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.

$\begin{array}{|clc|}\hline \text{No}& \text{Data Dispersi}& \text{Simbol}\\ 1.& \text{Jangkauan}& R\text{atau}J\\ 2.& \text{Jangkauan}& H\\ & \text{antarkuartil}& \\ 3.& \text{Simpangan}& Q_d\\ & \text{kuartil}& \\ 4.& \text{Langkah}& L\\ 5.& \text{Pagar dalam}& Q_1-L\\ 6.& \text{Pagar luar}& Q_3-L\\ 7.& \text{Simpangan}& SR\\ & \text{rata-rata}& \\ 8.& \text{Ragam/variansi}& S^2\\ 9& \text{Simpangan baku}& S\\ 10.& \text{Koefisien variansi}& V\\ \hline\end{array}$.

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

$\begin{array}{|clc|}\hline \text{No}& \text{Data}& \text{Formula}\\ 1.& R\text{atau}J& x_{max}-x_{min}\\ 2.& H& Q_3-Q_1\\ 3.& Q_d& {\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ 4.& L& {\displaystyle \frac{3}{2}(Q_3-Q_1)}\\ 5.& Q_1-L& Q_1-L\\ 6.& Q_3-L& Q_3-L\\ 7.& SR& {\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ 8.& S^2& {\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ 9& S& \sqrt{S2}\\ 10.& V& {\displaystyle \frac{S}{\bar{x}}\times 100\mathrm{\%}}\\ \hline\end{array}$.

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

$\text{CONTOH SOAL}$.

$\begin{array}{ll}1.& \text{Diberikan data berikut}\\ & \begin{array}{llllll}30& 32& 32& 43& 50& 51\\ 53& 53& 58& 58& 58& 60\\ 63& 64& 66& 67& 68& 69\\ 70& 72& 75& 78& 80& 82\\ 84& 85& 86& 86& 83& 83\end{array}\\ & \text{Tentukan}\\ & \text{a}.\text{Jangkauan}\\ & \text{b}.Q_1,Q_2,\text{dan}{Q}_3\\ & \text{c}.\text{Jangkauan Antarkuartil}\\ & \text{d}.\text{Simpangan Kuartil}\\ & \text{e}.\text{Pagar Dalam}\\ & \text{f}.\text{Pagar Luar}\\ & \text{g}.\text{Pencilan}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan sajian data dalam bentuk}\\ & \text{diagram}\mathbf{\text{batang daun}}\text{berikut}\\ & \begin{array}{|cl|}\hline \mathbf{\text{Batang}}& \mathbf{\text{Daun}}\\ 3& 022\\ 4& 3\\ 5& 0133888\\ 6& 0346789\\ 7& 0258\\ 8& 02334566\\ \hline\end{array}\\ & \begin{array}{rl}\text{Diketa}& \text{hui}n=30\\ \text{a}.J& =x_{max}-x_{min}=86-30=56\\ \text{b}.Q_1& =\left(x_{{.}_{\frac{1}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{1}{4}.30+\frac{1}{2}}}\right)=x_{.8}=53\\ Q_2& =\left(x_{{.}_{\frac{2}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{2}{4}.30+\frac{1}{2}}}\right)\\ & ={\displaystyle \frac{x_{.15}+x_{.16}}{2}={\displaystyle \frac{66+67}{2}=66,7}}\\ Q_3& =\left(x_{{.}_{\frac{3}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{3}{4}.30+\frac{1}{2}}}\right)=x_{.23}=80\\ \text{c}.H& =Q_3-Q_1\\ & =x_{{.}_{23}}-x_{{.}_8}=80-53=27\end{array}\end{array}$

$.\begin{array}{rl}\text{d}.Q_d& ={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(H)={\displaystyle \frac{1}{2}\left(27\right)=13,5}}\\ \text{e}.L& ={\displaystyle \frac{3}{2}(H)={\displaystyle \frac{3}{2}(27)=40,5}}\\ \text{P}& \text{agar dalam}:\\ & =Q_1-L=53-40,5=12,5\\ \text{P}& \text{agar luar}:\\ & =Q_1-L=80+40,5=120,5\\ \text{g}.\text{D}& \text{ari fakta yang ada data ukuran}\\ & \text{yang besarnya kurang dari}\\ & \text{pagar dalam dan lebih besar dari}\\ & \text{pagar luar tidak ada, maka}\\ & \text{tidak ada}\mathbf{\text{data pencilan}}\end{array}$.

$\begin{array}{ll}2.& \text{Diberikan data berikut}\\ & \begin{array}{llllll}73& 74& 66& 65& 68& 65\\ 60& 64& 78& 79& 81& 61\\ 72& 74& 71& 68& 75& 76\\ 96& 56& 64& 80& 84& 43\end{array}\\ & \text{Tentukan}\\ & \text{a}.\text{Jangkauan}\\ & \text{b}.Q_1,Q_2,\text{dan}{Q}_3\\ & \text{c}.\text{Jangkauan Antarkuartil}\\ & \text{d}.\text{Simpangan Kuartil}\\ & \text{e}.\text{Pagar Dalam}\\ & \text{f}.\text{Pagar Luar}\\ & \text{g}.\text{Pencilan}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan sajian data dalam bentuk}\\ & \text{diagram}\mathbf{\text{batang daun}}\text{berikut}\\ & \begin{array}{|cl|}\hline \mathbf{\text{Batang}}& \mathbf{\text{Daun}}\\ 4& 3\\ 5& 6\\ 6& 014455688\\ 7& 123445689\\ 8& 013\\ 9& 6\\ \hline\end{array}\\ & \begin{array}{rl}\text{Diketa}& \text{hui}n=24\\ \text{a}.J& =x_{max}-x_{min}=96-43=53\\ \text{b}.Q_1& =\left(x_{{.}_{\frac{1}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{1}{4}.24+\frac{1}{2}}}\right)=x_{{}_{6,5}}\\ & ={\displaystyle \frac{1}{2}(x_{{.}_6}+x_{{.}_7})={\displaystyle \frac{64+65}{2}=64,5}}\\ Q_2& =\left(x_{{.}_{\frac{2}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{2}{4}.24+\frac{1}{2}}}\right)=x_{{}_{12,5}}\\ & ={\displaystyle \frac{x_{.12}+x_{.13}}{2}={\displaystyle \frac{71+72}{2}=71,5}}\\ Q_3& =\left(x_{{.}_{\frac{3}{4}n+\frac{1}{2}}}\right)=\left(x_{{.}_{\frac{3}{4}.24+\frac{1}{2}}}\right)=x_{{}_{18,5}}\\ & ={\displaystyle \frac{x_{{}_{18}}+x_{{}_{19}}}{2}={\displaystyle \frac{76+78}{2}=77}}\\ \text{c}.H& =Q_3-Q_1\\ & =77-64,5=12,5\end{array}\end{array}$.

$.\begin{array}{rl}\text{d}.Q_d& ={\displaystyle \frac{1}{2}(Q_3-Q_1)}\\ & ={\displaystyle \frac{1}{2}(H)={\displaystyle \frac{1}{2}(12,5)=6,26}}\\ \text{e}.L& ={\displaystyle \frac{3}{2}(H)={\displaystyle \frac{3}{2}(12,5)=18,75}}\\ \text{P}& \text{agar dalam}:\\ & =Q_1-L=64,5-18,75=45,75\\ \text{P}& \text{agar luar}:\\ & =Q_1-L=77+18,75=95,75\\ \text{g}.\text{D}& \text{ari fakta di atas terdapat}\mathbf{\text{pencilan}}\\ & \text{yaitu}:43\text{dan}96\end{array}$.

$\begin{array}{ll}3.& \text{Diberikan data berikut}\\ & \begin{array}{llllll}\text{a}.& 3& 4& 5& 6& 7\\ \text{b}.& 1& 2& 5& 8& 9\end{array}\\ & \text{Tentukan}\\ & \text{a}.\text{Simpangan rata-rata}\\ & \text{b}.\text{Ragam}\\ & \text{c}.\text{Simpangan baku}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Untuk data}:3,4,5,6,7\\ & \begin{array}{rl}\text{Diketahu}& \text{i}n=5\\ \text{a}.\bar{x}=& {\displaystyle \frac{3+4+5+6+7}{5}=\frac{25}{5}=5}\\ \text{sel}& \text{anjutnya}\\ \text{SR}& ={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{5}(|3-5|+|4-5|+|5-5|+|6-5|+|7-5|)}\\ & ={\displaystyle \frac{1}{5}(|-2|+|-1|+\left|0\right|+\left|1\right|+\left|2\right|)}\\ & ={\displaystyle \frac{1}{5}(2+1+0+1+2)}\\ & ={\displaystyle \frac{6}{5}=1,2}\end{array}\\ & \begin{array}{rl}\text{b}.{\text{S}}^2& ={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & ={\displaystyle \frac{1}{5}((3-5{)}^2+(4-5{)}^2+(5-5{)}^2+(6-5{)}^2+(7-5{)}^2)}\\ & ={\displaystyle \frac{1}{5}(4+1+0+1+4)}\\ & ={\displaystyle \frac{1}{5}(8)}\\ & ={\displaystyle \frac{8}{5}=1,6}\\ \text{c}.S& =\sqrt{S^2}\\ & =\sqrt{1,6}\approx 1,26\end{array}\\ \\ & \text{Dan untuk data}:1,2,5,8,9\\ & \begin{array}{rl}\text{Diketahu}& \text{i}n=5\\ \text{a}.\bar{x}=& {\displaystyle \frac{1+2+5+8+9}{5}=\frac{25}{5}=5}\\ \text{sel}& \text{anjutnya}\\ \text{SR}& ={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n|x_i-\bar{x}|}}\\ & ={\displaystyle \frac{1}{5}(|1-5|+|2-5|+|5-5|+|8-5|+|9-5|)}\\ & ={\displaystyle \frac{1}{5}(|-4|+|-3|+\left|0\right|+\left|3\right|+\left|4\right|)}\\ & ={\displaystyle \frac{1}{5}(4+3+0+3+4)}\\ & ={\displaystyle \frac{14}{5}=2,8}\end{array}\\ & \begin{array}{rl}\text{b}.{\text{S}}^2& ={\displaystyle \frac{1}{n}{\displaystyle \sum _{i=1}^n{(x_i-\bar{x})}^2}}\\ & ={\displaystyle \frac{1}{5}((1-5{)}^2+(2-5{)}^2+(5-5{)}^2+(8-5{)}^2+(9-5{)}^2)}\\ & ={\displaystyle \frac{1}{5}(16+9+0+9+16)}\\ & ={\displaystyle \frac{1}{5}(50)}\\ & ={\displaystyle \frac{50}{5}=10}\\ \text{c}.S& =\sqrt{S^2}\\ & =\sqrt{10}\approx 3,16\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{ll}1.& \text{Tentukan nilai Jangkauan},Q_1,Q_2,Q_3\\ & hamparan,\text{simpangan kuartil, langkah}\\ & \text{pagar dalam, pagar luar, dan pencilan}\\ & \text{dari data berikut}\\ & \begin{array}{ll}\text{a}.& 3,5,7,9,1,2,8,2,3,4,3,5,7\\ \text{b}.& 10,11,12,13,8,9,4,5,7,5\end{array}\end{array}$.

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

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

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

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

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.