Contoh Soal 13 Statistika

$\begin{array}{ll}56.& \text{Simpangan baku dari data berikut}:\\ & 6,7,4,5,3\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& \sqrt{2}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{2}}& \text{c}.& {\displaystyle \frac{1}{2}\sqrt{3}}& \text{e}.& \sqrt{3}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 6,7,4,5,3\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{6+7+4+5+3}{5}={\displaystyle \frac{25}{5}=5}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((6-5{)}^2+(7-5{)}^2+(4-5{)}^2+(5-5{)}^2+(3-5{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(1^2+2^2+1^2+0+2^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(1+4+1+4)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(10)}}=\sqrt{{\displaystyle \frac{10}{5}}}=\sqrt{2}\end{array}\end{array}$.

$\begin{array}{ll}57.& \mathbf{\text{UN 2010}}\\ & \text{Simpangan baku dari data berikut}:\\ & 2,3,4,5,6\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& \sqrt{15}& & & \text{d}.& \sqrt{3}\\ \text{b}.& {\displaystyle \sqrt{10}}& \text{c}.& {\displaystyle \sqrt{5}}& \text{e}.& \sqrt{2}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 2,3,4,5,6\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{2+3+4+5+6}{5}={\displaystyle \frac{20}{5}=4}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((2-5{)}^2+(3-5{)}^2+(4-5{)}^2+(5-5{)}^2+(6-5{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(3^2+2^2+1^2+0+1^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(9+4+1+1)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(15)}}=\sqrt{{\displaystyle \frac{15}{5}}}=\sqrt{3}\end{array}\end{array}$.

$\begin{array}{ll}58.& \text{Simpangan baku dari data berikut}:\\ & 7,9,11,13,15\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& 2,4& & & \text{d}.& 2,8\\ \text{b}.& {\displaystyle 2,5}& \text{c}.& {\displaystyle 2,7}& \text{e}.& 2,9\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 7,9,11,13,15\\ & \text{Simpangan bakuya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{7+9+11+13+15}{5}={\displaystyle \frac{55}{5}=11}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{5}((7-11{)}^2+(9-11{)}^2+(11-11{)}^2+(13-11{)}^2+(15-11{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(4^2+2^2+0+2^2+4^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(16+4+4+16)}}\\ & =\sqrt{{\displaystyle \frac{1}{5}(40)}}=\sqrt{{\displaystyle \frac{40}{5}}}=\sqrt{8}=2,82..\end{array}\end{array}$.

 $\begin{array}{ll}59.& \text{Simpangan baku dari data berikut}:\\ & 2,4,4,5,6,6,7,8,9,9\text{ adalah}....\\ \\ & \begin{array}{llllll}\text{a}.& 4\sqrt{3}& & & \text{d}.& {\displaystyle \frac{2}{5}\sqrt{30}}\\ \text{b}.& 2{\displaystyle \frac{2}{5}}& \text{c}.& {\displaystyle \sqrt{5}}& \text{e}.& 2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\text{d}\\ & \begin{array}{rl}& \text{Diketahui data sebagai berikut}\\ & 2,4,4,5,6,6,7,8,9,9\\ & \text{Simpangan bakunya adalah}:\\ & S=\sqrt{S^2}=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & \text{dengan}\\ & \bar{x}={\displaystyle \frac{2+4+4+5+6+6+7+8+9+9}{10}={\displaystyle \frac{60}{10}=6}}\\ & \text{maka nilai}\\ & S=\sqrt{{\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}}\\ & =\sqrt{{\displaystyle \frac{1}{10}((2-6{)}^2+2(4-6{)}^2+(5-6{)}^2+2(6-6{)}^2+(7-6{)}^2+(8-6{)}^2+2(9-6{)}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(4^2+{2.2}^2+1^2+0+1^2+2^2+{2.3}^2)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(16+8+1+1+4+18)}}\\ & =\sqrt{{\displaystyle \frac{1}{10}(48)}}=\sqrt{{\displaystyle \frac{48}{10}}}=\sqrt{{\displaystyle \frac{120}{25}}}={\displaystyle \frac{2}{5}\sqrt{30}}\end{array}\end{array}$.