{bentuk histogram seperti pada gambar di atas}\\ & \text{Rata-rata berat badan tersebut adalah}....\text{Kg}\\ & \begin{array}{l}\\ \text{a}.& 64,5& & & \text{d}.& 66\\ \text{b}.& 65& \text{c}.& 65,5& \text{e}.& 66,5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \begin{array}{|cccc|}\hline \begin{array}{rl}& \text{Berat}\\ & \text{Badan}\end{array}& {\text{f}}_1& \begin{array}{rl}& \text{Nilai Tengah}\\ & ({\text{x}}_1)\end{array}& {\text{f}}_1{\text{x}}_1\\ 50-54& 4& 52& 208\\ 55-59& 6& 57& 342\\ 60-64& 8& 62& 496\\ 65-69& 10& 67& 670\\ 70-74& 8& 72& 576\\ 75-79& 4& 77& 308\\ & \sum {\text{f}}_1=40& & \sum {\text{f}}_1{\text{x}}_1=2600\\ \hline\end{array}\\ & \text{maka nilai dari}\bar{\text{x}}\text{adalah}:\\ & \bar{\text{x}}={\displaystyle \frac{\sum {\text{f}}_1{\text{x}}_1}{\sum {\text{f}}_1}={\displaystyle \frac{2600}{40}=65}}\end{array}\end{array}$.

$\begin{array}{ll}48.& \text{Diketahui tabel distribusi frekuensi berikut}\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 145-149& 3\\ 150-154& 5\\ 155-159& 17\\ 160-164& 15\\ 165-169& 8\\ 170-174& 2\\ \hline\end{array}\\ & \text{Kuartil bawah dapat dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& 149,5+\left({\displaystyle \frac{12,5-3}{8}}\right).5\\ \text{b}.& 150+\left({\displaystyle \frac{12,5-3}{8}}\right).5\\ \text{c}.& 155+\left({\displaystyle \frac{12,5-8}{17}}\right).5\\ \text{d}.& 154,5+\left({\displaystyle \frac{12,5-8}{17}}\right).5\\ \text{e}.& 155,5+\left({\displaystyle \frac{12,5-8}{17}}\right).5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui}n=50\\ & \text{Ditanyakan kuartil bawah, maka hal ini}\\ & =Q_1\Rightarrow x_{{.}_{\frac{1}{4}n}}=x_{{.}_{\frac{1}{4}50}}=x_{{.}_{12,5}}\\ & \text{Perhatikan tabelnya lagi}\\ & \text{dengan penambahan warna berikut ini}\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 145-149& 3\\ 150-154& 5\\ 155-159& 17\\ 160-164& 15\\ 165-169& 8\\ 170-174& 2\\ \hline\end{array}\\ & \text{maka nilai dari}{Q}_1\text{adalahh}:\\ & Q_1=L+\left({\displaystyle \frac{\frac{1}{4}n-f_k}{f}}\right).c\\ & Q_1=154,5+\left({\displaystyle \frac{12,5-8}{17}}\right).5\end{array}\end{array}$.

$\begin{array}{ll}49.& \text{Diketahui tabel distribusi frekuensi berikut}\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 5-9& 2\\ 10-14& 8\\ 15-19& 10\\ 20-24& 7\\ 25-29& 3\\ \hline\end{array}\\ & \text{Median dari tabel di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& 15,0& & & \text{d}.& 16,5\\ \text{b}.& 15,5& \text{c}.& 16,0& \text{e}.& 17,0\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{e}\\ & \begin{array}{rl}& \text{Diketahui}n=30\\ & \text{Ditanyakan median, maka formulanya}\\ & =Q_2\Rightarrow x_{{.}_{\frac{2}{4}n}}=x_{{.}_{\frac{1}{2}.30}}=x_{{.}_{15}}\\ & \text{Perhatikan tabelnya lagi}\\ & \text{dengan penambahan warna berikut ini}\\ & \begin{array}{|cc|}\hline \text{Ukuran}& \text{Frekuensi}\\ 5-9& 2\\ 10-14& 8\\ 15-19& 10\\ 20-24& 7\\ 25-29& 3\\ \hline\end{array}\\ & \text{maka nilai dari}{Q}_2\text{adalah}:\\ & Q_2=L+\left({\displaystyle \frac{\frac{2}{4}n-f_k}{f}}\right).c\\ & Q_1=14,5+\left({\displaystyle \frac{15-10}{10}}\right).5\\ & =14,5+\frac{25}{10}=14,5+2,5=17,0\end{array}\end{array}$.

$\begin{array}{ll}50.& \text{Jangkauan antarkuartil dari data}\\ & \text{berikut}:\\ & 3625564055424364\\ & 82702835384554\\ & \text{adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{20}& & & \text{d}.& \text{5}\\ \text{b}.& \text{10}& \text{c}.& \text{8}& \text{e}.& \text{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{a}\\ & \begin{array}{rl}& \text{Data mula-mula}n=15\\ & 3625564055424364\\ & 82702835384554\\ & \text{data durutkan}\\ & 2528353638404243\\ & 45545556647082\\ & \text{Jangkauan antarkuartil adalah}H,\\ & H=Q_3-Q_1\\ & =x_{{.}_{\frac{3}{4}(n+1)}}-x_{{.}_{\frac{1}{4}(n+1)}}\\ & =x_{{.}_{\frac{3}{4}(15+1)}}-x_{{.}_{\frac{1}{4}(15+1)}}\\ & =(x_{{.}_{12}}-x_{{.}_4})\\ & =56-36\\ & \text{Jadi},H=Q_3-Q_1=20\end{array}\end{array}$.