Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 3

 D. Persentil

$\begin{array}{rl}& \text{Dilambangkan dengan}{P}_i\\ & \text{dibaca: persentil ke}-i\\ & \text{Rumus data tunggal}:\\ & \mathbf{\text{Dengan rumus pendekatan interpolasi linear}}\\ & P_i=\text{datum ke}-{\displaystyle \frac{i}{100}(n+1)}\\ & \text{jika}{\displaystyle \frac{i}{100}(n+1)\text{tidak bulat, gunakan rumus}}\\ & \text{interpolasi linear, yaitu}:\\ & P_i=x_k+d(x_{k+1}-x_k)\\ & \text{dengan}d\text{adalah nilai desimalnya}\\ & \mathbf{\text{Dengan tanpa rumus interpolasi linear}}\\ & \begin{array}{|cc|}\hline n\text{ganjil}& n\text{genap}\\ P_1=x_{{.}_{\frac{1}{100}(n+1)}}& P_1={\displaystyle \frac{1}{2}(x_{{.}_{\frac{1}{100}n}}+x_{{.}_{\frac{1}{100}n+1}})}\\ P_2=x_{{.}_{\frac{2}{100}(n+1)}}& P_2={\displaystyle \frac{1}{2}(x_{{.}_{\frac{2}{100}n}}+x_{{.}_{\frac{2}{100}n+1}})}\\ ⋮& ⋮\\ P_{99}=x_{{.}_{\frac{99}{100}(n+1)}}& P_{99}={\displaystyle \frac{1}{2}(x_{{.}_{\frac{99}{100}n}}+x_{{.}_{\frac{99}{100}n+1}})}\\ \hline\end{array}\\ & \text{Catatan: sesuaikan dengan kondisi soal}\\ & \text{Rumus data berkelompok/berfrekuensi}:\\ & P_i=L_i+\left({\displaystyle \frac{{\displaystyle \frac{i}{100}n-f_k}}{f}}\right)\times c\\ & \text{Penjelasan}\\ & \begin{array}{rl}{D}_i& =\text{persentil ke}-i\\ i& =1,2,3\\ L_i& =\text{tepi bawah kelas persentil ke}-i\\ f_k& =\text{frekuensi kumulatif sebelum}\\ & \text{sebelum kelas persentil ke}-i\\ f& =\text{frekuensi kelas persentil ke}-i\\ c& =\text{panjang kelas interval}\\ n& =\text{banyak data/kelas interval}\end{array}\end{array}$.

Catatan :untuk bahasan interpolasi linear ada di sini

$\text{CONTOH SOAL}$.

$\begin{array}{ll}1.& \text{Tentukan persentil ke-29 dan ke-75 dari data berikut}\\ & 4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ \\ & \text{Jawab}:\\ & \text{Banyak datum}=15\\ & \mathbf{\text{dengan rumus pendekatan interpolasi linear}}\\ & \text{Data mula-mula}\\ & :4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ & \text{Data setelah diurutkan}\\ & :2,3,4,4,4,5,5,6,6,6,7,7,8,8,9\\ & \begin{array}{rl}{P}_{29}& ={\displaystyle \frac{29}{100}(15+1)}\\ & ={\displaystyle \frac{464}{100}=4,64}\\ & =x_4+0,64(x_5-x_4)\\ & =4+0,64(5-5)=4+0=4\\ P_{75}& ={\displaystyle \frac{75}{100}(15+1)}\\ & ={\displaystyle \frac{1200}{100}=12}\\ & =x_{12}\\ & =7\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Persentil ke-32}\left(P_{32}\right)\text{dari data berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 41-45& 7\\ 46-50& 12\\ 51-55& 9\\ 56-60& 8\\ 61-65& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 46\\ \text{b}.& 47\\ \text{c}.& 48\\ \text{d}.& 51\\ \text{e}.& 52\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \text{persentil ke}-32=P_{32},\\ & \text{dengan}n=\sum f=40\\ P_i& =L_i+c\left({\displaystyle \frac{{\displaystyle \frac{i\times n}{100}-f_k}}{f}}\right)\\ P_{32}& =\text{datum ke}-\left({\displaystyle \frac{32n}{100}}\right)\\ & =x_{\frac{32\times 40}{100}}=x_{12,8}\\ \text{Dan}{x}_{12}& \text{terletak di kelas interval}:46-50\\ P_{32}& =545,5+5\left({\displaystyle \frac{12,8-7}{12}}\right)\\ & =45,5+0,48333...\\ & =45.9833...\approx 46\end{array}\end{array}$

Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 2

 C. Desil

$\begin{array}{rl}& \text{Dilambangkan dengan}{D}_i\\ & \text{dibaca: desil ke}-i\\ & \text{Rumus data tunggal}:\\ & \mathbf{\text{Dengan rumus pendekatan interpolasi linear}}\\ & D_i=\text{datum ke}-{\displaystyle \frac{i}{10}(n+1)}\\ & \text{jika}{\displaystyle \frac{i}{10}(n+1)\text{tidak bulat, gunakan rumus}}\\ & \text{interpolasi linear, yaitu}:\\ & D_i=x_k+d(x_{k+1}-x_k)\\ & \text{dengan}d\text{adalah nilai desimalnya}\\ & \mathbf{\text{Dengan tanpa rumus interpolasi linear}}\\ & \begin{array}{|cc|}\hline n\text{ganjil}& n\text{genap}\\ D_1=x_{{.}_{\frac{1}{10}(n+1)}}& D_1={\displaystyle \frac{1}{2}(x_{{.}_{\frac{1}{10}n}}+x_{{.}_{\frac{1}{10}n+1}})}\\ D_2=x_{{.}_{\frac{2}{10}(n+1)}}& D_2={\displaystyle \frac{1}{2}(x_{{.}_{\frac{2}{10}n}}+x_{{.}_{\frac{2}{10}n+1}})}\\ ⋮& ⋮\\ D_9=x_{{.}_{\frac{9}{10}(n+1)}}& D_9={\displaystyle \frac{1}{2}(x_{{.}_{\frac{9}{10}n}}+x_{{.}_{\frac{9}{10}n+1}})}\\ \hline\end{array}\\ & \text{Catatan: sesuaikan dengan kondisi soal}\\ & \text{Rumus data berkelompok/berfrekuensi}:\\ & D_i=L_i+\left({\displaystyle \frac{{\displaystyle \frac{i}{10}n-f_k}}{f}}\right)\times c\\ & \text{Penjelasan}\\ & \begin{array}{rl}{D}_i& =\text{desil ke}-i\\ i& =1,2,3\\ L_i& =\text{tepi bawah kelas desil ke}-i\\ f_k& =\text{frekuensi kumulatif sebelum}\\ & \text{sebelum kelas desil ke}-i\\ f& =\text{frekuensi kelas desil ke}-i\\ c& =\text{panjang kelas interval}\\ n& =\text{banyak data/kelas interval}\end{array}\end{array}$.

Catatan :untuk bahasan interpolasi linear ada di sini

$\text{CONTOH SOAL}$.

$\begin{array}{ll}1.& \text{Tentukanlah}{D}_1,D_2,D_3,D_4,D_5,D_6\\ & D_7,D_8,D_9\text{dari data berikut}\\ & 2,3,8,9,2,4,5,8,9\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Total datum}=9\\ & \text{Data mula-mula}:2,3,8,9,2,4,5,8,9\\ & \text{Setelah data diurutkan menjadi}\\ & :2,2,3,4,5,8,8,9,9\\ & D_i={\displaystyle \frac{i}{10}(n+1).\text{Jika hasilnay tidak bulat}}\\ & \text{maka dihitung dengan}{D}_i=x_k+d.(x_{k+1}-x_k)\\ & \text{Sehingga}\\ & \begin{array}{rl}{D}_1& ={\displaystyle \frac{1}{10}(9+1)=\frac{10}{10}=1}\\ & =x_1=2\\ D_2& ={\displaystyle \frac{2}{10}(9+1)=\frac{20}{10}=2}\\ & =x_2=2\\ D_3& ={\displaystyle \frac{3}{10}(9+1)=\frac{30}{10}=3}\\ & =x_3=3\\ D_4& ={\displaystyle \frac{4}{10}(9+1)=\frac{40}{10}=4}\\ & =x_4=4\\ D_5& ={\displaystyle \frac{5}{10}(9+1)=\frac{50}{10}=5}\\ & =x_5=5\\ D_6& ={\displaystyle \frac{6}{10}(9+1)=\frac{60}{10}=6}\\ & =x_6=8\\ D_7& ={\displaystyle \frac{7}{10}(9+1)=\frac{70}{10}=7}\\ & =x_7=8\\ D_8& ={\displaystyle \frac{8}{10}(9+1)=\frac{80}{10}=8}\\ & =x_8=9\\ D_9& ={\displaystyle \frac{9}{10}(9+1)=\frac{90}{10}=9}\\ & =x_9=9\end{array}\end{array}\end{array}$.

$\begin{array}{ll}2.& \text{Tentukan desil ke-4 dan ke-6 dari data berikut}\\ & 4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ \\ & \text{Jawab}:\\ & \text{Banyak datum}=15\\ & \mathbf{\text{dengan rumus pendekatan interpolasi linear}}\\ & \text{Data mula-mula}\\ & :4,7,5,6,6,7,8,4,9,5,2,3,6,4,8\\ & \text{Data setelah diurutkan}\\ & :2,3,4,4,4,5,5,6,6,6,7,7,8,8,9\\ & \begin{array}{rl}{D}_4& ={\displaystyle \frac{4}{10}(15+1)}\\ & ={\displaystyle \frac{64}{10}=6,4}\\ & =x_6+0,4(x_7-x_6)\\ & =5+0,4(5-5)=5\\ D_6& ={\displaystyle \frac{6}{10}(15+1)}\\ & ={\displaystyle \frac{96}{10}=9,6}\\ & =x_9+0,6(x_{10}-x_9)\\ & =6+0,6(6-6)=6\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Desil ke-8}\left(D_8\right)\text{dari data berikut adalah}....\\ & \begin{array}{|cc|}\hline \text{Nilai}& f\\ 41-45& 7\\ 46-50& 12\\ 51-55& 9\\ 56-60& 8\\ 61-65& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 58\\ \text{b}.& 57,5\\ \text{c}.& 57\\ \text{d}.& 56,75\\ \text{e}.& 56,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \text{desil ke}-8=D_8,\text{dengan}n=\sum f=40\\ D_i& =L_i+c\left({\displaystyle \frac{{\displaystyle \frac{i\times n}{10}-f_k}}{f}}\right)\\ D_8& =\text{datum ke}-\left({\displaystyle \frac{8n}{10}}\right)=x_{\frac{8\times 40}{10}}=x_{32}\\ \text{Dan}{x}_{32}& \text{terletak di kelas interval}:56-60\\ D_8& =55,5+5\left({\displaystyle \frac{32-28}{8}}\right)\\ & =55,5+2,5\\ & =58\end{array}\end{array}$

Ukuran Letak Data (Materi Kelas XII Matematika Wajib) Bagian 1

A. Pendahuluan

Sebelumnya telah dipelajari tentang salah satu bentuk ukuran pemusatan data yaitu median yang membagi sebuah data menjadi dua bagian yang sama. Selain median ada juga istilah lain yang dapat membagi sebuah data menjadi beberapa bagian yang sama pula, yaitu kuartl yang membagi sebuah data menjadi 4 bagian yang sama. Kemudian selain kuartil, ada juga desil yang memabgi sebuah data menjadi 10 bagian yang sama serta persentil yang membagi sebuah data menjadi 100 bagian yang sama pula.

B. Kuartil

$\begin{array}{rl}& \text{Dilambangkan dengan}{Q}_i\\ & \text{dibaca: kuartil ke}-i\\ & \text{Rumus data tunggal}:\\ & \mathbf{\text{Dengan rumus pendekatan interpolasi linear}}\\ & Q_i=\text{datum ke}-{\displaystyle \frac{i}{4}(n+1)}\\ & \text{jika}{\displaystyle \frac{i}{4}(n+1)\text{tidak bulat, gunakan rumus}}\\ & \text{interpolasi linear, yaitu}:\\ & Q_i=x_k+d(x_{k+1}-x_k)\\ & \text{dengan}d\text{adalah nilai desimalnya}\\ & \mathbf{\text{Dengan tanpa rumus interpolasi linear}}\\ & \begin{array}{|cc|}\hline n\text{ganjil}& n\text{genap}\\ Q_1=x_{{.}_{\frac{1}{4}(n+1)}}& Q_1=x_{{.}_{\frac{1}{4}n+\frac{1}{2}}}\\ Q_2=x_{{.}_{\frac{2}{4}(n+1)}}& Q_2=x_{{.}_{\frac{2}{4}n+\frac{1}{2}}}\\ Q_3=x_{{.}_{\frac{3}{4}(n+1)}}& Q_3=x_{{.}_{\frac{3}{4}n+\frac{1}{2}}}\\ \hline\end{array}\\ & \text{Catatan: sesuaikan dengan kondisi soal}\\ & \text{Rumus data berkelompok/berfrekuensi}:\\ & Q_i=L_i+\left({\displaystyle \frac{{\displaystyle \frac{i}{4}n-f_k}}{f}}\right)\times c\\ & \text{Penjelasan}\\ & \begin{array}{rl}{Q}_i& =\text{kuartil ke}-i\\ i& =1,2,3\\ L_i& =\text{tepi bawah kelas kuartil ke}-i\\ f_k& =\text{frekuensi kumulatif sebelum}\\ & \text{sebelum kelas kuartil ke}-i\\ f& =\text{frekuensi kelas kuartil ke}-i\\ c& =\text{panjang kelas interval}\\ n& =\text{banyak data/kelas interval}\end{array}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{ll}1.& \text{Tentukanlah}{Q}_1,Q_2,Q_3\text{dari data}\\ & \text{berikut}\\ & \text{a}.3,5,7,1,2,4,9,7\\ & \text{b}.2,3,8,9,2,4,5,8,9\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Dengan rumus pendekatan interpolasi linear}}\\ & \begin{array}{rl}\text{a}.& \text{Total datum}=8\\ & \text{Data mula-mula}:3,5,7,1,2,4,9,7\\ & \text{Setelah data diurutkan menjadi}\\ & :1,2,3,4,5,7,7,9\\ & Q_i={\displaystyle \frac{i}{4}(n+1)\{\begin{array}{ll}{Q}_1& ={\displaystyle \frac{1}{4}(8+1)=2{\displaystyle \frac{1}{4}}}\\ \\ Q_2& ={\displaystyle \frac{2}{4}(8+1)=4{\displaystyle \frac{1}{2}}}\\ \\ Q_3& ={\displaystyle \frac{3}{4}(8+1)=6{\displaystyle \frac{3}{4}}}\end{array}}\\ & Q_1=x_2+{\displaystyle \frac{1}{4}(x_3-x_2)=2+{\displaystyle \frac{1}{4}=2{\displaystyle \frac{1}{4}}}}\\ & Q_2=x_4+{\displaystyle \frac{1}{2}(x_5-x_4)=4+{\displaystyle \frac{1}{2}=4{\displaystyle \frac{1}{2}}}}\\ & Q_3=x_6+{\displaystyle \frac{3}{4}(x_7-x_6)=7+0=7}\end{array}\\ & \begin{array}{rl}\text{b}.& \text{Total datum}=9\\ & \text{Data mula-mula}:2,3,8,9,2,4,5,8,9\\ & \text{Setelah data diurutkan menjadi}\\ & :2,2,3,4,5,8,8,9,9\\ & Q_i={\displaystyle \frac{i}{4}(n+1)\{\begin{array}{ll}{Q}_1& ={\displaystyle \frac{1}{4}(9+1)=2{\displaystyle \frac{1}{2}}}\\ \\ Q_2& ={\displaystyle \frac{2}{4}(9+1)=5}\\ \\ Q_3& ={\displaystyle \frac{3}{4}(9+1)=7{\displaystyle \frac{1}{2}}}\end{array}}\\ & Q_1=x_2+{\displaystyle \frac{1}{2}(x_3-x_2)=2+{\displaystyle \frac{1}{1}=2{\displaystyle \frac{1}{2}}}}\\ & Q_2=x_5=5\\ & Q_3=x_7+{\displaystyle \frac{1}{2}(x_8-x_7)=8+{\displaystyle \frac{1}{2}=8{\displaystyle \frac{1}{2}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Data penjualan suatu barang setiap bulan}\\ & \text{di sebuah toko pada tahun 2019 adalah}:\\ & 20,3,9,11,4,12,1,9,9,12,8,10.\\ & \text{Median, kuartil bawah, dan kuartil atasnya}\\ & \text{berturut-turut adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6{\displaystyle \frac{1}{2},3\frac{1}{2},\text{dan}9\frac{1}{2}}\\ \text{b}.& 9,6,\text{dan}11{\displaystyle \frac{1}{2}}\\ \text{c}.& 6{\displaystyle \frac{1}{2},9,\text{dan}12}\\ \text{d}.& 9,4,\text{dan}12\\ \text{e}.& 9,3{\displaystyle \frac{1}{2},\text{dan}12}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \mathbf{\text{Dengan tanpa rumus interpolasi linear}}\\ & \begin{array}{rl}\text{Data}& \text{mula-mula}\\ :& 20,3,9,11,4,12,1,9,9,12,8,10\\ \text{Sete}& \text{lah data diurutkan}\\ :& 1,3,4,8,9,9,9,10,11,12,12,20\\ \text{Dike}& \text{tahui}n=12\mathbf{\text{genap}}\\ Q_1& =x_{\frac{1}{4}n+\frac{1}{2}}=x_{\frac{1}{4}.12+\frac{1}{2}}=x_{3,5}=6\\ Q_2& =x_{\frac{2}{4}n+\frac{1}{2}}=x_{\frac{2}{4}.12+\frac{1}{2}}=x_{6,5}=9=M_e\\ Q_3& =x_{\frac{3}{4}n+\frac{1}{2}}=x_{\frac{3}{4}.12+\frac{1}{2}}=x_{9,5}=11{\displaystyle \frac{1}{2}}\\ \text{Sela}& \text{njutnya data dapat dituliskan}\\ & 1,3,\underset{\begin{array}{c}\Downarrow \\ Q_1\end{array}}{\underbrace{4,8}},9,\underset{\begin{array}{c}\Downarrow \\ Q_2=M_e\end{array}}{\underbrace{9,9}},10,\underset{\begin{array}{c}\Downarrow \\ Q_3\end{array}}{\underbrace{11,12}},12,20\end{array}\end{array}$

$\begin{array}{l}\\ 3.& (\mathbf{\text{UN IPA 2014}})\\ & \text{Kuartil atas dari data pada tabel berikut}\\ & \text{adalah}....\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ 44-49& 12\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & \begin{array}{l}\\ \text{a}.& 49,25\\ \text{b}.& 48,75\\ \text{c}.& 48,25\\ \text{d}.& 47,75\\ \text{e}.& 47,25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Kuartil atas}=Q_3,\text{dengan}n=\sum f=50\\ & \text{Kita sertakan lagi tabel di atas berikut}\\ & \begin{array}{|cc|}\hline \text{Data}& f\\ 20-25& 4\\ 26-31& 6\\ 32-37& 6\\ 38-43& 10\\ \text{44-49}& \text{12}\\ 50-55& 8\\ 56-61& 4\\ \hline\end{array}\\ & Q_3=\text{Datum ke}-\left({\displaystyle \frac{3n}{4}}\right)=x_{\frac{3.50}{4}}=x_{37,5}\\ & \text{dan}{x}_{37,5}\text{terletak di kelas interval}44-49\\ & Q_3=L_3+c\left({\displaystyle \frac{{\displaystyle \frac{3n}{4}-f_k}}{f}}\right)\\ & =43,5+6\left({\displaystyle \frac{37,5-26}{12}}\right)\\ & =43,5+{\displaystyle \frac{11,5}{2}}\\ & =49,5+5,75\\ & =49,25\end{array}\end{array}$