A. Pendahuluan
Dalam statistik deskriptif ada nilai yang yang dapat mewakili pengukuran baik dalam pengukuran secara tunggal ataupun berkelompok dan nilai ini selanjutnya dinamakan sebagai ukuran pemusatan data dikarenakan akan memiliki kecenderungan nilai yang sama. Bahasan yang dimasukkan dalam kelompok ukuran pemusatan data ini ada 3 macam, yaitu: mean, median, dan modus.
B. Ukuran Pemusatan Data
Jika diketahui terdapat data dengan datum-datum: $x_{1},x_{2},x_{3},x_{4},...,x_{n}$, maka
1. Mean(rata-rata)
$\begin{aligned}&\textrm{Dilambangkan dengan}\: \: \overline{x}\\ &\textrm{dibaca: x bar}.\\ &\color{red}\textrm{Rumus data tunggal}:\\ &\overline{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}+...+x_{n}}{n}\\ &\textrm{atau}\\ &\overline{x}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}x_{i}}{n}\\ &\color{red}\textrm{Rumus data berkelompok/berfrekuensi}:\\ &\overline{x}=\displaystyle \frac{f_{1}.x_{1}+f_{2}.x_{2}+f_{3}.x_{3}+...+f_{_{n}}.x_{n}}{f_{1}+f_{2}+f_{3}+...+f_{n}}\\ &\textrm{atau}\\ &\overline{x}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.x_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}&=\textrm{mean/rataan hitung}\\ f_{i}&=\textrm{frekuensi kelas interval ke}-i\\ x_{i}&=\textrm{nilai tengah kelas interval ke}-i\\ n&=\textrm{banyak data/kelas interval} \end{aligned} \end{aligned}$.
$\begin{aligned}&\color{red}\textrm{Rumus data berkelompok}:\\ &\color{red}\textrm{dengan menggunakan rataan sementara}\\ &\overline{x}=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}&=\textrm{mean/rataan hitung}\\ x_{s}&=\textrm{rataan sementara dipilih}\\ x_{i}&=\textrm{nilai tengah kelas interval ke}-i\\ f_{i}&=\textrm{frekuensi kelas interval ke}-i\\ d_{i}&=\textrm{simpangan/deviasi}=x_{i}-x_{s}\\ \end{aligned}\\\\ &\color{red}\textrm{Rumus gabungan}:\\ &\overline{x}_{g}=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.\overline{x}_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &\color{blue}\textrm{Penjelasan}\\ &\qquad \begin{aligned}\overline{x}_{g}&=\textrm{rataan gabungan}\\ f_{i}&=\textrm{frekuensi kelompok ke}-i\\ \overline{x}_{i}&=\textrm{rata-rata kelompok ke}-i\\ \end{aligned} \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah rata-rata dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{purple}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ \overline{x}&=\displaystyle \frac{5.1+6.4+7.6+8.4+9.1}{16}\\ &=\color{black}\frac{112}{16}=\color{red}7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 2-6&2\\\hline 7-11&3\\\hline 12-16&3\\\hline 17-21&6\\\hline 22-26&6\\\hline \end{array}\\ &\textrm{Tentukan rataannya}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\begin{array}{|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&\color{red}f_{i}&\color{black}x_{i}.f_{i}\\\hline 2-6&4&2&8\\\hline 7-11&9&3&27\\\hline 12-16&14&3&42\\\hline 17-21&19&6&114\\\hline 22-26&24&6&144\\\hline &\color{black}\sum_{i=1}^{5}&\color{red}20&\color{black}335\\\hline \end{array}\\ &\color{black}\textrm{maka rata-ratanya}\\ &\overline{x}=\displaystyle \frac{\sum x_{i}.f_{i}}{\sum f_{i}}=\frac{335}{20}=\color{red}16,75 \end{aligned} \end{array}$.
$.\qquad\begin{aligned}&\textrm{Dengan rumus rataan sementara},\: \textrm{yaitu}:\\ &\begin{array}{|c|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&d_{i}=x_{i}-x_{s}&\color{red}f_{i}&\color{black}f_{i}.d_{i}\\\hline 2-6&4&-15&2&-30\\\hline 7-11&9&-10&3&-30\\\hline 12-16&14&-5&3&-15\\\hline 17-21&\color{red}19&\color{blue}0&6&0\\\hline 22-26&24&5&6&30\\\hline &&&\color{red}20&-45\\\hline \end{array}\\ &\begin{aligned}\overline{x}&=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}.d_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\\ &=19+\displaystyle \frac{-45}{20}\\ &=19-2,25\\ &=\color{red}16,75 \end{aligned} \end{aligned}$.
$\begin{array}{ll}\\ 3.&\textrm{Nilai rata-rata dari 20 bilangan adalah}\: \: 14,2\\ &\textrm{Jika rata-rata dari 12 bilangan pertama adalah}\\ &\textrm{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ &\textrm{adalah 18,2, tentukan rata-rata 2 bilangan}\\ &\textrm{tersisa}\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\color{black}\displaystyle \frac{\overline{x}_{12}\times (12)+\overline{x}_{6}\times (6)+\overline{x}_{2}\times (2)}{20}\\ 14,2&=\color{black}\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+\overline{x}_{2}\times (2)}{20}\\ 284&=\color{black}151,2+109,2+2\overline{x}_{2}\\ 2\overline{x}_{2}&=\color{black}284-(151,2+109,2)=\color{red}23,6\\ \overline{x}_{2}&=\displaystyle \frac{23,6}{2}\\ &=\color{red}11,8 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&(\textbf{SPMB 04})\\ &\textrm{Nilai rata-rata tes Matematika dari kelompok}\\ &\textrm{siswa dan kelompok siswi di suatu kelas berturut-}\\ &\textrm{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ &\textrm{tersebut adalah 6,2, maka perbandingan banyaknya}\\ &\textrm{siswa dan siswi adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2:3\\ \textrm{b}.&3:4\\ \textrm{c}.&2:5\\ \textrm{d}.&3:5\\ \textrm{e}.&4:5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\displaystyle \frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}\\ \color{black}6,2&\color{blue}=\displaystyle \frac{n_{1}(5)+n_{2}(7)}{n_{1}+n_{2}}\\ 6,2(n_{1}+n_{2})&=5n_{1}+7n_{2}\\ 6,2n_{1}-5n_{1}&=7n_{2}-6,2n_{2}\\ 1,2n_{1}&=0,8n_{2}\\ \displaystyle \frac{n_{1}}{n_{2}}&=\displaystyle \frac{0,8}{1,2}=\color{red}\frac{2}{3} \end{aligned} \end{array}$.
2. Median
$\begin{aligned}&\textrm{Dilambangkan dengan}:\: Me\\ &\color{red}\textrm{Rumus data tunggal}\\ &\bullet \: \: \textrm{data ganjil},\: Me=x_{._{\frac{n+1}{2}}}\\ &\bullet \: \: \textrm{data genap},\: Me=\displaystyle \frac{1}{2}\left ( x_{._{\frac{n}{2}}}+x_{._{\frac{n}{2}+1}} \right )\\ &\color{red}\textrm{Rumus data berkelompok}\\ &Me=L+\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\times c \\ &\color{blue}\textrm{Penjelasan}\\ &\begin{aligned}L&=\textrm{tepi bawah kelas median}\\ f_{k}&=\textrm{frekuensi kumulatif sebelum}\\ &\: \, \quad \textrm{kelas median}\\ c&=\textrm{lebar kelas interval}\\ n&=\textrm{total datum} \end{aligned} \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah median dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{purple}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ Me&=\displaystyle \frac{1}{2}\left ( x_{._{\frac{n}{2}}}+x_{._{\frac{n}{2}+1}} \right )\\ &=\displaystyle \frac{1}{2}\left ( x_{._{\frac{16}{2}}}+x_{._{\frac{16}{2}+1}} \right )=\displaystyle \frac{1}{2}\left ( x_{8}+x_{9} \right )\\ &=\displaystyle \frac{7+7}{2}\\ &=\color{black}\frac{14}{2}=\color{red}7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Dari data berikut yang memiliki}\\ &\textbf{mean}\: \: 7\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \textbf{median}\: 7\: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2,5,6,9,7,8,5,14,8,11\\ \textrm{b}.&6,3,7,8,6,4,11,8,9,8\\ \textrm{c}.&3,7,10,7,9,5,10,2,14,11\\ \textrm{d}.&4,1,6,12,8,11,4,5,8,2\\ \color{red}\textrm{e}.&2,3,4,3, 10,8,12,6,15,12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{array}{|l|l|}\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \textrm{Median}&2,5,6,9,7,8,5,14,8,11\\\hline \quad \textrm{a}&\color{blue}2,5,5,6,\color{red}7,8,\color{blue}8,9,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{70}{10}=7\\\hline \textrm{Median}&6,3,7,8,6,4,11,8,9,8\\\hline \quad \textrm{b}&\color{blue}3,4,6,6,\color{red}7,8,\color{blue}8,8,9,11\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{78}{10}=7,8\\\hline \textrm{Median}&3,7,10,7,9,5,10,2,14,11\\\hline \quad \textrm{c}&\color{blue}2,3,5,7,\color{red}7,9,\color{blue}10,10,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{61}{10}=6,1\\\hline \textrm{Median}&4,1,6,12,8,11,4,5,8,2\\\hline \quad \textrm{d}&\color{blue}1,2,4,4,\color{red}5,6,\color{blue}8,8,11,12\\\hline \color{blue}\textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \color{blue}\textrm{Median}&2,3,4,3, 10,8,12,6,15,12\\\hline \quad \color{red}\textrm{e}&\color{blue}2,3,3,4,\color{red}6,8,\color{blue}10,12,12,15\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 51-60&5\\\hline 61-70&7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline \end{array}\\ &\textrm{Tentukanlah mediannya}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&\textrm{Diketahui},\: n=\sum f=40,\: \: \textrm{Perhatikan tabel}\\ &\textrm{berikut ini}\\ &\begin{array}{|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}f_{i}\\\hline 51-60&\color{red}5\\\hline 61-70&\color{red}7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline &\color{red}40\\\hline \end{array}\\ &\color{red}Median=\textrm{Datum ke}-\left ( \displaystyle \frac{40}{2} \right )=x_{._{20}}\\ &x_{._{20}}\: \: \textrm{terletak pada kelas interval}:\: \: 71-80\\ &\textrm{dengan}\: \: f=14,\: \: f_{k}\: \: \textrm{sebelum}\: \: Me=12,\\ &L=70,5,\: \: \textrm{serta}\: \: c=10\\ &\textrm{maka mediannya}\\ &Q_{2}=\color{purple}L+c\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\\ &=\color{black}70,5+10\left ( \displaystyle \frac{20-12}{14} \right )\\ &=\color{black}70,5+5,714=\color{red}76,21 \end{aligned} \end{array}$.
3. Modus
Modus dikatakan ada jika sekelompok data memiliki datum yang paling banyak muncul. Jika frekuensi munculnya datum terjadi kesamaan, maka dikatakan tidak ada modus.
$\begin{aligned}&\textrm{Dilambangkan dengan}:\: Mo\\ &\color{red}\textrm{Rumus data tunggal}\\ &\textrm{hanya untuk datum yang sering muncul}\\ &\color{red}\textrm{Rumus data berkelompok}\\ &Mo=L+\left ( \displaystyle \frac{\triangle _{1}}{\triangle _{1}+\triangle _{2}} \right )\times c \\ &\color{blue}\textrm{Penjelasan}\\ &\begin{aligned}L&=\textrm{tepi bawah kelas modus}\\ \triangle _{1}&=\textrm{frekuensi kelas modus dengan}\\ &\: \, \quad \textrm{frekuensi kelas sebelumnya}\\ \triangle _{2}&=\textrm{frekuensi kelas modus dengan}\\ &\: \, \quad \textrm{frekuensi kelas setelahnya}\\ c&=\textrm{lebar kelas interval} \end{aligned} \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah modus dari data berikut}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &\color{black}6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &\color{red}5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ Mo&=\color{black}\textrm{berupa datum yang sering muncul}\\ &=\color{red}7 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Skor}&f\\\hline 40-49&8\\\hline 50-59&9\\\hline 60-69&22\\\hline 70-79&15\\\hline 80-89&6\\\hline \end{array}\\ &\textrm{Modusnya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&65,50\\ \color{red}\textrm{b}.&66,00\\ \textrm{c}.&66,50\\ \textrm{d}.&67,00\\ \textrm{e}.&85,50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: n=\sum f=60,\: \: \color{red}modusnya\\ &\textrm{terdapat pada kelas dengan frekuensi terbanyak}\\ & \textrm{yaitu}:\: 60-69,\: \: \textrm{dengan}\: \: p=10\\ &\begin{cases} \triangle_{1} & =f-f_{1}=22-9=13 \\ \triangle_{2} & =f-f_{2}=22-15=7 \end{cases}\\ &\color{black}\textrm{Sehingga}\\ &M_{0}=\color{purple}L+c\left ( \displaystyle \frac{\triangle_{1}}{\triangle_{1}+\triangle_{2}} \right )\\ &M_{0}=\color{black}59,5+10\left ( \displaystyle \frac{22-9}{(22-9)+(22-15)} \right )\\ &\: \: \: =\color{black}59,5+\displaystyle \frac{10.13}{13+7}\\ &\: \: \: =\color{black}59,5+6,5=\color{red}66,0 \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Johanes, Kastolan, Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Alam KBK. Jakarta: YUDHISTIRA.
- Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: Srikandi Empat Widya Utama.
- Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA
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