Statistika
A. Pendahuluan
Metode statistika banyak dijumpai dalam kehidupan kita sehari-hari. Pernyataan-pernyataan berikut terkait dengan statistika:rRata-rata nilai ulangan matematika siswa kelas XI SMA Ceria adalah 7,8; Jenis mobil yang banyak dibeli masyarakat triwulan tahun 2024 ini adalah jenis MPV, dan lain-lain.
B. Statitik dan Statistika
B. 1 Pengertian Statistika dan Statistik
Statistika adalah ilmu yang mempelajari metode-metode ilmiah terkait pengumpulan data, pengorganisasian dat, penyajian data, pengolahan data, serta interpretasi dan penarikan kesimpulan. Metode yang berkaitan dengan pengumpulan, pengorganisasian, penyajian serta pengolahan data disebut statistika deskriptif, sedangkan metode yang berkaitan dengan pengujian hipotesis, penarikan kesimpulan, pendugaan dan lain-lainnya yang semisal disebut statistika inferensia.
Dalam kehidupan sehari-hari kita sering melihat atau membaca berbagai macam laporan baik dalam bentuk angka maupun diagram. Laporan dalam bentuk diagram atau angka ini selanjutnya diamakan statistik.
B. 2 Populasi dan Sampel
Misalkan suatu hari tertentu seorang karyawan dari sebuah pabrik lampu ingin mengetahui berapa persen produksi lampu yang mengalami cacat produksi. Untuk keperluan tersebut tentunya karyawan tersebut tidak akan mengecek seluruh lampu yang telah diproduksi tersebut, tetapi cukup mengambil secara acak/random untuk diteliti. Dalam hal ini bagian dari total produksi yang diambil secara acak tadi disebut sebagai sampel dari keselurhan produksi lampu tadi yang selanjutnya disebut sebagai contoh populasi.
C. Penyajian Data
Data yang telah dikumpulkan semuanya dapat disajikan dalam bentuk daftar atau tabel. Untuk data yang besar biasanya akan disusun dalam suatu daftar atau tabel yang disebut daftar distriusi frekuensi atau daftar sebaran frekuensi. Adapun dafar distribusi dapat dibedakan menjadi dua macam, yaitu daftar distribusi frekuensi tunggal dan satunya daftar distribusi frekuensi berkelompok.
D. Ukuran Pemusatan Data
D.1 Data Tunggal
$\begin{array}{|c|l|}\hline 1.& \textrm{Mean}\quad \left ( \overline{X} \right )\\\hline &\begin{aligned}\overline{x}&=\displaystyle \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n} \end{aligned}\\\hline 2.&\textrm{Median}\quad \left ( M_{e} \right )\\\hline &\begin{aligned}&\textrm{Nilai datum tengah}\\ &\begin{aligned}M_{e}&=x_{\textrm{ganjil}}=x_{\frac{n+1}{2}}\\ M_{e}&=x_{\textrm{genap}}=\displaystyle \frac{1}{2}\left ( x_{\frac{1}{2}}+x_{\frac{1}{2}+1} \right ) \end{aligned} \end{aligned}\\\hline 3.&\textrm{Modus}\quad \left ( M_{o} \right )\\\hline &\textrm{Datum dengan frekuensi terbesar}\\\hline \end{array}$.
$\LARGE\colorbox{magenta}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Tentukan rata-rata dari}\: \: 75,60,62,87,83,65\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\overline{x}&=\displaystyle \frac{75+60+62+87+83+65}{6}\\ &=\frac{432}{6}=72 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Diketahui nilai rata-rata 10 siswa adalah }\: \: 7,0\\ &\textrm{Jika ditambah sejumlah siswa dengan nilai 8,}\\ &\textrm{nilai rata-ratanya menjadi}\: 7\displaystyle \frac{1}{3}.\: \textrm{Berapakah }\\ &\textrm{banyak siswa yang ditambahkan}?\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\overline{x}_{baru}&=\displaystyle \frac{\overline{x}_{lama}\times 10+8\times n}{10+n}\\ 7\displaystyle \frac{1}{3}&=\displaystyle \frac{7\times 10+8\times n}{10+n}\\ \displaystyle \frac{22}{3}&=\displaystyle \frac{70+8n}{10+n}\\ &\begin{aligned}\Leftrightarrow \quad &220+22n=210+24n\\ \Leftrightarrow \quad &22n-24n=210-220\\ \Leftrightarrow \quad &-2n=-10\\ \Leftrightarrow \quad &n=\displaystyle \frac{-10}{-2}=5 \end{aligned}\\ &\textrm{Jadi, banyak siswa yang ditambahkan sebanyak 5 orang} \end{aligned} \end{array}$
D. 2 Data Berkelompok
$\begin{array}{|c|l|}\hline 1.& \textrm{Mean}\quad \left ( \overline{X} \right )\\\hline &\begin{aligned}\overline{x}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}x_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}}\quad \textbf{atau}\quad \overline{X}=\overline{x}_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}d_{i}x_{i}}{\displaystyle \sum_{i=1}^{n}f_{i}} \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ \overline{x}&=\textrm{rataan sementara}\\ x_{i}&=\textrm{titik tengah interval kelas ke}-i\\ d_{i}&=x_{i}-\overline{x}_{s}\\ f_{i}&=\textrm{frekuensi kelas ke}-i \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|l|}\hline 2.& \textrm{Median}\quad \left ( M_{e}=Q_{2} \right )\\\hline &\begin{aligned}M_{e}&=L_{2}+c\left ( \displaystyle \frac{\displaystyle \frac{1}{2}n-F_{2}}{f_{3}} \right ) \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ L_{2}&=\textrm{tepi bawah kelas kuartil tengah}\: \left (Q_{2} \right )\\ n&=\textrm{ukuran data=total datum=total frekuensi}\\ f_{2}&=\textrm{frekuensi pada kelas kuartil tengah}\: \left (Q_{2} \right )\\ F_{2}&=\textrm{frekuensi kumulatif sebelum kelas kuartil tengah}\\ c&=\textrm{panjang kelas} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|l|}\hline 3.& \textrm{Modus}\quad \left ( M_{o} \right )\\\hline &\begin{aligned}M_{o}&=L+c\left ( \displaystyle \frac{d_{1}}{d_{1}+d_{2}} \right ) \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ L&=\textrm{tepi bawah kelas modus}\\ d_{1}&=\textrm{frekuensi kelas modus dengan kelas sebelumnya}\\ d_{2}&=\textrm{frekuensi kelas modus dengan kelas setelahnya}\\ c&=\textrm{panjang kelas} \end{aligned}\\\hline \end{array}$
E. Ukuran Letak Data
$\begin{array}{|c|l|}\hline 1.& \textrm{Kuartil}\quad \left ( Q_{i} \right )\\\hline &\begin{aligned}Q_{i}&=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{i}{4}n-F_{i}}{f_{i}} \right ) \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ L_{i}&=\textrm{tepi bawah kelas kuartil ke}-i\\ n&=\textrm{ukuran data=total datum=total frekuensi}\\ f_{i}&=\textrm{frekuensi pada kelas kuartil ke}-i\\ F_{i}&=\textrm{frekuensi kumulatif sebelum kelas kuartil ke}-i\\ c&=\textrm{panjang kelas} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|l|}\hline 2.& \textrm{Desil}\quad \left ( D_{i} \right )\\\hline &\begin{aligned}D_{i}&=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{i}{10}n-F_{i}}{f_{i}} \right ) \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ L_{i}&=\textrm{tepi bawah kelas desil ke}-i\\ n&=\textrm{ukuran data=total datum=total frekuensi}\\ f_{i}&=\textrm{frekuensi pada kelas desil ke}-i\\ F_{i}&=\textrm{frekuensi kumulatif sebelum kelas desil ke}-i\\ c&=\textrm{panjang kelas} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|l|}\hline 3.& \textrm{Persentil}\quad \left ( P_{i} \right )\\\hline &\begin{aligned}P_{i}&=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{i}{100}n-F_{i}}{f_{i}} \right ) \end{aligned}\\\hline &\begin{aligned}\textrm{Keter}&\textrm{angan}:\\ L_{i}&=\textrm{tepi bawah kelas persentil ke}-i\\ n&=\textrm{ukuran data=total datum=total frekuensi}\\ f_{i}&=\textrm{frekuensi pada kelas persentil ke}-i\\ F_{i}&=\textrm{frekuensi kumulatif sebelum kelas persentil ke}-i\\ c&=\textrm{panjang kelas} \end{aligned}\\\hline \end{array}$.
F. Ukuran Penyebaran Data
$\begin{array}{|c|l|c|}\hline \textrm{No}&\: \: \: \: \textrm{Data Dispersi}&\textrm{Simbol}\\\hline 1.&\textrm{Jangkauan}&R\: \: \textrm{atau}\: \: J\\\hline 2.&\textrm{Jangkauan}&H\\ &\textrm{antarkuartil}&\\\hline 3.&\textrm{Simpangan}&Q_{d}\\ &\textrm{kuartil}&\\\hline 4.&\textrm{Langkah}&L\\\hline 5.&\textrm{Pagar dalam}&Q_{1}-L\\\hline 6.&\textrm{Pagar luar}&Q_{3}-L\\\hline 7.&\textrm{Simpangan}&SR\\ &\textrm{rata-rata}&\\\hline 8.&\textrm{Ragam/variansi}&S^{2}\\\hline 9&\textrm{Simpangan baku}&S\\\hline 10.&\textrm{Koefisien variansi}&V\\\hline \end{array}$.
G. Histogram
- Diagram batang
- Diagram garis
- Line plot
- Histogram dan poligon
Daftar Pustaka
- Johanes, Kastolan, & Sulasim. 2005. Kompetensi Matematika Program Basaha SMA Kelas XI. JAkarta: YUDHISTIRA.
- Muklis. Ngapiningsih, & Astuti, A.Y. 2022. Buku Interaktif Matematika untuk SMA/MA/SMK/MAK Kelas X. Yogyakarta:PENERBIT INTAN PARIWARA.
- Noormandiri. 2022. Matematika untuk SMA/MA Kelas X Kurikulum Merdeka. Jakarta: ERLANGGA.
- Tim Penyusun. ....... Belajar Praktis Matematika Mata Pelajaran Wajib untuk SMA/MA Kelas XII. Klaten. VIVA PAKARINDO
Contoh Soal 13 Statistika
$\begin{array}{ll} 56.&\textrm{Simpangan baku dari data berikut}:\\ &6,7,4,5,3\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{2}\quad &\textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\quad&\textrm{e}.&\sqrt{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,4,5,3\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+4+5+3}{5}=\displaystyle \frac{25}{5}=\color{red}5 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((6-5)^{2}+(7-5)^{2}+(4-5)^{2}+(5-5)^{2}+(3-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(1^{2}+2^{2}+1^{2}+0+2^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 1+4+1+4 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(10)}=\sqrt{\displaystyle \frac{10}{5}}=\color{red}\sqrt{2}\\ \end{aligned} \end{array}$.
$\begin{array}{ll} 57.&\textbf{UN 2010}\\ &\textrm{Simpangan baku dari data berikut}:\\ &2,3,4,5,6\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\sqrt{15}&&&\textrm{d}.&\sqrt{3}\\ \textrm{b}.&\displaystyle \sqrt{10}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&\sqrt{2} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,3,4,5,6\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+3+4+5+6}{5}=\displaystyle \frac{20}{5}=\color{red}4 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((2-5)^{2}+(3-5)^{2}+(4-5)^{2}+(5-5)^{2}+(6-5)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(3^{2}+2^{2}+1^{2}+0+1^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 9+4+1+1 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(15)}=\sqrt{\displaystyle \frac{15}{5}}=\color{red}\sqrt{3}\\ \end{aligned} \end{array}$.
$\begin{array}{ll} 58.&\textrm{Simpangan baku dari data berikut}:\\ &7,9,11,13,15\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&2,4&&&\textrm{d}.&2,8\\ \textrm{b}.&\displaystyle 2,5\quad &\textrm{c}.&\displaystyle 2,7\quad&\textrm{e}.&2,9 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &7,9,11,13,15\quad \\ &\textrm{Simpangan bakuya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{7+9+11+13+15}{5}=\displaystyle \frac{55}{5}=\color{red}11 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left((7-11)^{2}+(9-11)^{2}+(11-11)^{2}+(13-11)^{2}+(15-11)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left(4^{2}+2^{2}+0+2^{2}+4^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}\left( 16+4+4+16 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{5}(40)}=\sqrt{\displaystyle \frac{40}{5}}=\color{red}\sqrt{8}=2,82..\\ \end{aligned} \end{array}$.
$\begin{array}{ll} 59.&\textrm{Simpangan baku dari data berikut}:\\ &2,4,4,5,6,6,7,8,9,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&4\sqrt{3}&&&\textrm{d}.&\displaystyle \frac{2}{5}\sqrt{30}\\ \textrm{b}.&2\displaystyle \frac{2}{5}\quad &\textrm{c}.&\displaystyle \sqrt{5}\quad&\textrm{e}.&2 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &2,4,4,5,6,6,7,8,9,9\quad \\ &\textrm{Simpangan bakunya adalah}: \\ &S=\sqrt{S^{2}}=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{2+4+4+5+6+6+7+8+9+9}{10}=\displaystyle \frac{60}{10}=\color{red}6 \\ &\textrm{maka nilai}\\ &S=\sqrt{\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} }\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left((2-6)^{2}+2(4-6)^{2}+(5-6)^{2}+2(6-6)^{2}+(7-6)^{2}+(8-6)^{2}+2(9-6)^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left(4^{2}+2.2^{2}+1^{2}+0+1^{2}+2^{2}+2.3^{2} \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}\left( 16+8+1+1+4+18 \right)}\\ & \quad =\sqrt{\displaystyle \frac{1}{10}(48)}=\sqrt{\displaystyle \frac{48}{10}}=\sqrt{\displaystyle \frac{120}{25}}=\color{red}\displaystyle \frac{2}{5}\sqrt{30}\\ \end{aligned} \end{array}$.
Contoh Soal 12 Statistika
$\begin{array}{ll} 51.&\textrm{Simpangan kuartil dari data}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \: \textrm{adalah}\: \: 1\displaystyle \frac{1}{2}.\: \textrm{Jika median datanya}\\ & \textrm{adalah}\: \: 5\displaystyle \frac{1}{2},\: \textrm{maka rata-rata data}\: \: \textrm{tersebut adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{4}&&&\textrm{d}.&\textrm{5}\displaystyle \frac{1}{2}\\ \textrm{b}.&\textrm{4}\displaystyle \frac{1}{2}\quad &\textrm{c}.&\textrm{5}\quad&\textrm{e}.&\textrm{6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &5\: \: 6\: \: a\: \: 3\: \: 7\: \: 8\: \Rightarrow \: n=6\\&\color{red}\textrm{karena mediannya}=M_{e}=Q_{2}=5\frac{1}{2}=\frac{11}{2}\\&\color{blue}\textrm{ data menjadi}\: :\: \color{black}a\: \: 3\: \: 5\: \: 6\: \: 7\: \: 8\: \: \color{blue}\textrm{atau}\color{black}\: \: 3\: \: a\: \: 5\: \: 6\: \: 7\: \: 8\\ &\textrm{Simpangan kuartilnya adalah}\: 1\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2}\\ &Q_{d}=\displaystyle \frac{1}{2}\left (Q_{3}-Q_{1} \right ) =\displaystyle \frac{3}{2}\: \Rightarrow\: Q_{3}-Q_{1}=3\\&\textrm{maka}\\ &3=Q_{3}-Q_{1}=x_{._{\frac{3}{4}n+\frac{1}{2}}}-x_{._{\frac{1}{4}n+\frac{1}{2}}}=x_{._{\frac{3}{4}6+\frac{1}{2}}}-x_{._{\frac{1}{4}6+\frac{1}{2}}} \\ &\quad\: =\left (x_{._{5}}-x_{._{2}} \right )=7-x_{._{2}}=\color{red}3\: \color{black}\Rightarrow\: x_{._{2}}=4\\ &\textrm{Jadi, rata-ratanya adalah}:\\ &\overline{x}=\frac{3+4+5+6+7+8}{6}=\color{red}5,5 \end{aligned} \end{array}$.
$\begin{array}{ll} 52.&\textrm{Simpangan kuartil dari data}\\ &\textrm{berikut}\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{10}&&&\textrm{d}.&\textrm{6}\\ \textrm{b}.&\textrm{9}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{-}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &61,61,50,50,53,53,70,61\\ &53,70,53,61,50,61,70\\ &\color{blue}\textrm{data durutkan}\\ &50,50,50,53,53,53,53\\ &61,61,61,61,61,70,70,70\\ &\textrm{Simpangan kuartil adalah}\: Q_{d},\\ &Q_{d}=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =\displaystyle \frac{1}{2}\left ( 61-53 \right )=\displaystyle \frac{1}{2}\times 8=4\\ &\textrm{Jadi},\: Q_{d}=4 \end{aligned} \end{array}$.
$\begin{array}{ll} 53.&\textrm{Simpangan rata-rata dari data berikut}:\\ &6\quad 4\quad 2\quad 8\quad 10\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{2}&&&\textrm{d}.&\textrm{3},0\\ \textrm{b}.&\textrm{2},4\quad &\textrm{c}.&\textrm{2},5\quad&\textrm{e}.&\textrm{3},5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6\quad 4\quad 2\quad 8\quad 10\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+4+2+8+10}{5}=\displaystyle \frac{30}{5}=\color{red}6 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 6-6 \right|+\left| 4-6 \right|+\left| 2-6 \right|+\left| 8-6 \right|+\left| 10-6 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( \left| 0 \right|+\left| -2 \right|+\left| -4 \right|+\left| 2 \right|+\left| 4 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}\left( 0+2+4+2+4 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{5}(12)=\displaystyle \frac{12}{5}=\color{red}2,4\\ \end{aligned} \end{array}$.
$\begin{array}{ll} 54.&\textrm{Simpangan rata-rata dari data berikut}:\\ &10,8,7,10,7,5,8,6,10,9\quad\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1,0}&&&\textrm{d}.&\textrm{8,0}\\ \textrm{b}.&\textrm{1,4}\quad &\textrm{c}.&\textrm{6,0}\quad&\textrm{e}.&\textrm{14,0} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &10,8,7,10,7,5,8,6,10,9\quad \\ &\textrm{Simpangan rata-ratanya adalah}: \\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{10+8+7+10+7+5+8+6+10+9}{10}\\ &\: \: \: \, =\displaystyle \frac{80}{10}=\color{red}8 \\ &\textrm{maka nilai}\\ &SR=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n}\left | x_{i}-\overline{x} \right |\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| 5-8 \right|+\left| 6-8 \right|+2\left| 7-8 \right|+2\left| 8-8 \right|+\left| 9-8 \right|+3\left| 10-8 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( \left| -3 \right|+\left| -2 \right|+2\left| -1 \right|+2\left| 0 \right|+\left| 1 \right|+3\left| 2 \right| \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}\left( 3+2+2+0+1+6 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{10}(14)=\displaystyle \frac{14}{10}=\color{red}1,4\\ \end{aligned} \end{array}$.
$\begin{array}{ll} 55.&\textrm{Nilai variansi dari data}\\ &6,7,7,8,8,8,8,12\textrm{ adalah}\: ....\\\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{1}&&&\textrm{d}.&\textrm{8}\\ \textrm{b}.&\displaystyle \frac{26}{8}\quad &\textrm{c}.&\displaystyle \frac{11}{4}\quad&\textrm{e}.&\textrm{22} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{c}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &6,7,7,8,8,8,8,12\quad \\ &\textrm{Variannya adalah}: \\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\&\textrm{dengan}\\ & \overline{x}=\displaystyle \frac{6+7+7+8+8+8+8+12}{8}=\displaystyle \frac{64}{8}=\color{red}8 \\ &\textrm{maka nilai}\\ &S^{2}=\displaystyle \frac{1}{n}\displaystyle \sum_{i=1}^{n} \left( x_{i}-\overline{x} \right)^{2} \\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( (6-8)^{2}+2(7-6)^{2}+4(8-8)^{2}+(12-8)^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 2^{2}+2.1+4.0+4^{2} \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}\left( 4+2+0+16 \right)\\ &\: \: \: \quad =\displaystyle \frac{1}{8}(22)=\color{red}\displaystyle \frac{11}{4}=\color{black}2,75\\ \end{aligned} \end{array}$.
Contoh Soal 11 Statistika
$\begin{array}{ll} 46.&\textrm{Rata-rata dari data yang disajikan }\\ &\textrm{dengan hitogram berikut adalah}\: ....\\ \end{array}$.
$.\qquad\begin{array}{ll} &\begin{array}{lllllll}\\ \textrm{a}.&41,372&&&\textrm{d}.&43,135\\ \textrm{b}.&42,150\quad&\textrm{c}.&43,125\quad&\textrm{e}.&44,250 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned} &\begin{array}{|c|c|c|c|}\hline \begin{aligned}&\textrm{Berat}\\ &\textrm{Badan} \end{aligned}&\color{red}\textrm{f}_{1}&\begin{aligned}&\textrm{Nilai Tengah}\\ &\qquad(\textrm{x}_{1}) \end{aligned}&\color{red}\textrm{f}_{1}\textrm{x}_{1}\\\hline 30-34&5&32&160\\ 35-39&7&37&259\\ 40-44&12&42&504\\ 45-49&9&47&423\\ 50-54&4&52&208\\ 55-59&3&57&171\\\hline &\sum \textrm{f}_{1}=\color{red}40&&\sum \textrm{f}_{1}\textrm{x}_{1}=\color{red}1725\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: \overline{\textrm{x}}\: \: \textrm{adalah}:\\ &\overline{\textrm{x}}=\displaystyle \frac{\sum \textrm{f}_{1}\textrm{x}_{1}}{\sum \textrm{f}_{1}}=\displaystyle \frac{1725}{40}=\color{red}43,125 \end{aligned} \end{array}$.
$\begin{array}{ll} 47.&(\textbf{UN Mat IPA 2006})\\ &\textrm{Perhatikan gambar berikut}\end{array}$.
$\begin{array}{ll} 48.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&3\\ 150-154&5\\ 155-159&17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{Kuartil bawah dapat dinyatakan dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&149,5+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{b}.&150+\left( \displaystyle \frac{12,5-3}{8} \right).5\\ \textrm{c}.&155+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{d}.&154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5\\ \textrm{e}.&155,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned} &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{Ditanyakan kuartil bawah, maka hal ini}\\ &=Q_{1}\: \Rightarrow \: x_{._{\frac{1}{4}n}}=x_{._{\frac{1}{4}50}}=x_{._{12,5}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 145-149&\color{red}3\\ 150-154&\color{red}5\\ \color{blue}155-159&\color{blue}17\\ 160-164&15\\ 165-169&8\\ 170-174&2\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{1}\: \: \textrm{adalahh}:\\ &Q_{1}=L+\left( \displaystyle \frac{\frac{1}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}154,5+\left( \displaystyle \frac{12,5-8}{17} \right).5 \end{aligned} \end{array}$.
$\begin{array}{ll} 49.&\textrm{Diketahui tabel distribusi frekuensi berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&2\\ 10-14&8\\ 15-19&10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{Median dari tabel di atas adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&15,0&&&\textrm{d}.&16,5\\ \textrm{b}.&15,5\quad&\textrm{c}.&16,0\quad&\textrm{e}.&17,0 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{e}\\ &\begin{aligned} &\textrm{Diketahui}\: n=\color{blue}30\\ &\textrm{Ditanyakan median, maka formulanya}\\ &=Q_{2}\: \Rightarrow \: x_{._{\frac{2}{4}n}}=x_{._{\frac{1}{2}.30}}=x_{._{15}}\\ &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 5-9&\color{red}2\\ 10-14&\color{red}8\\ \color{blue}15-19&\color{blue}10\\ 20-24&7\\ 25-29&3\\\hline \end{array}\\ &\textrm{maka nilai dari}\: \: Q_{2}\: \: \textrm{adalah}:\\ &Q_{2}=L+\left( \displaystyle \frac{\frac{2}{4}n-f_{k}}{f} \right).c\\ &Q_{1}=\color{red}14,5+\left( \displaystyle \frac{15-10}{10} \right).5\\ &\quad\: \: =14,5+\frac{25}{10}=14,5+2,5=\color{red}17,0 \end{aligned} \end{array}$.
$\begin{array}{ll} 50.&\textrm{Jangkauan antarkuartil dari data}\\ &\textrm{berikut}:\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{20}&&&\textrm{d}.&\textrm{5}\\ \textrm{b}.&\textrm{10}\quad &\textrm{c}.&\textrm{8}\quad&\textrm{e}.&\textrm{3} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{a}\\ &\begin{aligned} &\color{red}\textrm{Data mula-mula}\: \: \color{black}n=15\\ &36\: \: 25\: \: 56\: \: 40\: \: 55\: \: 42\: \: 43\: \: 64\\ &82\: \: 70\: \: 28\: \: 35\: \: 38\: \: 45\: \: 54\\ &\color{blue}\textrm{data durutkan}\\ &25\: \: 28\: \: 35\: \: 36\: \: 38\: \: 40\: \: 42\: \: 43\\ &45\: \: 54\: \: 55\: \: 56\: \: 64\: \: 70\: \: 82\\ &\textrm{Jangkauan antarkuartil adalah}\: H,\\ &H= Q_{3}-Q_{1} \\ &\quad\: =x_{._{\frac{3}{4}(n+1)}}-x_{._{\frac{1}{4}(n+1)}}\\ &\quad\: =x_{._{\frac{3}{4}(15+1)}}-x_{._{\frac{1}{4}(15+1)}}\\ &\quad\: =\left ( x_{._{12}}-x_{._{4}} \right )\\ &\quad\: =56-36\\ &\textrm{Jadi},\: H=Q_{3}-Q_{1}=\color{red}20 \end{aligned} \end{array}$.
Contoh Soal 10 Statistika
Soal sebelumnya (yaitu Contoh Soal 9 Statistika) klik di sini
$\begin{array}{ll} 41.&\textrm{Median dan modus dari data berikut}\\ &\color{red}3,6,7,8,4,5,9,6\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7 dan 5}&&&\textrm{d}.&\textrm{5 dan 6}\frac{1}{2}\\ \textrm{b}.&\textrm{6 dan 6} &\textrm{c}.&\textrm{6 dan 7}&\textrm{e}.&\textrm{5 dan 6} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\textrm{Diketahui}\: \: \color{blue} n=8\\ &\textrm{Datum diurutkan dari kecil ke besar}\\ &:3,4,5,\color{red}6,6\color{black},7,8,9\\ &\bullet \: \: \textbf{modus} =M_{o}=6,\\ &\bullet \: \: \textbf{mean} =\overline{x}=6 \end{aligned} \end{array}$.
$\begin{array}{ll} 42.&\textrm{Hasil tes matematika di suatu kelas yang}\\ &\textrm{diikuti tes 49 siswa menghasilkan nilai}\\ &\textrm{rata-rata 7}.\: \textrm{Jika Andi ikut tes susulan}\\ &7,04.\: \textrm{Nilai Andi adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{7,5}&&&\textrm{d}.&\textrm{9}\\ \textrm{b}.&\textrm{8} &\textrm{c}.&\textrm{8,5}&\textrm{e}.&\textrm{9,5} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}\textrm{Misalkan}&\: \: \color{red}y\color{black}=\textrm{besar nilai Andi}\\ \overline{x}_{gabungan}&=\displaystyle \frac{n.\overline{x}+\color{red}y}{n+1}\\ 7,04&=\displaystyle \frac{49\times 7+\color{red}y}{49+1}\\ 7,04&=\displaystyle \frac{343+\color{red}y}{50}\\ 343+\color{red}y&=50\times \left ( 7,04 \right )\\ 343+\color{red}y&=352\\ y&=352-343\\ &=\color{red}9 \end{aligned} \end{array}$.
$\begin{array}{ll} 43.&\textrm{Mean dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&10\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{60,5}&&&\textrm{d}.&\textrm{58,5}\\ \textrm{b}.&\textrm{60} &\textrm{c}.&\textrm{59,5}&\textrm{e}.&\textrm{57} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{d}\\ &\begin{aligned}&\begin{array}{|c|c|c|c|c|}\hline \textrm{Ukuran}&x_{i}&d_{i}=x_{i}-x_{s}&f_{i}&f_{i}\times d_{i}\\\hline 50-54&52&-10&4&-40\\\hline 55-59&57&-5&6&-30\\\hline 60-64&62&0&10&0\\\hline \textrm{Jumlah}&&&20&-70\\\hline \end{array}\\ &\overline{x}=x_{s}+\displaystyle \frac{\displaystyle \sum_{i=1}^{n}f_{i}\times d_{1}}{\displaystyle \sum_{i=0}^{n}f_{i}}=62+\displaystyle \frac{-70}{20}\\ &\: \: =62-3,5=\color{red}58,5 \end{aligned} \end{array}$.
$\begin{array}{ll} 44.&\textrm{Median dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&1\\ 50-52&6\\ 53-55&6\\ 56-58&7\\ 59-61&4\\\hline \end{array}\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{55,5}&&&\textrm{d}.&\textrm{53,5}\\ \textrm{b}.&\textrm{55} &\textrm{c}.&\textrm{54,5}&\textrm{e}.&\textrm{53} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned}&\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 47-49&\color{red}1\\ 50-52&\color{red}6\\ \color{blue}53-55&\color{blue}6\\ 56-58&7\\ 59-61&4\\\hline &24\\\hline \end{array}\\ &\textrm{Median posisi datumnya}:\\ &\textrm{datum ke}-\left ( \displaystyle \frac{24}{2} \right )=x_{._{12}}\\ &\textrm{dan terletak di interval}\\ &53-55,\: \: \textrm{dengan}\: \: f_{k}=1+6=7\\ &\textrm{serta}\: \: c=3\\ &M_{e}=Q_{2}=L_{i}+c\left ( \displaystyle \frac{\displaystyle \frac{n}{2}-f_{k}}{f} \right )\\ &M_{e}=Q_{2}=52,2+3\left ( \displaystyle \frac{\displaystyle \frac{24}{2}-7}{6} \right )\\ &\: \quad=52,5+3\left ( \displaystyle \frac{12-7}{6} \right )\\ &\: \quad=52,5+\left ( \displaystyle \frac{5}{2} \right )\\ &\: \quad=52,5+2,5\\ &\: \quad=\color{red}55 \end{aligned} \end{array}$.
$\begin{array}{ll} 45.&\textrm{Modus dari tabel berikut adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&8\\ 65-69&16\\ 70-74&10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\begin{array}{lllllll}\\ \textrm{a}.&\textrm{67,32}&&&\textrm{d}.&\textrm{70,12}\\ \textrm{b}.&\textrm{67,36} &\textrm{c}.&\textrm{67,56}&\textrm{e}.&\textrm{70,36} \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\textrm{Perhatikan tabelnya lagi}\\ &\textrm{dengan penambahan warna berikut ini}\\ &\begin{array}{|c|c|}\hline \textrm{Ukuran}&\textrm{Frekuensi}\\\hline 50-54&4\\ 55-59&6\\ 60-64&\color{red}8\\ \color{blue}65-69&\color{blue}16\\ 70-74&\color{red}10\\ 75-79&4\\ 80-84&2\\\hline \end{array}\\ &\textrm{Diketahui}\: n=\color{blue}50\\ &\textrm{modus terletak pada kelas}\\ &\textrm{interval dengan frekuensi}\\ &\textrm{terbesar, yaitu}:16.\: \textrm{Kelas intervalnya}\\ &65-69,\: \: \textrm{dengan}\: \: c=5\\ &\textrm{serta}\: \: \begin{cases} \triangle _{1} & =f-f_{1}=16-8=8 \\ \triangle _{2} & =f-f_{2}=16-10=6 \end{cases}\\ &M_{o}=L+c\left ( \displaystyle \frac{\triangle _{1}}{\triangle _{1}+\triangle _{2}} \right )\\ &\quad\: \: =64,5+5\left ( \displaystyle \frac{8}{8+6} \right )\\ &\quad\: \: =64,5+\displaystyle \frac{40}{14}\\ &\quad\: \: =64,5+2,857\\ &\quad\: \: =\color{red}67,36 \end{aligned} \end{array}$.
Contoh Soal 9 Statistika
$\begin{array}{ll}\\ 37.&(\textbf{SMBTN 2013})\\ &\textrm{Median dan rata-rata dari data yang}\\ &\textrm{terdiri dari empat bilangan asli yang}\\ &\textrm{telah diurutkan mulai dari yang terkecil}\\ &\textrm{adalah 7. Jika data tersebut tidak}\\ &\textrm{memiliki modus dan selisih antara data}\\ &\textrm{dan data terkecil adalah 8, maka hasil}\\ &\textrm{kali terbesar dari datum kedua dan}\\ &\textrm{keempat adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&39\\ \textrm{b}.&44\\ \textrm{c}.&48\\ \textrm{d}.&55\\ \color{red}\textrm{e}.&66 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui data}\: :\: \color{black}x_{1},x_{2},x_{3},\: \: \textrm{dan}\: \: x_{4}\\ &\bullet \quad\textrm{Modus}\: :\: \textrm{tidak ada}\\ &\textrm{berarti}\: \color{red}\textrm{semua datumnya berbeda}\\ &\bullet \quad \textrm{Median} \: :\: 7\\ &\textrm{maka}\: \: \displaystyle \frac{x_{2}+x_{3}}{2}=7\color{black}\Rightarrow x_{2}+x_{3}=14\: ...\color{red}(1)\\ &\bullet \quad\textrm{Rata-rata (Mean)}\: :\: 7\\ &\textrm{berarti}\: \: \displaystyle \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=7\\ &\color{black}\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}=28\: ..........\color{red}(2)\\ &\bullet \quad\textrm{Jangkauan}\: :\: x_{4}-x_{1}=8\: .....\color{red}(3)\\ &\color{black}\textrm{SUBSTITUSI}\\ &\textrm{Dari persamaan (1) ke (2) diperoleh}:\\ &x_{1}+x_{4}=14\: ...........\color{red}(5)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (3) dan (4) diperoleh}:\\ &x_{1}=3\: \: \textrm{dan}\: \: x_{4}=11\\ &\color{black}\textrm{KEMUNGKINAN}\\ &\textrm{nilai}\: x_{2}\: \textrm{dan}\: x_{3}\: \textrm{adalah}:\: 3< x_{2}\: ;\: x_{3}<11\\ &\blacklozenge \: x_{2}=4\rightarrow x_{3}=10<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=5\rightarrow x_{3}=9<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=6\rightarrow x_{3}=8<11\: \: \color{red}\checkmark\\ &\blacklozenge \: x_{2}=7\rightarrow x_{3}=7<11\: \: \color{red}\times \\ &\color{black}\textrm{KESIMPULAN}\\ &\textrm{Hasil kali terbesar}\: \: \color{red}x_{2}\times x_{3}=6\times 11=66 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 38.&(\textbf{SIMAK UI 2012})\\ &\textrm{Diketahui bahwa jika Deni mendapatkan}\\ &\textrm{nilai 75 pada ulangan yang akan datang}\\ &\textrm{maka rata-rata nilai ulangannya adalah 82.}\\ &\textrm{Jika Deni mendapatkan nilai 93, maka}\\ &\textrm{rata-rata nilai ulangannya adalah 85.}\\ &\textrm{Banyak ulangan yang telah diikuti Deni}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \color{red}\textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\color{black}\textrm{MISAL}\\ &n=\textrm{banyak ulangan yang dijalani oleh Deni}\\ &x=\textrm{Total nilai ulangannya Deni}\\ &\color{black}\textrm{MODEL MATEMATIKA}\\ &\bullet \quad \displaystyle \frac{x+75}{n+1}=82\\ &\Leftrightarrow\: x+75=82(n+1)\Leftrightarrow x=82n+82-75\\ &\Leftrightarrow \: x=82n+7\: ..................\color{red}(1)\\ &\bullet \quad \displaystyle \frac{x+93}{n+1}=85\\ &\Leftrightarrow\: x+93=85(n+1)\Leftrightarrow x=85n+85-93\\ &\Leftrightarrow \: x=85n-8\: ..................\color{red}(2)\\ &\color{black}\textrm{KESAMAAN}\\ &\textrm{Dari persamaan (1) dan (2), maka}\\ &\qquad x=x\\ &85n-8=82n+7\\ &\color{red}85n-82n=7+8\Leftrightarrow 3n=15\Leftrightarrow n=5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 39.&(\textbf{SIMAK UI})\\ &\textrm{Jika rata-rata 20 bilangan bulat nonnegatif}\\ &\textrm{berbeda adalah 20, maka bilangan terbesar}\\ &\textrm{yang mungkin adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&210\\ \textrm{b}.&229\\ \textrm{c}.&230\\ \textrm{d}.&239\\ \textrm{e}.&240 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\color{black}\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}}{20}=\color{red}20\\ &x_{1}+x_{2}+x_{3}+...+x_{18}+x_{19}+x_{20}=\color{red}400\\ &\color{black}\textrm{Yang mungkin bilangan bulat nonnegatif}\\ &\textrm{dan berbeda adalah}\: :\: \color{red}0, 1,2,3,....dst\\ &0+1+2+..+18+x_{20}=\color{red}400\\ &\displaystyle \frac{18\times 19}{2}+x_{20}=\color{red}400\\ &171+x_{20}=\color{red}400\\ &x_{20}=\color{red}400-171=229 \end{aligned} \end{array}$.
$\begin{array}{ll} 40.&\textbf{SPMB 2006}\\ &\textrm{Jika jangkauan dari data terurut}:x-1,\: 2x-1,\: 3x,\\ &5x-3,\: 4x+3,\: 6x+2\: \: \textrm{adalah}\: \: 18,\: \textrm{maka mediannya}\\ &\textrm{ adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\textrm{9}&&&\textrm{d}.&\textrm{21}\\ \textrm{b}.&\textrm{10},5\quad &\textrm{c}.&\textrm{12}\quad&\textrm{e}.&\textrm{24},8 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textrm{b}\\ &\begin{aligned} &\color{blue}\textrm{Diketahui data sebagai berikut}\\ &x-1,\: 2x-1,\: 3x,\: 5x-3,\: 4x+3,\: 6x+2\\ &\textrm{Dan diketahui pula nilai jangkauannya}\: =18\\ &J=x_{_{max}}-x_{_{min}}=18\\ &\Leftrightarrow (6x+2)-(x-1)=18\\ &\Leftrightarrow 5x+3=18\\ &\Leftrightarrow 5x=15\\ &\Leftrightarrow x=3\\ &\textrm{Sehingga datanya}:\: \color{red}2,3,9,12,15,20\\ &\textrm{Selanjutnya ditentukan mediannya}=M_{e}=Q_{_{2}}\\ &\textrm{karena}\: \: n=6,\: \textrm{datum ke}-\displaystyle \frac{2}{4}(6+1)=3,5\\ & M_{e}=Q_{2}=x_{._{3}}+0,5(x_{._{4}}-x_{._{3}})=9+0,5(12-9)\\ &\: \: \: \quad =9+0,5(3)=9+1,5=10,5\\ &\textrm{Jadi},\: J=\color{red}10,5 \end{aligned} \end{array}$.
Soal lanjutannya (yaitu Contoh Soal 10 Statistika) silahkan klik di sini
DAFTAR PUSTAKA
- Kanginan, M., Terzalgi, Y. 2013. Matematika Wajib untuk SMA/MA Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
Contoh Soal 8 Statistika
$\begin{array}{ll}\\ 36.&\textrm{Diketahui nilai statistik lima serangkai}\\ &\textrm{dari empat kelompok data seperti terlihat}\\ &\textrm{dalam tabel berikut}\\ &\begin{array}{|l|c|c|c|c|c|}\hline \textrm{Data}&\textrm{min}&Q_{1}&Q_{2}&Q_{3}&\textrm{mak}\\\hline \: \: \textrm{I}&74&80&88&92,5&99\\ \: \: \textrm{II}&66&81,5&86&90,5&96\\ \: \: \textrm{III}&70&77,5&85&92,5&100\\ \: \: \textrm{IV}&55&80&88&90&97,5\\\hline \end{array}\\ &\textrm{Data yang memuat pencilan terdapat pada}\\ &\textrm{tabel}\\ &\begin{array}{llll}\\ \textrm{a}.&\textrm{I dan II}\\ \textrm{b}.&\textrm{II dan III}\\ \textrm{c}.&\textrm{I dan III}\\ \textrm{d}.&\textrm{III dan IV}\\ \color{red}\textrm{e}.&\textrm{II dan IV} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}\: \: \color{red}\textbf{pencilan}\: \: \color{blue}\textrm{adalah datum yang}\\ \color{black}\textrm{bernil}&\textrm{ai kurang dari pagar dalam dan lebih besar}\\ \textrm{dari p}&\textrm{agar luar}\\ \textrm{Rumu}&\textrm{s pagar dalam}\: =\: Q_{1}-L\\ &=Q_{1}-\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )=\color{red}\displaystyle \frac{5}{2}Q_{1}-\frac{3}{2}Q_{3}\\ \textrm{Rumu}&\textrm{s pagar luar}\: =\: Q_{3}+L\\ &=Q_{3}+\displaystyle \frac{3}{2}\left ( Q_{3}-Q_{1} \right )=\color{red}\displaystyle \frac{5}{2}Q_{3}-\frac{3}{2}Q_{1}\\ \textrm{Data I}&\: \textrm{pagar dalamnya}=\displaystyle \frac{5}{2}(80)-\frac{3}{2}(92,5)\\ &=200-138,75=\color{black}61,25\\ \textrm{Data I}&\: \textrm{pagar luarnya}=\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(80)\\ &=231,25-120=\color{black}111,25\\ \textrm{Jadi, d}&\textrm{ata I tidak ada pencilan}\\ \textrm{Data I}&\textrm{I pagar dalamnya}=\displaystyle \frac{5}{2}(81,5)-\frac{3}{2}(90,5)\\ &=203,75-135,75=\color{black}68\\ \textrm{Data I}&\textrm{I pagar luarnya}=\displaystyle \frac{5}{2}(90,5)-\frac{3}{2}(81,5)\\ &=226,25-122,25=\color{black}104\\ \textrm{Jadi, d}&\textrm{ata II ada pencilan, yaitu}\: \: \color{red}66<\color{black}68\\ &\textrm{karena}\: \color{red}66\: \color{blue}\textrm{adalah datum terkecil data II}\\ \textrm{Data I}&\textrm{II pagar dalamnya}=\displaystyle \frac{5}{2}(77,5)-\frac{3}{2}(92,5)\\ &=193,75-138,75=\color{black}55\\ \textrm{Data I}&\textrm{II pagar luarnya}=\displaystyle \frac{5}{2}(92,5)-\frac{3}{2}(77,5)\\ &=231,25-116,25=\color{black}115\\ \textrm{Jadi, d}&\textrm{ata III tidak ada pencilan}\\ \textrm{Data I}&\textrm{V pagar dalamnya}=\displaystyle \frac{5}{2}(80)-\frac{3}{2}(90)\\ &=200-135=\color{black}65\\ \textrm{Data I}&\textrm{V pagar luarnya}=\displaystyle \frac{5}{2}(90)-\frac{3}{2}(70)\\ &=225-105=\color{black}120\\ \textrm{Jadi, d}&\textrm{ata IV ada pencilan, yaitu}\: \: \color{red}55<\color{black}65\\ &\textrm{karena}\: \color{red}55\: \color{blue}\textrm{adalah datum terkecil data IV} \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Johanes, Kastolan, & Sulasim. 2005. Kompetensi Matematika SMA Kelas 2 Semester 1 Program IPA Kurikulum Berbasis Kompetensi. Jakarta: YUDHISTIRA.
- Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung: YRAMA WIDYA.
- Sharma, dkk. 2017. Jelajah Matematika SMA Kelas XII Program Wajib. Jakarta: YUDHISTIRA.
- Tim Supermat. 2007. Cara Mudah MenghadapiS SMBB TELKOM. Jakarta: LITERATUR MEDIA SUKSES.
Contoh Soal 7 Statistika
Contoh Soal 6 Statistika
$\begin{array}{ll}\\ 26.&\textrm{Jika rata-rata dari}\: \: x_{1},x_{2},x_{3},x_{4},...,x_{10}\\ &\textrm{adalah}\: \: x_{0},\: \textrm{maka rata-rata dari data}\\ &(x_{1}-1),(x_{2}+2),(x_{3}-3),(x_{4}+4),..\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&x_{0}+5,5\\ \textrm{b}.&x_{0}+25\\ \color{red}\textrm{c}.&x_{0}+0,5\\ \textrm{d}.&x_{0}-0,5\\ \textrm{e}.&x_{0}-2,5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Rata}-&\textrm{ratanya adalah}:\\ \overline{x}=x_{0}&=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{10}}{10}\\ \Leftrightarrow 10x_{0}&=x_{1}+x_{2}+x_{3}+...+x_{10}\\ \color{black}\textrm{Selanju}&\color{black}\textrm{tnya penghitungan rata-rata yang data baru},\\ \overline{x}_{baru}&=\displaystyle \frac{(x_{1}-\color{red}1)\color{blue}+(x_{2}+\color{red}2)\color{blue}+(x_{3}-\color{red}3)\color{blue}+...+(x_{10}+\color{red}10)}{10}\\ &=\displaystyle \frac{x_{1}+x_{2}+...+x_{10}+\color{red}(2-1+4-3+..+10-9)}{10}\\ &=\displaystyle \frac{10x_{0}+5}{10}\\ &=\color{red}x_{0}+0,5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 27.&\textrm{Nilai rata-rata dari 20 bilangan adalah}\: \: 14,2\\ &\textrm{Jika rata-rata dari 12 bilangan pertama adalah}\\ &\textrm{12,6 dan rata-rata dari 6 bilangan berikutnya}\\ &\textrm{adalah 18,2, rata-rata 2 bilangan tersisa}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&10,4\\ \color{red}\textrm{b}.&11,8\\ \textrm{c}.&12,2\\ \textrm{d}.&12,8\\ \textrm{e}.&13,8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\overline{x}_{total}&=\displaystyle \frac{\overline{x}_{12}\times (12)+\overline{x}_{6}\times (6)+\overline{x}_{2}\times (2)}{20}\\ 14,2&=\displaystyle \frac{(12,6\times 12)+(18,2\times 6)+\overline{x}_{2}\times (2)}{20}\\ 284&=151,2+109,2+2\overline{x}_{2}\\ 2\overline{x}_{2}&=284-(151,2+109,2)=\color{red}23,6\\ \overline{x}_{2}&=\displaystyle \frac{23,6}{2}\\ &=\color{red}11,8 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 28.&\textrm{Dari 3 bilangan yang terkecil adalah 39}\\ &\textrm{dan terbesarnya adalah 75, maka rata-rata}\\ &\textrm{hitung ketiga bilangan tersebut tidak}\\ &\textrm{mungkin sama dengan}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&49\\ \textrm{b}.&52\\ \textrm{c}.&53\\ \textrm{d}.&59\\ \textrm{e}.&60 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{3}&=\displaystyle \frac{39+\color{red}A\color{blue}+75}{3}\\ \textrm{Se}&\textrm{lanjutnya rentang nilai}\: \: A\: \: \textrm{akan berada di}\\ :\: &39\leq \color{red}A\color{blue}\leq 75\\ \textrm{Se}&\textrm{hingga},\\ \textrm{un}&\textrm{tuk}\: \: \color{red}A=39,\: \: \color{blue}\textrm{maka}\\ \overline{x}_{3}&=\displaystyle \frac{39+39+75}{3}=\frac{153}{3}=\color{red}51,\: \: \color{black}\textrm{dan}\\ \textrm{un}&\textrm{tuk}\: \: \color{red}A=75,\: \: \color{blue}\textrm{maka}\\ \overline{x}_{3}&=\displaystyle \frac{39+75+75}{3}=\frac{189}{3}=\color{red}63 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 29.&(\textbf{SPMB 04})\\ &\textrm{Nilai rata-rata tes Matematika dari kelompok}\\ &\textrm{siswa dan kelompok siswi di suatu kelas berturut-}\\ &\textrm{turut adalah 5 dan 7. Jika nilai rata-rata di kelas}\\ &\textrm{tersebut adalah 6,2, maka perbandingan banyaknya}\\ &\textrm{siswa dan siswi adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2:3\\ \textrm{b}.&3:4\\ \textrm{c}.&2:5\\ \textrm{d}.&3:5\\ \textrm{e}.&4:5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\overline{x}_{\color{red}gabungan}&=\displaystyle \frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}\\ \color{black}6,2&\color{blue}=\displaystyle \frac{n_{1}(5)+n_{2}(7)}{n_{1}+n_{2}}\\ 6,2(n_{1}+n_{2})&=5n_{1}+7n_{2}\\ 6,2n_{1}-5n_{1}&=7n_{2}-6,2n_{2}\\ 1,2n_{1}&=0,8n_{2}\\ \displaystyle \frac{n_{1}}{n_{2}}&=\displaystyle \frac{0,8}{1,2}=\color{red}\frac{2}{3} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 30.&(\textbf{SPMB 05})\\ &\textrm{Nilai rata-rata ulangan kelas A adalah}\: \: \overline{x}_{A}\: \: \textrm{dan}\\ &\textrm{kelas B adalah}\: \: \overline{x}_{B}.\: \textrm{Setelah kedua kelas digabungkan}\\ &\textrm{nilai rata-ratanya adalah}\: \: \overline{x}.\: \textrm{Perbandingan nilai}\\ &\textrm{kelas A dan B adalah}\: \: 10:9.\: \textrm{Jika perbandingan nilai}\\ &\textrm{rata-rata kedua kelas dan kelas B adalah}\: \: 85:81,\\ &\textrm{maka perbandinganbanyaknya siswa kelas A dan B}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&8:9\\ \color{red}\textrm{b}.&4:5\\ \textrm{c}.&3:4\\ \textrm{d}.&3:5\\ \textrm{e}.&9:10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\overline{x}_{A}:\overline{x}_{B}&=10:9=90:81\\ \overline{x}:\overline{x}_{B}&=85:81,\: \textrm{maka}\\ \color{red}\overline{x}:\overline{x}_{A}:\overline{x}_{B}&=\color{red}85:90:81\\ (n_{A}+n_{B})\overline{x}&=n_{A}\times \overline{x}_{A}+n_{B}\times \overline{x}_{B}\\ \displaystyle \frac{n_{A}}{n_{B}}&=\displaystyle \frac{\overline{x}-\overline{x}_{B}}{\overline{x}_{A}-\overline{x}}\\ &=\displaystyle \frac{\displaystyle \frac{85}{81}\overline{x}_{B}-\overline{x}_{B}}{\displaystyle \frac{10}{9}\overline{x}_{B}-\displaystyle \frac{85}{81}\overline{x}_{B}}\\ &=\displaystyle \frac{\displaystyle \frac{4}{81}\overline{x}_{B}}{\displaystyle \frac{5}{81}\overline{x}_{B}}\\ &=\color{red}\displaystyle \frac{4}{5} \end{aligned} \end{array}$
Contoh Soal 5 Statistika
$\begin{array}{ll}\\ 21.&\textrm{Simpangan kuartil dari data}\\ &71,70,68,40,45,48,52,53,53,67,62\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&8\\ \color{red}\textrm{b}.&10\\ \textrm{c}.&15\\ \textrm{d}.&18\\ \textrm{e}.&20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\: (\color{red}\textrm{degan total datum ganjil})\\ :\: &71,70,68,40,45,48,52,53,53,67,62\\ \textrm{Sete}&\textrm{lah data diurutkan menjadi}\\ :\: &\color{cyan}40,45,48,52,53,53,62,67,68,70,71\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=11\: \: \color{black}\textbf{ganjil}\\ Q_{1}&=x_{\frac{1}{4}(n+1)}=x_{\frac{1}{4}.12}=x_{3}=\color{red}48\\ Q_{2}&=x_{\frac{2}{4}(n+1)}=x_{\frac{2}{4}.12}=x_{6}=53\\ Q_{3}&=x_{\frac{3}{4}(n+1)}=x_{\frac{3}{4}.12}=x_{9}=\color{red}68\\ \textrm{Sela}&\textrm{njutnya data dapat dituliskan}\\ &40,45,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{1} \end{matrix}}{\underbrace{\color{red}48}},52,53,53,62,67,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{3} \end{matrix}}{\underbrace{\color{red}68}},70,71\\ \textrm{Simp}&\textrm{angan kuartil data tunggal adalah}:\\ &=\displaystyle \frac{1}{2}\left ( Q_{3}-Q_{1} \right )\\ &=\displaystyle \frac{1}{2}(68-48)\\ &=\displaystyle \frac{1}{2}.20=\color{red}10 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 22.&\textrm{Data penjualan suatu barang setiap bulan}\\ &\textrm{di sebuah toko pada tahun 2019 adalah}:\\ &20,3,9,11,4,12,1,9,9,12,8,10.\\ &\textrm{Median, kuartil bawah, dan kuartil atasnya}\\ &\textrm{berturut-turut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&6\displaystyle \frac{1}{2},3\frac{1}{2},\: \textrm{dan}\: \: 9\frac{1}{2}\\ \color{red}\textrm{b}.&9,6,\: \textrm{dan}\: \: 11\displaystyle \frac{1}{2}\\ \textrm{c}.&6\displaystyle \frac{1}{2},9,\: \textrm{dan}\: \: 12\\ \textrm{d}.&9,4,\: \textrm{dan}\: \: 12\\ \textrm{e}.&9,3\displaystyle \frac{1}{2},\: \textrm{dan}\: \: 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &20,3,9,11,4,12,1,9,9,12,8,10\\ \textrm{Sete}&\textrm{lah data diurutkan}\\ :\: &1,3,4,8,9,9,9,10,11,12,12,20\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=12\: \: \color{black}\textbf{genap}\\ Q_{1}&=x_{\frac{1}{4}n+\frac{1}{2}}=x_{\frac{1}{4}.12+\frac{1}{2}}=x_{3,5}=\color{red}6\\ Q_{2}&=x_{\frac{2}{4}n+\frac{1}{2}}=x_{\frac{2}{4}.12+\frac{1}{2}}=x_{6,5}=\color{red}9=\color{black}M_{e}\\ Q_{3}&=x_{\frac{3}{4}n+\frac{1}{2}}=x_{\frac{3}{4}.12+\frac{1}{2}}=x_{9,5}=\color{red}11\displaystyle \frac{1}{2}\\ \textrm{Sela}&\textrm{njutnya data dapat dituliskan}\\ &1,3,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{1} \end{matrix}}{\underbrace{\color{red}4,8}},9,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{2}=M_{e} \end{matrix}}{\underbrace{\color{red}9,9}},10,\underset{\begin{matrix} \Downarrow\\ \color{black}Q_{3} \end{matrix}}{\underbrace{\color{red}11,12}},12,20\\ \end{aligned} \end{array}$
$\begin{array}{ll}\\ 23.&\textrm{Ragam(varians) dari data}\\ &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&1\\ \textrm{b}.&1\displaystyle \frac{3}{8}\\ \textrm{c}.&1\displaystyle \frac{1}{8}\\ \textrm{d}.&\displaystyle \frac{7}{8}\\ \textrm{e}.&\displaystyle \frac{5}{8} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &6,8,6,7,8,7,9,7,7,6,7,8,6,5,8,7\\ \textrm{Sete}&\textrm{lah data diurutkan(untuk memudahkan)}\\ :\: &5,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=16.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ \overline{x}&=\displaystyle \frac{5.1+6.4+7.6+8.4+9.1}{16}\\ &=\frac{112}{16}=7\\ \textrm{Dan}\: &\textrm{rumus untuk menghitung ragam adalah}:\\ S^{2}&=\displaystyle \frac{\displaystyle \sum_{i=1}^{16}\left ( x_{i}-\overline{x} \right )^{2}}{n}\\ &=\displaystyle \frac{(5-7)^{2}+4(6-7)^{2}+6(7-7)^{2}+4(8-7)^{2}+(9-7)^{2}}{16}\\ &=\displaystyle \frac{4+4.1+6.0+4.1+4}{16}\\ &=\displaystyle \frac{16}{16}\\ &=\color{red}1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 24.&\textrm{Diketahui}\: \: x_{1}=2,\: x_{2}=3,5,\: x_{3}=5,\: x_{4}=7,\\ &\textrm{dan}\: \: x_{5}=7,5.\: \textrm{Deviasi rata-rata data di atas}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \color{red}\textrm{c}.&1,8\\ \textrm{d}.&2,6\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &2,3\displaystyle \frac{1}{2},5,7,7\displaystyle \frac{1}{2}\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=5.\: \: \color{black}\textrm{Selanjutnya kita cari}\\ \overline{x}&=\displaystyle \frac{2+3,5+5+7+7,5}{5}=\frac{25}{5}=5\\ \textrm{Dan}\: &\textrm{rumus simpangan rata-rata adalah}:\\ SR&=\displaystyle \frac{\displaystyle \sum_{i=1}^{5} \left |x_{i}-\overline{x} \right |}{n}\\ &=\displaystyle \frac{\left | 2-5 \right |+\left | 3,5-5 \right |+\left | 5-5 \right |+\left | 7-5 \right |+\left | 7,5-5 \right |}{5}\\ &=\displaystyle \frac{\left | -3 \right |+\left | -1,5 \right |+\left | 0 \right |+\left | 2 \right |+\left | 2,5 \right |}{5}\\ &=\displaystyle \frac{3+1,5+0+2+2,5}{5}=\frac{9}{5}\\ &=\color{red}1,8 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 25.&\textrm{Jumlah rataan dan median dari}\\ &(x-6),(x+5),(x+4),(x-7),\\ &(x+9),\: \textrm{dan}\: \: (x-2)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2x-1,5\\ \textrm{b}.&2x-0,5\\ \color{red}\textrm{c}.&2x+1,5\\ \textrm{d}.&2x+2,5\\ \textrm{e}.&2x+3,5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Data}&\: \textrm{mula-mula}\\ :\: &(x-6),(x+5),(x+4),(x-7),(x+9),(x-2)\\ \color{black}\textrm{Dike}&\color{black}\textrm{tahui}\: \: \color{red}n=6.\: \: \color{black}\textrm{Selanjutnya kita urutkan datanya}\\ &(x-7),(x-6),(x-2),(x+4),(x+5),(x+9)\\ \textrm{Rata}&\textrm{annya}:\\ \overline{x}&=\displaystyle \frac{6x+3}{6}=\color{red}x+0,5\\ \textrm{Medi}&\textrm{annya}:\\ M_{e}&=x_{3,4}=\displaystyle \frac{x_{3}+x_{4}}{2}=\frac{(x-2)+(x+4)}{2}\\ &=\displaystyle \frac{2x+2}{2}=\color{red}x+1\\ \textrm{Rata}&\textrm{an+median}=\color{black}x+0,5+x+1=\color{red}2x+1,5 \end{aligned} \end{array}$
Contoh Soal 4 Statistika
$\begin{array}{ll}\\ 16.&(\textbf{UN IPA 2014})\\ &\textrm{Kuartil atas dari data pada tabel berikut}\\ &\textrm{adalah}\: ....\\ &\begin{array}{|c|c|}\hline \textrm{Data}&f\\\hline 20-25&4\\\hline 26-31&6\\\hline 32-37&6\\\hline 38-43&10\\\hline 44-49&12\\\hline 50-55&8\\\hline 56-61&4\\\hline \end{array}\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&49,25\\ \textrm{b}.&48,75\\ \textrm{c}.&48,25\\ \textrm{d}.&47,75\\ \textrm{e}.&47,25 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\textrm{Kuartil atas}=\color{black}Q_{3},\: \textrm{dengan}\: \: n=\sum f=50\\ &\textrm{Kita sertakan lagi tabel di atas berikut}\\ &\begin{array}{|c|c|}\hline \color{black}\textrm{Data}&\color{red}f\\\hline 20-25&4\\\hline 26-31&6\\\hline 32-37&6\\\hline 38-43&10\\\hline \colorbox{magenta}{44-49}&\colorbox{magenta}{12}\\\hline 50-55&8\\\hline 56-61&4\\\hline \end{array}\\ &Q_{3}=\textrm{Datum ke}-\left ( \displaystyle \frac{3n}{4} \right )=x_{\frac{3.50}{4}}=x_{37,5}\\ &\textrm{dan}\: \: x_{37,5}\: \: \textrm{terletak di kelas interval}\: \: 44-49\\ &Q_{3}=t_{b}+p\left ( \displaystyle \frac{\displaystyle \frac{3n}{4}-F}{f} \right )\\ &\: \: \: \, =43,5+6\left ( \displaystyle \frac{37,5-26}{12} \right )\\ &\: \: \: \, =43,5+\displaystyle \frac{11,5}{2}\\ &\: \: \: \, =49,5+5,75\\ &\: \: \: \, =\color{red}49,25 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 17.&(\textbf{UN IPA 2014})\\ &\textrm{Perhatikanlah histrogram berikut}\\ & \end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Median dari data}\\ &3,4,7,5,6,9,9,7,6, 5,8\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&5\\ \color{red}\textrm{b}.&6\\ \textrm{c}.&7\\ \textrm{d}.&8\\ \textrm{e}.&9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Diketah}&\textrm{ui data adalah ganjil}\\ \textrm{Median}&\: \textrm{ (datum tengah) data tunggal}:\\ \textrm{Data}&:\: \: \color{black}3,4,7,5,6,9,9,7,6, 5,8\\ \textrm{setel}&\textrm{ah diurutkan}\\ \textrm{Data}&:\: \: \color{red}3,4,5,5,6,6,7,7,8,9,9\\ \textrm{Data}&:\: \: \color{red}3,4,5,5,6,\color{blue}6,\color{red}7,7,8,9,9 \end{aligned}\end{array}$
$\begin{array}{ll}\\ 19.&\textrm{Dari data berikut yang memiliki}\\ &\textbf{mean}\: \: 7\displaystyle \frac{1}{2}\: \: \textrm{dan}\: \textbf{median}\: 7\: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&2,5,6,9,7,8,5,14,8,11\\ \textrm{b}.&6,3,7,8,6,4,11,8,9,8\\ \textrm{c}.&3,7,10,7,9,5,10,2,14,11\\ \textrm{d}.&4,1,6,12,8,11,4,5,8,2\\ \color{red}\textrm{e}.&2,3,4,3, 10,8,12,6,15,12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{array}{|l|l|}\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \textrm{Median}&2,5,6,9,7,8,5,14,8,11\\\hline \quad \textrm{a}&\color{blue}2,5,5,6,\color{red}7,8,\color{blue}8,9,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{70}{10}=7\\\hline \textrm{Median}&6,3,7,8,6,4,11,8,9,8\\\hline \quad \textrm{b}&\color{blue}3,4,6,6,\color{red}7,8,\color{blue}8,8,9,11\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{78}{10}=7,8\\\hline \textrm{Median}&3,7,10,7,9,5,10,2,14,11\\\hline \quad \textrm{c}&\color{blue}2,3,5,7,\color{red}7,9,\color{blue}10,10,11,14\\\hline \textrm{Mean}&\overline{x}=\displaystyle \frac{61}{10}=6,1\\\hline \textrm{Median}&4,1,6,12,8,11,4,5,8,2\\\hline \quad \textrm{d}&\color{blue}1,2,4,4,\color{red}5,6,\color{blue}8,8,11,12\\\hline \color{blue}\textrm{Mean}&\overline{x}=\displaystyle \frac{75}{10}=7,5\\\hline \color{blue}\textrm{Median}&2,3,4,3, 10,8,12,6,15,12\\\hline \quad \color{red}\textrm{e}&\color{blue}2,3,3,4,\color{red}6,8,\color{blue}10,12,12,15\\\hline \end{array} \end{array}$
$\begin{array}{ll}\\ 20.&\textrm{Berikut adalah daftar nilai matematika}\\ &\textrm{kelas XII IA1}\\ &\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \textrm{Nilai}&3&4&5&6&7&8&9&10\\\hline \textrm{Frekuensi}&3&5&5&9&8&6&2&2\\\hline \end{array}\\ &\textrm{Jika siswa yang nilainya di atas rata-rata}\\ &\textrm{akan diikutsertakan dalam seleksi}\\ &\textrm{olimpiade matematika, maka banyak}\\ & \textrm{siswa yang mengikuti seleksi olimpiade}\\ &\textrm{adalah}\: ...\: \textrm{siswa}\\ &\begin{array}{llll}\\ \textrm{a}.&6\\ \textrm{b}.&8\\ \textrm{c}.&10\\ \color{red}\textrm{d}.&18\\ \textrm{e}.&27 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\begin{array}{|l|c|c|c|c|c|c|c|c|}\hline x_{i}&3&4&5&6&7&8&9&10\\\hline f_{i}&3&5&5&9&8&6&2&2\\\hline x_{i}f_{i}&9&20&25&54&56&48&18&20\\\hline \end{array}\\ &\color{blue}\textrm{Rata-rata nilai matematikanya}\\ &\color{magenta}\overline{x}=\displaystyle \frac{\sum x_{i}f_{i}}{\sum f_{i}}\\ &=\displaystyle \frac{250}{40}=6,25\\ &\color{blue}\textrm{Jadi, nilai rata-ratanya}\: \: \color{red}6,25\\ &\textrm{Sehingga yang bisa ikut selesksi adalah}\\ &\textrm{nilai di atas rata-rata yaitu}:7,8,9,10\\ &\textrm{dan totalnya yang mendapatkan nilai}\\ &\textrm{itu sebanyak}:\: \color{blue}8+6+2+2=\color{red}18\: \: \color{black}\textrm{siswa} \end{aligned} \end{array}$
Contoh Soal 3 Statistika
$\begin{array}{ll}\\ 11.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 2-6&2\\\hline 7-11&3\\\hline 12-16&3\\\hline 17-21&6\\\hline 22-26&6\\\hline \end{array}\\ &\textrm{Rata-ratanya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&17,5\\ \textrm{b}.&17\\ \color{red}\textrm{c}.&16,75\\ \textrm{d}.&16,5\\ \textrm{e}.&15,5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\begin{array}{|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&\color{red}f_{i}&\color{black}x_{i}.f_{i}\\\hline 2-6&4&2&8\\\hline 7-11&9&3&27\\\hline 12-16&14&3&42\\\hline 17-21&19&6&114\\\hline 22-26&24&6&144\\\hline &\color{black}\sum_{i=1}^{5}&\color{red}20&\color{black}335\\\hline \end{array}\\ &\textrm{maka rata-ratanya}\\ &\overline{x}=\displaystyle \frac{\sum x_{i}.f_{i}}{\sum f_{i}}=\frac{335}{20}=16,75 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 12.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 51-60&5\\\hline 61-70&7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline \end{array}\\ &\textrm{Rata-ratanya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75,50\\ \color{red}\textrm{b}.&76,25\\ \textrm{c}.&76,50\\ \textrm{d}.&78,25\\ \textrm{e}.&80,50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\begin{array}{|c|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}x_{i}&\color{red}f_{i}&\color{black}x_{i}.f_{i}\\\hline 51-60&55,5&5&277,5\\\hline 61-70&65,5&7&458,5\\\hline 71-80&75,5&14&1057\\\hline 81-90&85,5&8&684\\\hline 91-100&95,5&6&573\\\hline &\color{black}\sum_{i=1}^{5}&\color{red}40&\color{black}3050\\\hline \end{array}\\ &\textrm{maka rata-ratanya}\\ &\overline{x}=\displaystyle \frac{\sum x_{i}.f_{i}}{\sum f_{i}}=\frac{3050}{40}=76,25 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 13.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut(data sama dengan no.12 di atas)}\\ &\begin{array}{|c|c|}\hline \textrm{Interval}&f\\\hline 51-60&5\\\hline 61-70&7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline \end{array}\\ &\textrm{Mediannya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&75,50\\ \textrm{b}.&76,20\\ \color{red}\textrm{c}.&76,21\\ \textrm{d}.&77,22\\ \textrm{e}.&78,23 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui},\: n=\sum f=40,\: \: \textrm{Perhatikan tabel}\\ &\textrm{berikut ini}\\ &\begin{array}{|c|c|c|}\hline \color{black}\textrm{Interval}&\color{black}f_{i}\\\hline 51-60&\color{red}5\\\hline 61-70&\color{red}7\\\hline 71-80&14\\\hline 81-90&8\\\hline 91-100&6\\\hline &\color{red}40\\\hline \end{array}\\ &Q_{k}=\textrm{Datum ke}-\left ( \displaystyle \frac{kn}{4} \right )\\ &\color{red}Median=Q_{2}=\textrm{Datum ke}-\left ( \displaystyle \frac{2.40}{4} \right )=x_{20}\\ &x_{20}\: \: \textrm{terletak pada kelas interval}:\: \: 71-80\\ &\textrm{dengan}\: \: f=14,\: \: F\: \: \textrm{sebelum}\: \: Q_{2}=12,\\ &t_{b}=70,5,\: \: \textrm{serta}\: \: p=10\\ &\textrm{maka mediannya}\\ &Q_{2}=t_{b}+p\left ( \displaystyle \frac{\displaystyle \frac{2n}{4}-F}{f} \right )\\ &=70,5+10\left ( \displaystyle \frac{20-12}{14} \right )\\ &=70,5+5,714=76,21 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 14.&\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Skor}&f\\\hline 40-49&8\\\hline 50-59&9\\\hline 60-69&22\\\hline 70-79&15\\\hline 80-89&6\\\hline \end{array}\\ &\textrm{Modusnya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&65,50\\ \color{red}\textrm{b}.&66,00\\ \textrm{c}.&66,50\\ \textrm{d}.&67,00\\ \textrm{e}.&85,50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Diketahui}\: \: n=\sum f=60,\: \: \color{red}modusnya\\ &\textrm{terdapat pada kelas dengan frekuensi terbanyak}\\ & \textrm{yaitu}:\: 60-69,\: \: \textrm{dengan}\: \: p=10\\ &\begin{cases} \triangle f_{i} & =f-f_{1}=22-9=13 \\ \triangle f_{ii} & =f-f_{2}=22-15=7 \end{cases}\\ &\textrm{Sehingga}\\ &M_{0}=t_{b}+p\left ( \displaystyle \frac{\triangle f_{1}}{\triangle f_{1}+\triangle f_{2}} \right )\\ &M_{0}=59,5+10\left ( \displaystyle \frac{22-9}{(22-9)+(22-15)} \right )\\ &\: \: \: =59,5+\displaystyle \frac{10.13}{13+7}\\ &\: \: \: =\color{black}59,5+6,5=\color{red}66,0 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 15.&(\textbf{UN 2013})\\ &\textrm{Perhatikan tabel distribusi frekuensi}\\ &\textrm{berikut}\\ &\begin{array}{|c|c|}\hline \textrm{Berat Badan (Kg)}&f\\\hline 45-49&3\\\hline 50-54&6\\\hline 55-59&10\\\hline 60-64&12\\\hline 65-69&15\\\hline 70-74&6\\\hline 75-79&4\\\hline \end{array}\\ &\textrm{Kuartil atasnya adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&66\displaystyle \frac{5}{6}\\ \textrm{b}.&67\displaystyle \frac{1}{6}\\ \textrm{c}.&67\displaystyle \frac{5}{6}\\ \color{red}\textrm{d}.&68\displaystyle \frac{1}{6}\\ \textrm{e}.&68\displaystyle \frac{4}{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Kuartil atas}=\color{black}Q_{3},\: \: \textrm{dengan}\: \: n=\sum f=56\\ &Q_{3}=\textrm{Datum ke}-\left ( \displaystyle \frac{3n}{4} \right )=x_{\frac{3.56}{4}}=x_{42}\\ &\textrm{dan}\: \: x_{42}\: \: \textrm{terletak di kelas interval}\: \: 65-69\\ &Q_{3}=t_{b}+p\left ( \displaystyle \frac{\displaystyle \frac{3n}{4}-F}{f} \right )\\ &\: \: \: =64,5+5\left ( \displaystyle \frac{42-(3+6+10+12)}{15} \right )\\ &\: \: \: =64\displaystyle \frac{1}{2}+\displaystyle \frac{11}{3}\\ &\: \: \: =\color{purple}64\displaystyle \frac{3}{6}+3\displaystyle \frac{4}{6}=\color{red}68\displaystyle \frac{1}{6} \end{aligned} \end{array}$
Contoh Soal 2 Statistika
$\begin{array}{ll}\\ 6.&\textrm{Tabel berikut adalah nilai tes matematika}\\ &\qquad\qquad \begin{array}{|c|c|}\hline \textrm{Nilai}&f\\\hline 21-30&1\\\hline 31-40&2\\\hline 41-50&5\\\hline 51-60&7\\\hline 61-70&8\\\hline 71-80&8\\\hline 81-90&6\\\hline 91-100&1\\\hline \end{array}\\ &\textrm{Banyak siswa yang mendapatkan nilai 71}\\ &\textrm{atau lebih adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&16\\ \color{red}\textrm{b}.&15\\ \textrm{c}.&12\\ \textrm{d}.&12\\ \textrm{e}.&10 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}&\textrm{Perhatikan kembali tabelnya}\\ &\qquad\qquad \color{black}\begin{array}{|c|c|}\hline \textrm{Nilai}&f\\\hline 21-30&1\\\hline 31-40&2\\\hline 41-50&5\\\hline 51-60&7\\\hline 61-70&8\\\hline \color{blue}71-80&\color{red}8\\\hline \color{blue}81-90&\color{red}6\\\hline \color{blue}91-100&\color{red}1\\\hline \end{array}\\ &\textrm{Nilai yang lebih dari 71 adalah}:\\ &\color{blue}8+6+1=17 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 7.&\textrm{Jangkauan dari tabel distribusi}\\ &\textrm{frekuensi pada no.6 di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&60\\ \color{red}\textrm{b}.&70\\ \textrm{c}.&79\\ \textrm{d}.&89\\ \textrm{e}.&100 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}&\textrm{Karena data berkelompok, maka}\\ &\textrm{Jangkauan}=\left ( \textrm{Nilai tengah kelas tertinggi} \right )\\ &\qquad\quad - \left ( \textrm{Nilai tengah kelas pertama} \right )\\ &=\displaystyle \frac{1}{2}\left ( (100+91)-(30+21) \right )=70 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: n=\textrm{banyak data},\: k=\textrm{banyak interval}\\ &\textrm{kelas, maka menurut aturan Sturges, rumus}\\ &\textrm{untuk menentukan nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&k=\log \left (10.n^{3,3} \right )\\ \color{red}\textrm{b}.&k=1+3,3\log n\\ \textrm{c}.&k=1-3,3\log (n-1)\\ \textrm{d}.&k=\log \left ( 10^{3,3}.n \right )\\ \textrm{e}.&k=\log n^{3,3}+2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 9.&\textrm{Rata-rata data soal no.6}\\ &\textrm{di atas adalah}\: ....\\ &\begin{array}{ll}\\ \color{red}\textrm{a}&\displaystyle 64,45\\ \textrm{b}&\displaystyle 64,55\\ \textrm{c}&\displaystyle 65,45\\ \textrm{d}&\displaystyle 65,55\\ \textrm{e}&\displaystyle 66\\ \end{array} \\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&\begin{array}{|c|c|c|c|}\hline \color{black}\textrm{Nilai}&\color{black}x_{i}&\color{red}f_{i}&\color{black}x_{i}.f_{i}\\\hline 21-30&25,5&1&25,5\\\hline 31-40&35,5&2&71\\\hline 41-50&45,5&5&227,5\\\hline 51-60&55,5&7&388,5\\\hline 61-70&65,5&8&524\\\hline 71-80&75,5&8&604\\\hline 81-90&85,5&6&513\\\hline 91-100&95,5&1&95,5\\\hline &\color{black}\sum_{i=1}^{8}&\color{red}38&\color{black}2449\\\hline \end{array}\\ &\textrm{Sehingga, rata-rata data nilai di atas adalah}\\ &\overline{x}=\displaystyle \frac{\sum x_{i}f_{i}}{\sum f_{i}}=\frac{2449}{38}=64,447368421\approx 64,45 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 10.&\textrm{Rumus untuk menentukan}\: \: \%f_{\textrm{rel}}=....\\ &\begin{array}{ll}\\ \textrm{a}&\displaystyle \frac{f_{i}+1}{\sum f}\times 100\%\\ \textrm{b}&\displaystyle \frac{f_{i}-1}{\sum f}\times 100\%\\ \color{red}\textrm{c}&\displaystyle \frac{f_{i}}{\sum f}\times 100\%\\ \textrm{d}&\displaystyle \frac{f_{i}}{\sum f +1}\times 100\%\\ \textrm{e}&\displaystyle \frac{f_{i}}{\sum f -1}\times 100\%\\ \end{array} \\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Cukup jelas} \end{aligned} \end{array}$
Contoh Soal 1 Statistika
$\begin{aligned}&\textrm{Perhatikanlah tabel berikut}\\ &\textrm{Nilai ulangan kelas XII}\\ &\textrm{untuk menjawab soal no. 1 sampai 5}\\ &\begin{array}{|c|c|c|}\hline \colorbox{magenta}{Nilai}&\colorbox{yellow}{Matematika}&\colorbox{cyan}{Bahasa Inggris}\\\hline 30-39&1&0\\\hline 40-49&4&2\\\hline 50-59&6&7\\\hline 60-69&17&18\\\hline 70-79&10&12\\\hline 80-89&7&6\\\hline \end{array} \end{aligned}$
$\begin{array}{ll}\\ 1.&\textrm{Total datum pada data tabel}\\ &\textrm{distribusi frekuensi di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&30\\ \textrm{b}.&35\\ \textrm{c}.&40\\ \color{red}\textrm{d}.&45\\ \textrm{e}.&50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Total datum pada tabel di atas}\\ &\textrm{sama dengan total frekuensi yaitu}: \\ &=1+4+6+17+10+7=\color{black}45,\\ &\color{red}\textrm{atau}\\ &=0+2+7+18+12+6=\color{black}45 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 2.&\textrm{Banyak kelas interval pada tabel}\\ &\textrm{distribusi frekuensi tersebut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&5\\ \color{red}\textrm{c}.&6\\ \textrm{d}.&7\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Data di atas terbagai dalam 6}\\ &\textrm{kelas intervalnya, yaitu}:\\ &\textrm{kelas pertama : 30-39}\\ &\textrm{kelas kedua : 40-49}\\ &\textrm{kelas ketiga : 50-59}\\ &\textrm{kelas keempat : 60-69}\\ &\textrm{kelas kelima : 70-79}\\ &\textrm{kelas keenam : 80-89}\\ \end{aligned} \end{array}$
$\begin{array}{ll}\\ 3.&\textrm{Panjang kelas interval pada tabel}\\ &\textrm{di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9\\ \color{red}\textrm{b}.&10\\ \textrm{c}.&11\\ \textrm{d}.&12\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Panjnag kelas interval pada}\\ &\textrm{tabel distribusi frekuensi di atas} \\ &\textrm{ambil contoh kelas pertama yaitu}:\\ &\textrm{pada}\: \: 30-39\: \: \textrm{ada}\\ &\color{red}=(39-30)+1=10 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 4.&\textrm{Titik tengah dari kelas interval ke enam}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&84\\ \color{red}\textrm{b}.&84,5\\ \textrm{c}.&85\\ \textrm{d}.&85,5\\ \textrm{e}.&86 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Titik tengah interval kelas keenam}\\ &\textrm{pada tabel distribusi frekuensi di atas} \\ &\textrm{yaitu}:\\ &\color{red}=\displaystyle \frac{1}{2}(80+89)=\frac{169}{2}=84,5 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 5.&\textrm{Tepi bawah dan tepi atas dari kelas}\\ &\textrm{interval dari tabel distribusi frekuensi}\\ &\textrm{di atas yang tepat adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&30,5\: \: \textrm{dan}\: \: 39,5\\ \color{red}\textrm{b}.&39,5\: \: \textrm{dan}\: \: 49,5\\ \textrm{c}.&50,5\: \: \textrm{dan}\: \: 59,5\\ \textrm{d}.&60,5\: \: \textrm{dan}\: \: 70,5\\ \textrm{e}.&79,05\: \: \textrm{dan}\: \: 89,05\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}&\textrm{Tepi bawah dan tepi atas dari kelas}\\ &\textrm{interval pada tabel distribusi frekuensi}\\ &\textrm{di atas yaitu}:\\ &\begin{cases} \textrm{tepi bawah} & =x_{i}-0,5 \\ \textrm{tepi atas} & =x_{i}+0,5 \end{cases}\\ &\textrm{Berikut tabelnya}\\ &\begin{aligned} &\color{blue}\begin{array}{|c|c|c|}\hline \textrm{Nilai}&\textrm{tepi bawah}&\textrm{tepi atas}\\\hline 30-39&30-0,5=29,5&39+0,5=39,5\\\hline \color{red}40-49,5&\color{red}40-0,5=39,5&\color{red}49+0,5=49,5\\\hline 50-59&.....=49,5&....=59,5\\\hline 60-69&....=59,5&....=69,5\\\hline 70-79&....=69,5&....=79,5\\\hline 80-89&....=79,5&\: \, ....=89,5\\\hline \end{array} \end{aligned} \end{aligned} \end{array}$
Statistika (Matematika Wajib kelas XII)
$\color{blue}\textrm{A. Pendahuluan}$
$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\: \: \: \textrm{Istilah}&\textrm{Pengertian}\\\hline 1.&\textrm{Statistika}&\textrm{Cabang ilmu tentang cara mengumpulkan,} \\ &&\textrm{menyusun, penyajian, dan}\\ &&\textrm{penganalisaan dari suatu data}\\\hline 2.&\textrm{Statistik}&\textrm{Data yang telah tersusun ke dalam}\\ &&\textrm{daftar atau diagram}\\\hline 3.&\textrm{Populasi}&\textrm{Keseluruhan objek dari hasil penelitian}\\ &&\textrm{yang memenuhi syarat tertentu}\\\hline 4.&\textrm{Sampel}&\textrm{Bagian dari populasi yang dapat mewakili}\\ &&\textrm{seluruh populasi}\\\hline \end{array}$
Sebagai tambahan penjelasan
$\color{purple}\begin{array}{|l|l|}\hline .\: \: \: \: \: \qquad \textrm{Istilah}&\textrm{Pengertian dan atau Penjelasan}\\\hline \textrm{Statistika}&\textrm{Lihat pengertian di atas}\\ \textrm{Statistik}&\textrm{Hasil pengolahan data}\\ \textrm{Statistika deskriptif}&\textrm{Statistika baik yang berkenaan dengan}\\ &\textrm{kegiatan pengumpulan, penyajian},\\ & \textrm{penyederhanaan atau penganalisaan},\\ & \textrm{serta penentuan khusus dari suatu data}\\ & \textrm{tanpa penarikan suatu kesimpulan}\\ \textrm{populasi}&\textrm{Keseluruhan objek yang akan diteliti}\\ \color{magenta}\textrm{Sampel (Contoh)}&\color{magenta}\textrm{Bagian dari populasi yang diamati}\\ \textrm{Data}&\textrm{Kumpulan dari datum}\\ \textrm{Datum}&\textrm{Informasi atau catatan keterangan dari}\\ & \textrm{penelitian}\\ \textrm{Data kualitatif}&\textrm{Data yang menunjukkan sifat atau}\\ & \textrm{kondisi objek}\\ \textrm{Data kuantitatif}&\textrm{Data yang menunjukkan jumlah objek}\\ \textrm{Data ukuran}&\textrm{Data yang diperoleh dengan cara}\\ \textrm{(Data kontinu)}& \textrm{mengukur besaran objek}\\ \textrm{Data cacahan}&\textrm{Data yang diperoleh dengan cara}\\ \textrm{(Data diskrit)}& \textrm{mencacah, membilang atau menghitung}\\ &\textrm{banyak objek}\\\hline \end{array}$
$\color{blue}\textrm{B. Penyajian Data}$
$\color{purple}\begin{cases} 1.&\textrm{Daftar bilangan} \\ 2.&\textrm{Tabel distribusi frekuensi} \\ 3.&\textrm{Diagram batang} \\ 4.&\textrm{Diagram garis} \\ 5.&\textrm{Diagram lingkaran} \\ 6.&\textrm{Piktogram} \\ 7.&\textrm{Histogram} \\ 8.&\textrm{Poligon distribusi frekuensi} \\ 9.&\textrm{Ogive} \end{cases}$
$\color{blue}\textrm{C. Data Tunggal}$
$\color{blue}\textrm{C. 1 Ukuran Pemusatan Data (Tendesi Sentral)}$
$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Nilai}&\textrm{Ukuran Pemusatan Data}\\\hline 1.&\textrm{Mean}\: \: \left ( \bar{x} \right )&\bar{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}\\\hline 2.&\textrm{Median}\: \: \left (M_{e} \right )&\begin{cases} \textrm{Ganjil} & M_{e}=x_{\frac{n+1}{2}} \\ \textrm{Genap} & M_{e}=\displaystyle \frac{1}{2}\left ( x_{\frac{n}{2}}+x_{\frac{n}{2}+1} \right ) \end{cases}\\\hline 3.&\textrm{Modus}\: \: \left ( M_{o} \right )&\textrm{Nilai yang sering muncul}\\\hline 4.&\textrm{Kuartil}\: \: \left ( Q \right )&\begin{cases} \textrm{Ganjil} &\begin{cases} Q_{1}= & x_{\frac{1}{4}(n+1)}\\ Q_{2}= & x_{\frac{2}{4}(n+1)}\\ Q_{3}= & x_{\frac{3}{4}(n+1)} \end{cases} \\\\ \textrm{Genap} & \begin{cases} Q_{1}= & x_{\frac{1}{4}n+\frac{1}{2}}\\ Q_{2}= & x_{\frac{2}{4}n+\frac{1}{2}}\\ Q_{3}= & x_{\frac{3}{4}n+\frac{1}{2}} \end{cases} \end{cases}\\\hline \end{array}$
$\color{blue}\textrm{C. 2 Ukuran Penyebaran Data (Dispersi)}$
$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Nilai}&\textrm{Ukuran Penyebaran Data}\\\hline 1.&\textrm{Jangkauan}\: \: \left ( J \right )&J=R\\ &\textrm{atau Rentang}\: \: (R)&=x_{datum\: max}-x_{datum\: min}\\\hline 2.&\textrm{Hamparan}\: \: (H)&\\ &\textrm{Atau Jangkauan}&H=Q_{3}-Q_{1}\\ &\textrm{antar kuartil}&\\\hline 3.&\textrm{Simpangan}&Q_{d}=\displaystyle \frac{1}{2}H\\ &\textrm{Kuartil}\: \: \left (Q_{d} \right )&\\\hline 4.&\textrm{Langkah}\: \: \left ( L \right )&L=\displaystyle \frac{3}{2}H\\\hline 5.&\textrm{Pagar}&\begin{cases} \textrm{Dalam} &=Q_{1}-L \\ \textrm{Luar} &=Q_{3}+L \end{cases}\\\hline 6&\textrm{Data}&\begin{cases} &\textrm{Normal} \\ &:Q_{1}-L\leq x_{i}\leq Q_{3}+L \\ &\textrm{Tidak Normal} \\ &:\begin{cases} x_{i}<Q_{1}-L \\ x_{i}>Q_{3}+L \end{cases} \end{cases}\\\hline 7.&\textrm{Simpangan}&SR=\displaystyle \frac{\sum \left | x_{i}-\bar{x} \right |}{n}\\ &\textrm{Rata-rata}\: \: (SR)&\\\hline 8.&\textrm{Ragam}\: \: \left ( s^{2} \right )&s^{2}=\displaystyle \frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}\\ &\textrm{atau Varian}&\\\hline 9.&\textrm{Simpangan}&s=\sqrt{s^{2}}\\ &\textrm{Baku}\: \: (s)&\\\hline \end{array}$
$\color{blue}\textrm{D. Data Berkelompok}$
Untuk tipe ini antara lain
$\begin{cases} \textrm{(1) Mean},& \bar{\textrm{x}}=\bar{x}_{s}+\displaystyle \frac{\sum f_{i}.d_{i}}{\sum f_{i}} \\\\ \textrm{(2) Modus},& \textrm{M}_{o}=t_{p}+\displaystyle p\left ( \frac{d_{1}}{d_{1}+d_{2}} \right ) \\\\ \textrm{(3) Kuartil},& \textrm{Q}_{i}=t_{p}+\displaystyle p\left ( \frac{\displaystyle \frac{i.n}{4}-\sum f_{i}}{f_{b}} \right ) \end{cases}$
Berikut keterangannya untuk beberapa istilah pada formula di atas baik poin 1, poin 2, maupun poin 3
$\begin{cases} \begin{aligned}&(1)\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}\bar{x}_{s}=&\textrm{rataan sementara}\\ x_{i}=&\textrm{titik tengahinterval}\\ &\textrm{kelas ke}-i\\ d_{i}=&x_{i}-\bar{x}_{s}\\ f_{i}=&\textrm{frekuensi kelas ke}-i\\ \end{aligned}\\\hline \end{array} \\\\ \begin{aligned}&(2)\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}t_{p}=&\textrm{tepi bawah kelas modus}\\ p=&\textrm{panjang interval kelas}\\ d_{1}=&f_{0}-f_{-1}\\ d_{2}=&f_{0}-f_{+1}\\ f_{0}=&\textrm{frekuensi kelas modus}\\ f_{-1}=&\textrm{frekuensi sebelum kelas modus}\\ f_{+1}=&\textrm{frekuensi setelah kelas modus} \end{aligned}\\\hline \end{array} \\\\ \begin{aligned}&(3)\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}tp=&\textrm{tepi bawah kuartil ke}-i\\ p=&\textrm{panjang interval kelas}\\ n=&\textrm{banyaknya data}\\ \sum f_{i}=&\textrm{jumlah semua frekuensi}\\ &\textrm{sebelum kelas kurtil ke}-i\\ f_{q}=&\textrm{frekuensi kelas kuartil ke}-i\\ Q_{i}=&\textrm{kuartil ke}-i\\ Q_{1}=&\textrm{kuartil bawah}\\ Q_{2}=&\textrm{kuatil tengah/median}\\ Q_{3}=&\textrm{kuatil atas} \end{aligned}\\\hline \end{array} \end{cases}$
DAFTAR PUSTAKA
- Johanes, Kastolan, dan Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Sosial Kurikulum Berbasis Kompetensi 2004. Jakarta: Yudistira
- Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung SEWU.
- Sobirin. 2006. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 2 IPA. Jakarta: Kawan Pustaka.
- Tampomas, H. 1999. Seribu Pena Matematika SMU Jilid 2 Kelas 2. Jakarta: Erlangga.
- Wirodikromo, S. 2007. Matematika untuk SMA Kelas XI. Jakarta: Erlangga.