Ukuran Penyebaran Data Tunggal (Materi Kelas XII Matematika Wajib) (Bagian 1)

A. Pengertian

Ukuran penyebaran data adalah nilai dari ukuran yang memberikan gambaran sejauh mana data menyebar atau menyimpang (dispersi/deviasi) dari ukuran pemusatan data. Dalam hal ini bagian yang akan disinggung dalam materi ini adalah: Jangkauan (Range), Jangkauan antar kuartil, Simpangan kuartil, Simpangan rata-rata, Ragam (Variansi), Simpangan baku (Deviasi Standar), Koefisien variansi.

$\begin{array}{|c|l|c|}\hline \textrm{No}&\: \: \: \: \textrm{Data Dispersi}&\textrm{Simbol}\\\hline 1.&\textrm{Jangkauan}&R\: \: \textrm{atau}\: \: J\\\hline 2.&\textrm{Jangkauan}&H\\ &\textrm{antarkuartil}&\\\hline 3.&\textrm{Simpangan}&Q_{d}\\ &\textrm{kuartil}&\\\hline 4.&\textrm{Langkah}&L\\\hline 5.&\textrm{Pagar dalam}&Q_{1}-L\\\hline 6.&\textrm{Pagar luar}&Q_{3}-L\\\hline 7.&\textrm{Simpangan}&SR\\ &\textrm{rata-rata}&\\\hline 8.&\textrm{Ragam/variansi}&S^{2}\\\hline 9&\textrm{Simpangan baku}&S\\\hline 10.&\textrm{Koefisien variansi}&V\\\hline \end{array}$.

Sebagai catatan bahwa $H$ selain disebut jangkauan antarkuartil sebagaian ada yang menyebut dengan istilah rentang antar kuartil dan terkadang pula dengan sebutan jangkauan interkuartil (Inter Quartile Range) dan juga terkadang menyebutnya dengan hamparan. Untuk $Q_{d}$  selanjutnyanya ada yang buku yang menyebutnya dengan istilah simpangan kuartil terkadang juga rentang semi interkuartil atau jangkauan antarkuartil.

Perhatikan gambar distribusi frekuensi suatu data berikut

B. Ukuran Penyebaran Data

B. 1 Data Tunggal

$\begin{array}{|c|l|c|}\hline \textrm{No}&\quad \textrm{Data}&\textrm{Formula}\\\hline 1.&R\: \: \textrm{atau}\: \: J&x_{max}-x_{min}\\\hline 2.&H&Q_{3}-Q_{1}\\\hline 3.&Q_{d}&\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 4.&L&\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )\\\hline 5.&Q_{1}-L&Q_{1}-L\\\hline 6.&Q_{3}-L&Q_{3}-L\\\hline 7.&SR&\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\\hline 8.&S^{2}&\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2}\\\hline 9&S&\sqrt{S2}\\\hline 10.&V&\displaystyle \frac{S}{\overline{x}}\times 100 \%\\\hline \end{array}$.

Catata: Data ukuran yang kurang dari pagar dalam dan atau lebih besar dari pagar luar dinamakan pencilan.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 1.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} 30&32&32&43&50&51\\ 53&53&58&58&58&60\\ 63&64&66&67&68&69\\ 70&72&75&78&80&82\\ 84&85&86&86&83&83 \end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Jangkauan}\\ &\textrm{b}.\quad Q_{1},\, Q_{2},\, \textrm{dan}\: \: Q_{3}\\ &\textrm{c}.\quad \textrm{Jangkauan Antarkuartil}\\ &\textrm{d}.\quad \textrm{Simpangan Kuartil}\\ &\textrm{e}.\quad \textrm{Pagar Dalam}\\ &\textrm{f}.\quad \textrm{Pagar Luar}\\ &\textrm{g}.\quad \textrm{Pencilan}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan sajian data dalam bentuk}\\ &\textrm{diagram}\: \textbf{batang daun}\: \textrm{berikut}\\ &\begin{array}{|c|l|}\hline \textbf{Batang}&\: \quad\quad\textbf{Daun}\\\hline \color{red}3&0\: \: 2\: \: 2\\ \color{red}4&3\\ \color{red}5&0\: \: 1\: \: 3\: \: 3\: \: 8\: \: 8\: \: 8\\ \color{red}6&0\: \: 3\: \: 4\: \: 6\: \: 7\: \: 8\: \: 9\\ \color{red}7&0\: \: 2\: \: 5\: \: 8\\ \color{red}8&0\: \: 2\: \: 3\: \: 3\: \: 4\: \: 5\: \: 6\: \: 6\\\hline \end{array}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\: \: \color{blue}n=30\\ \textrm{a}.\quad \: J&=x_{max}-x_{min}=86-30=\color{red}56\\ \textrm{b}.\: \: \: Q_{1}&=\left ( x_{._{\frac{1}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{1}{4}.30+\frac{1}{2}}} \right )=x_{.8}=53\\ Q_{2}&=\left ( x_{._{\frac{2}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{2}{4}.30+\frac{1}{2}}} \right )\\ &=\displaystyle \frac{x_{.15}+x_{.16}}{2}=\displaystyle \frac{66+67}{2}=66,7\\ Q_{3}&=\left ( x_{._{\frac{3}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{3}{4}.30+\frac{1}{2}}} \right )=x_{.23}=80\\ \textrm{c}.\: \: \: H&=Q_{3}-Q_{1}\\ &=x_{._{23}}-x_{._{8}}=80-53=27\\ \end{aligned} \end{array}$

$.\qquad\begin{aligned}\textrm{d}.\: \: \: Q_{d}&=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &=\displaystyle \frac{1}{2}(H)=\displaystyle \frac{1}{2}\left ( 27 \right )=\color{red}13,5\\ \textrm{e}.\: \quad L&=\displaystyle \frac{3}{2}(H)=\displaystyle \frac{3}{2}(27)=\color{red}40,5\\ \textrm{P}&\textrm{agar dalam}:\\ &=Q_{1}-L=53-40,5=\color{red}12,5\\ \textrm{P}&\textrm{agar luar}:\\ &=Q_{1}-L=80+40,5=\color{red}120,5\\ \textrm{g}.\: \quad \textrm{D}&\textrm{ari fakta yang ada data ukuran}\\ &\textrm{yang besarnya kurang dari}\\ &\textrm{pagar dalam dan lebih besar dari}\\ &\textrm{pagar luar tidak ada, maka} \\ &\textrm{tidak ada}\: \color{red}\textbf{data pencilan} \end{aligned}$.

$\begin{array}{ll} 2.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} 73&74&66&65&68&65\\ 60&64&78&79&81&61\\ 72&74&71&68&75&76\\ 96&56&64&80&84&43\end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Jangkauan}\\ &\textrm{b}.\quad Q_{1},\, Q_{2},\, \textrm{dan}\: \: Q_{3}\\ &\textrm{c}.\quad \textrm{Jangkauan Antarkuartil}\\ &\textrm{d}.\quad \textrm{Simpangan Kuartil}\\ &\textrm{e}.\quad \textrm{Pagar Dalam}\\ &\textrm{f}.\quad \textrm{Pagar Luar}\\ &\textrm{g}.\quad \textrm{Pencilan}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan sajian data dalam bentuk}\\ &\textrm{diagram}\: \textbf{batang daun}\: \textrm{berikut}\\ &\begin{array}{|c|l|}\hline \textbf{Batang}&\: \quad\quad\textbf{Daun}\\\hline \color{red}4&3\\ \color{red}5&6\\ \color{red}6&0\: \: 1\: \: 4\: \: 4\: \: 5\: \: 5\: \: 6\: \: 8\: \: 8\\ \color{red}7&1\: \: 2\: \: 3\: \: 4\: \: 4\: \: 5\: \: 6\: \: 8\: \: 9\\ \color{red}8&0\: \: 1\: \: 3\\ \color{red}9&6\\\hline \end{array}\\ &\begin{aligned}\textrm{Diketa}&\textrm{hui}\: \: \color{blue}n=24\\ \textrm{a}.\quad \: J&=x_{max}-x_{min}=96-43=\color{red}53\\ \textrm{b}.\: \: \: Q_{1}&=\left ( x_{._{\frac{1}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{1}{4}.24+\frac{1}{2}}} \right )=x_{_{6,5}}\\ &=\displaystyle \frac{1}{2}\left ( x_{._{6}}+x_{._{7}} \right )=\displaystyle \frac{64+65}{2}=\color{red}64,5\\ Q_{2}&=\left ( x_{._{\frac{2}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{2}{4}.24+\frac{1}{2}}} \right )=x_{_{12,5}}\\ &=\displaystyle \frac{x_{.12}+x_{.13}}{2}=\displaystyle \frac{71+72}{2}=\color{red}71,5\\ Q_{3}&=\left ( x_{._{\frac{3}{4}n+\frac{1}{2}}} \right )=\left ( x_{._{\frac{3}{4}.24+\frac{1}{2}}} \right )=x_{_{18,5}}\\ &=\displaystyle \frac{x_{_{18}}+x_{_{19}}}{2}=\displaystyle \frac{76+78}{2}=\color{red}77\\ \textrm{c}.\: \: \: H&=Q_{3}-Q_{1}\\ &=77-64,5=12,5\\ \end{aligned} \end{array}$.

$.\qquad\begin{aligned}\textrm{d}.\: \: \: Q_{d}&=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &=\displaystyle \frac{1}{2}(H)=\displaystyle \frac{1}{2}\left ( 12,5 \right )=\color{red}6,26\\ \textrm{e}.\: \quad L&=\displaystyle \frac{3}{2}(H)=\displaystyle \frac{3}{2}(12,5)=\color{red}18,75\\ \textrm{P}&\textrm{agar dalam}:\\ &=Q_{1}-L=64,5-18,75=\color{red}45,75\\ \textrm{P}&\textrm{agar luar}:\\ &=Q_{1}-L=77+18,75=\color{red}95,75\\ \textrm{g}.\: \quad \textrm{D}&\textrm{ari fakta di atas terdapat}\: \textbf{pencilan}\\ &\textrm{yaitu}:\: \color{red}43 \: \color{black}\textrm{dan}\: \: \color{red}96 \end{aligned}$.

$\begin{array}{ll} 3.&\textrm{Diberikan data berikut}\\ &\color{purple}\begin{array}{lllllll} \color{black}\textrm{a}.&3&4&5&6&7\\ \color{black}\textrm{b}.&1&2&5&8&9\end{array}\\ &\textrm{Tentukan}\\ &\textrm{a}.\quad \textrm{Simpangan rata-rata}\\ &\textrm{b}.\quad \textrm{Ragam}\\ &\textrm{c}.\quad \textrm{Simpangan baku}\\\\ &\textbf{Jawab}:\\ &\textrm{Untuk data}:3,4,5,6,7\\ &\begin{aligned}\textrm{Diketahu}&\textrm{i}\: \: \color{blue}n=5\\ \textrm{a}.\quad \: \overline{x}=&\displaystyle \frac{3+4+5+6+7}{5}=\frac{25}{5}=\color{red}5\\ \textrm{sel}&\textrm{anjutnya}\\ \textrm{SR}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right | \\ &=\displaystyle \frac{1}{5}\left (\left | 3-5 \right | +\left | 4-5 \right |+\left | 5-5 \right |+\left |6-5 \right |+\left | 7-5 \right | \right )\\ &=\displaystyle \frac{1}{5}\left ( \left | -2 \right |+\left | -1 \right |+\left | 0 \right |+\left | 1 \right |+\left | 2 \right | \right )\\ &=\displaystyle \frac{1}{5}(2+1+0+1+2)\\ &=\displaystyle \frac{6}{5}=\color{red}1,2 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \: \textrm{S}^{2}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2} \\ &=\displaystyle \frac{1}{5}\left ((3-5)^{2} +(4-5)^{2}+(5-5)^{2}+(6-5)^{2}+(7-5)^{2} \right )\\ &=\displaystyle \frac{1}{5}\left ( 4+1+0+1+4 \right )\\ &=\displaystyle \frac{1}{5}(8)\\ &=\displaystyle \frac{8}{5}=\color{red}1,6\\ \textrm{c}.\quad \: \: \: S&=\sqrt{S^{2}}\\ &=\sqrt{1,6}\approx \color{red}1,26 \end{aligned}\\\\ &\textrm{Dan untuk data}:1,2,5,8,9\\ &\begin{aligned}\textrm{Diketahu}&\textrm{i}\: \: \color{blue}n=5\\ \textrm{a}.\quad \: \overline{x}=&\displaystyle \frac{1+2+5+8+9}{5}=\frac{25}{5}=\color{red}5\\ \textrm{sel}&\textrm{anjutnya}\\ \textrm{SR}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right | \\ &=\displaystyle \frac{1}{5}\left (\left | 1-5 \right | +\left | 2-5 \right |+\left | 5-5 \right |+\left |8-5 \right |+\left | 9-5 \right | \right )\\ &=\displaystyle \frac{1}{5}\left ( \left | -4 \right |+\left | -3 \right |+\left | 0 \right |+\left | 3 \right |+\left | 4 \right | \right )\\ &=\displaystyle \frac{1}{5}(4+3+0+3+4)\\ &=\displaystyle \frac{14}{5}=\color{red}2,8 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad \: \textrm{S}^{2}&=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left ( x_{i}-\overline{x} \right )^{2} \\ &=\displaystyle \frac{1}{5}\left ((1-5)^{2} +(2-5)^{2}+(5-5)^{2}+(8-5)^{2}+(9-5)^{2} \right )\\ &=\displaystyle \frac{1}{5}\left ( 16+9+0+9+16 \right )\\ &=\displaystyle \frac{1}{5}(50)\\ &=\displaystyle \frac{50}{5}=\color{red}10\\ \textrm{c}.\quad \: \: \: S&=\sqrt{S^{2}}\\ &=\sqrt{10}\approx \color{red}3,16 \end{aligned} \end{array}$.

$\LARGE\colorbox{aqua}{LATIHAN SOAL}$.

$\begin{array}{ll} 1.&\textrm{Tentukan nilai Jangkauan},Q_{1},Q_{2},Q_{3}\\ &hamparan,\: \textrm{simpangan kuartil, langkah}\\ &\textrm{pagar dalam, pagar luar, dan pencilan}\\ &\textrm{dari data berikut}\\ &\color{red}\begin{array}{llll} \color{black}\textrm{a}.&3,5,7,9,1,2,8,2,3,4,3,5,7\\ \color{black}\textrm{b}.&10,11,12,13,8,9,4,5,7,5\end{array} \end{array}$.

$\begin{array}{ll} 2.&\textrm{Tentukan simpangan rata-rata}\\ &\textrm{ragam, dan simpangan baku}\\ &\textrm{dari data berikut}\\ &\color{red}\begin{array}{llll} \color{black}\textrm{a}.&3,5,7,9,1\\ \color{black}\textrm{b}.&10,11,12,13,8,9,4,15,7,5\end{array} \end{array}$.

$\begin{array}{ll} 3.&\textrm{Empat buah bilangan memiliki mean,}\\ &\textrm{tentukanlah keempat bilangan tersebut}\\ \end{array}$.

$\begin{array}{ll} 4.&\textrm{Diketahui datum-datum}\\ &:x-4,x-2,x+1,x+2,x+4,x+5\\ &\textrm{tentukanlah}\\ &\textrm{a}.\quad\textrm{nilai simpangan baku(nyatakan dalam)\: }x\\ &\textrm{b}.\quad\textrm{nilai}\: \: x\: \: \textrm{dan simpangan baku jika mean}\\ & \: \: \: \: \quad\textrm{dari data di atas adalah 9} \end{array}$.

$\begin{array}{ll} 5.&\textrm{Diketahui simpangan baku}\\ &:2,4,7,11,9-n,9+n\: \: \textrm{adalah}\: \: \sqrt{11}\\ &\textrm{tentukanlah}\\ &\textrm{a}.\quad\textrm{mean}\\ &\textrm{b}.\quad\textrm{nilai}\: \: n\: \: \textrm{yang mungkin} \end{array}$.

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Alam Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
2. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: Srikandi Empat Widya Utama.
3. Sharma, S.N., dkk. 2017. Jelajah Matematika 3 SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.