Contoh Soal 12 Statistika

$\begin{array}{ll}51.& \text{Simpangan kuartil dari data}\\ & 56a378\text{adalah}1{\displaystyle \frac{1}{2}.\text{Jika median datanya}}\\ & \text{adalah}5{\displaystyle \frac{1}{2},\text{maka rata-rata data}\text{tersebut adalah}....}\\ & \begin{array}{llllll}\text{a}.& \text{4}& & & \text{d}.& \text{5}{\displaystyle \frac{1}{2}}\\ \text{b}.& \text{4}{\displaystyle \frac{1}{2}}& \text{c}.& \text{5}& \text{e}.& \text{6}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 56a378\Rightarrow n=6\\ & \text{karena mediannya}=M_e=Q_2=5\frac{1}{2}=\frac{11}{2}\\ & \text{ data menjadi}:a35678\text{atau}3a5678\\ & \text{Simpangan kuartilnya adalah}1{\displaystyle \frac{1}{2}={\displaystyle \frac{3}{2}}}\\ & Q_d={\displaystyle \frac{1}{2}(Q_3-Q_1)={\displaystyle \frac{3}{2}\Rightarrow Q_3-Q_1=3}}\\ & \text{maka}\\ & 3=Q_3-Q_1=x_{{.}_{\frac{3}{4}n+\frac{1}{2}}}-x_{{.}_{\frac{1}{4}n+\frac{1}{2}}}=x_{{.}_{\frac{3}{4}6+\frac{1}{2}}}-x_{{.}_{\frac{1}{4}6+\frac{1}{2}}}\\ & =(x_{{.}_5}-x_{{.}_2})=7-x_{{.}_2}=3\Rightarrow x_{{.}_2}=4\\ & \text{Jadi, rata-ratanya adalah}:\\ & \bar{x}=\frac{3+4+5+6+7+8}{6}=5,5\end{array}\end{array}$.

$\begin{array}{ll}52.& \text{Simpangan kuartil dari data}\\ & \text{berikut}\\ & 61,61,50,50,53,53,70,61\\ & 53,70,53,61,50,61,70\\ & \text{adalah}....\\ & \begin{array}{llllll}\text{a}.& \text{10}& & & \text{d}.& \text{6}\\ \text{b}.& \text{9}& \text{c}.& \text{8}& \text{e}.& \text{5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{-}\\ & \begin{array}{rl}& \text{Data mula-mula}n=15\\ & 61,61,50,50,53,53,70,61\\ & 53,70,53,61,50,61,70\\ & \text{data durutkan}\\ & 50,50,50,53,53,53,53\\ & 61,61,61,61,61,70,70,70\\ & \text{Simpangan kuartil adalah}{Q}_d,\\ & Q_d={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{\frac{3}{4}(n+1)}}-x_{{.}_{\frac{1}{4}(n+1)}})}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{\frac{3}{4}(15+1)}}-x_{{.}_{\frac{1}{4}(15+1)}})}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_{12}}-x_{{.}_4})}\\ & ={\displaystyle \frac{1}{2}(61-53)={\displaystyle \frac{1}{2}\times 8=4}}\\ & \text{Jadi},Q_d=4\end{array}\end{array}$.

$\begin{array}{ll}53.& \text{Simpangan rata-rata dari data berikut}:\\ & 642810\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{2}& & & \text{d}.& \text{3},0\\ \text{b}.& \text{2},4& \text{c}.& \text{2},5& \text{e}.& \text{3},5\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 642810\\ & \text{Simpangan rata-ratanya adalah}:\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+4+2+8+10}{5}={\displaystyle \frac{30}{5}=6}}\\ & \text{maka nilai}\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{5}(|6-6|+|4-6|+|2-6|+|8-6|+|10-6|)}\\ & ={\displaystyle \frac{1}{5}(\left|0\right|+|-2|+|-4|+\left|2\right|+\left|4\right|)}\\ & ={\displaystyle \frac{1}{5}(0+2+4+2+4)}\\ & ={\displaystyle \frac{1}{5}(12)={\displaystyle \frac{12}{5}=2,4}}\end{array}\end{array}$.

$\begin{array}{ll}54.& \text{Simpangan rata-rata dari data berikut}:\\ & 10,8,7,10,7,5,8,6,10,9\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{1,0}& & & \text{d}.& \text{8,0}\\ \text{b}.& \text{1,4}& \text{c}.& \text{6,0}& \text{e}.& \text{14,0}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{b}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 10,8,7,10,7,5,8,6,10,9\\ & \text{Simpangan rata-ratanya adalah}:\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{10+8+7+10+7+5+8+6+10+9}{10}}\\ & ={\displaystyle \frac{80}{10}=8}\\ & \text{maka nilai}\\ & SR={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{10}(|5-8|+|6-8|+2|7-8|+2|8-8|+|9-8|+3|10-8|)}\\ & ={\displaystyle \frac{1}{10}(|-3|+|-2|+2|-1|+2\left|0\right|+\left|1\right|+3\left|2\right|)}\\ & ={\displaystyle \frac{1}{10}(3+2+2+0+1+6)}\\ & ={\displaystyle \frac{1}{10}(14)={\displaystyle \frac{14}{10}=1,4}}\end{array}\end{array}$.

$\begin{array}{ll}55.& \text{Nilai variansi dari data}\\ & 6,7,7,8,8,8,8,12\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \text{1}& & & \text{d}.& \text{8}\\ \text{b}.& {\displaystyle \frac{26}{8}}& \text{c}.& {\displaystyle \frac{11}{4}}& \text{e}.& \text{22}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{c}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 6,7,7,8,8,8,8,12\\ & \text{Variannya adalah}:\\ & S^2={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+7+7+8+8+8+8+12}{8}={\displaystyle \frac{64}{8}=8}}\\ & \text{maka nilai}\\ & S^2={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & ={\displaystyle \frac{1}{8}((6-8{)}^2+2(7-6{)}^2+4(8-8{)}^2+(12-8{)}^2)}\\ & ={\displaystyle \frac{1}{8}(2^2+2.1+4.0+4^2)}\\ & ={\displaystyle \frac{1}{8}(4+2+0+16)}\\ & ={\displaystyle \frac{1}{8}(22)={\displaystyle \frac{11}{4}=2,75}}\end{array}\end{array}$.