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

$\text{CONTOH SOAL}$.

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

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

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