$$\bbox[yellow, 5px, border: 2px solid red] {\textbf{z} = \dfrac{x\:-\:\mu}{\sigma}}$$
\(\mu = \text{ mean}\)
\(\sigma = \text{ simpangan baku}\)
Transformasi Z diperlukan agar variabel acak X menjadi berdistribusi normal standar dengan mean 0 dan variansi 1.
$$\bbox[5px, border: 2px solid red] {\textbf{X}\sim \textbf{N}(\mu, \sigma^2)}$$
$$\bbox[5px, border: 2px solid red] {\textbf{X}\sim \textbf{N}(0, 1)}$$
Soal Latihan
Soal 01
Jika \(\text{X} \sim \text{N}(100, 64)\), tentukan \(\text{P(X ≤ 80)}\)
\(\text{X} \sim \text{N}(100, 64)\)
\(\mu = 100\)
\(\sigma^2 = 64 \rightarrow \sigma = 8\)
\(x = 80\)
\(\color{blue} \textbf{z} = \dfrac{x\:-\:\mu}{\sigma}\)
\(\textbf{z} = \dfrac{80\:-\:100}{8}\)
\(\textbf{z} =-2,5\)
\(\text{P(X ≤ 80)} = \text{P(Z ≤ −2,5)} \)
\(\text{P(Z ≤−2,5)} = 1\:-\:\text{P(Z ≤ 2,5)}\)
Gunakan tabel distribusi normal standar
\(\text{P(Z ≤ −2,5)} = 1\:-\:0,9938\)
\(\text{P(Z ≤ −2,5)} = 0,0062\)
Jadi, jika \(\text{X} \sim \text{N}(100, 64)\), \(\text{P(X ≤ 80)} = 0,0062\)
Soal 02
Jika \(\text{X} \sim \text{N}(100, 64)\), tentukan \(\text{P(X > 116)}\)
\(\text{X} \sim \text{N}(100, 64)\)
\(\mu = 100\)
\(\sigma^2 = 64 \rightarrow \sigma = 8\)
\(x = 116\)
\(\color{blue} \textbf{z} = \dfrac{x\:-\:\mu}{\sigma}\)
\(\textbf{z} = \dfrac{116\:-\:100}{8}\)
\(\textbf{z} = 2,00\)
\(\text{P(X > 116)} = \text{P(Z > 2,00)} \)
\(\text{P(Z > 2,00)} = 1\:-\:\text{P(Z ≤ 2,00)}\)
Gunakan tabel distribusi normal standar
\(\text{P(Z > 2,00)} = 1\:-\:0,9772\)
\(\text{P(Z > 2,00)} = 0,0228\)
Jadi, jika \(\text{X} \sim \text{N}(100, 64)\), \(\text{P(X > 116)} = 0,0228\)
Soal 03
Jika \(\text{X} \sim \text{N}(100, 64)\), tentukan \(\text{P(90 < X < 110)}\)
\(\text{X} \sim \text{N}(100, 64)\)
\(\mu = 100\)
\(\sigma^2 = 64 \rightarrow \sigma = 8\)
\(\text{P(90 < X < 110)} = \text{P}\left(\dfrac{90\:-\:100}{8} < \text{Z} < \dfrac{110\:-\:100}{8} \right)\)
\(\text{P(90 < X < 110)} = \text{P} \left(-1.25 < \text{Z} < 1,25 \right)\)
\(\text{P} \left(-1.25 < \text{Z} < 1,25 \right) = \text{P} (\text{Z} < 1,25) \:-\: \text{P}(\text{Z} < -1,25)\)
\(\text{P} (\text{Z} < 1,25)\:-\: [1\:-\: \text{P}(\text{Z} < 1,25)])\)
\(2\text{P} (\text{Z} < 1,25) \:-\:1\)
Gunakan tabel distribusi normal standar
\(2\text{P} (\text{Z} < 1,25) \:-\:1 = 2(0,8944)\:-\:1\)
\(2\text{P} (\text{Z} < 1,25 )\:-\:1 = 0,7888\)
Jadi, jika \(\text{X} \sim \text{N}(100, 64)\), \(\text{P(90 < X < 110)} = 0,7888\)
Soal 04
\(\text{X} \sim \text{N}(50, 25)\), tentukan nilai \(t\) jika \(\text{P} (\text{X} \leq t) = 0,8217\)
\(\text{X} \sim \text{N}(50, 25)\)
\(\mu = 50\)
\(\sigma^2 = 25 \rightarrow \sigma = 5\)
\(\text{P} (\text{X} \leq t) = 0,8217\)
Dengan membaca tabel distribusi normal standar
0,8217 berasal dari 0,8212 + 0,0005
\(\text{P} (\text{X} \leq t) = \text{P} (\text{Z} \leq 0,922)\)
\(\color{blue} \textbf{Z} = \dfrac{x\:-\:\mu}{\sigma}\)
\(\dfrac{t\:-\:50}{5} = 0,922\)
\(t \:-\: 50 = 4,61\)
\(t = 50 + 4,61\)
\(t = 54,61\)
Soal 05
\(\text{X} \sim \text{N}(50, 25)\), tentukan nilai \(t\) jika \(\text{P} (50\:-\:t <\text{X} < 50 + t) = 0,8230\)
\(\text{X} \sim \text{N}(50, 25)\)
\(\mu = 50\)
\(\sigma^2 = 25 \rightarrow \sigma = 5\)
\(\text{P} (50\:-\:t <\text{X} < 50 + t) = 0,8230\)
\(\text{P} (50\:-\:t <\text{X} < 50 + t) = \text{P} \left(\dfrac{50\:-\:t\:-\:50}{5} <\text{Z} < \dfrac{50 + t\:-\:50}{5}\right)\)
\(\text{P} (50\:-\:t <\text{X} < 50 + t) = \text{P} \left(\dfrac{-t}{5} <\text{Z} < \dfrac{ t}{5}\right)\)
\(\text{P} \left(\dfrac{-t}{5} <\text{Z} < \dfrac{ t}{5}\right) = \text{P}(\text{Z} < \dfrac{t}{5})\:-\: \text{P}(\text{Z} < \dfrac{-t}{5})\)
\(\text{P} \left(\dfrac{-t}{5} <\text{Z} < \dfrac{ t}{5}\right) = \text{P}(\text{Z} < \dfrac{t}{5})\:-\: [1\:-\: \text{P}(\text{Z} < \dfrac{t}{5})]\)
\(\text{P} \left(\dfrac{-t}{5} <\text{Z} < \dfrac{ t}{5}\right) = 2\text{P}(\text{Z} < \dfrac{t}{5})\:-\:1\)
\(2\text{P}(\text{Z} < \dfrac{t}{5})\:-\:1 = 0,8230\)
\(2\text{P}(\text{Z} < \dfrac{t}{5}) = 1,8230\)
\(\text{P}(\text{Z} < \dfrac{t}{5}) = 0,9115\)
Dengan membaca tabel distribusi normal standar
\(\dfrac{t}{5} = 1,35\)
\(t = 6,75\)
Jadi, nilai \(t\) adalah 6,75.
Soal 06
Diketahui nilai rata-rata ulangan harian statistika kelas XII adalah 74,50 dengan variansi 29,16 serta nilai ulangan tersebut terdistribusi secara normal. Jika dipilih seorang siswa dari kelas XII tersebut secara acak, maka tentukan peluang nilainya:
(A) kurang dari 75
(B) lebih dari 90
(C) antara 70 sampai 80
\(\mu = 74,50\)
\(\sigma^2 = 29,16 \rightarrow \sigma = 5,4\)
(A) Menghitung \(\text{P}(\text{X} < 75)\)
\(\text{P}(\text{X} < 75) = \text{P}\left(\text{Z} < \dfrac{75\:-\:74,50}{5,4}\right)\)
\(\text{P}(\text{X} < 75) = \text{P}(\text{Z} < 0,093)\)
Dengan membaca tabel distribusi normal standar
\(\text{P}(\text{X} < 75)\) = 0,5359 + 0,0012 = 0,5371
Jadi, peluang seorang siswa yang terpilih nilainya kurang dari 75 adalah 0,5371.
(B) Menghitung \(\text{P}(\text{X} > 90)\)
\(\text{P}(\text{X} > 90) = \text{P}\left(\text{Z} > \dfrac{90\:-\:74,50}{5,4}\right)\)
\(\text{P}(\text{X} > 90) = \text{P}(\text{Z} > 2,870)\)
\(\text{P}(\text{X} > 90) = 1\:-\: \text{P}(\text{Z} < 2,870)\)
Dengan membaca tabel distribusi normal standar
\(\text{P}(\text{X} > 90) = 1\:-\: \text{P}(\text{Z} < 2,870)\)
\(\text{P}(\text{X} > 90) = 1\:-\: 0,9979\)
\(\text{P}(\text{X} > 90) = 0,0021\)
Jadi, peluang seorang siswa yang terpilih nilainya lebih dari 90 adalah 0,0021.
(C) Menghitung \(\text{P}(70 < \text{X} < 80)\)
\(\text{P}(70 < \text{X} < 80) = \text{P} \left(\dfrac{70\:-\:74,50}{5,4} < \text{Z} < \dfrac{80\:-\:74,50}{5,4} \right)\)
\(\text{P}(70 < \text{X} < 80) = \text{P} (-0,83 < \text{Z} < 1,02)\)
\(\text{P} (-0,83 < \text{Z} < 1,02) = \text{P}(\text{Z} < 1,02) \:-\: \text{P}(\text{Z} < -0,83)\)
\(\text{P} (-0,83 < \text{Z} < 1,02) = \text{P}(\text{Z} < 1,02) \:-\: [1\:-\: \text{P}(\text{Z} < 0,83)]\)
\(\text{P} (-0,83 < \text{Z} < 1,02) = \text{P}(\text{Z} < 1,02) + \text{P}(\text{Z} < 0,83)\:-\:1\)
\(\text{P} (-0,83 < \text{Z} < 1,02) = 0,8461 + 0,7967 \:-\:1\)
\(\text{P} (-0,83 < \text{Z} < 1,02) =0,6428\)
Jadi, peluang seorang siswa yang terpilih nilainya antara 70 sampai 80 adalah 0,6428.
Soal 07
Sebuah perusahaan memproduksi lampu yang mempunyai ketahanan berdistribusi normal dengan rata-rata 4000 jam dan dengan simpangan baku 200 jam.
(A) Berapa persen lampu yang mempunyai ketahanan kurang dari 3800 jam?
(B) Berapa banyak lampu yang memiliki ketahanan lebih dari 3920 jam, jika diproduksi sebanyak 10.000 lampu?
\(\mu = 4000\)
\(\sigma = 200\)
(A) Menghitung \(\text{P}(\text{X} < 3800)\)
\(\text{P}(\text{X} < 3800) = \text{P}\left(\text{Z} < \dfrac{3800\:-\:4000}{200}\right)\)
\(\text{P}(\text{X} < 3800) = \text{P}(\text{Z} < -1)\)
\(\text{P}(\text{X} < 3800) = 1\:-\: \text{P}(\text{Z} < 1)\)
\(\text{P}(\text{X} < 3800) = 1\:-\: 0,8413\)
\(\text{P}(\text{X} < 3800) = 0,1587\)
(B) Menghitung \(\text{P}(\text{X} > 3920)\)
\(\text{P}(\text{X} > 3920) = \text{P}\left(\text{Z} > \dfrac{3920\:-\:4000}{200}\right)\)
\(\text{P}(\text{X} > 3920) = \text{P}(\text{Z} > -0,4)\)
\(\text{P}(\text{X} > 3920) = \text{P}(\text{Z} < 0,4)\)
\(\text{P}(\text{X} > 3920) = 0,6554\)
Banyak lampu yang memiliki ketahanan lebih dari 3920 jam, jika diproduksi sebanyak 10.000 lampu = 0,6554 × 10.000 = 6554 buah.
Soal 08
Dalam sebuah ujian fisika, didapatkan rata-rata 70 dengan simpangan bakunya 8, dan nilai ujian fisika terdistribusi normal.
(A) Jika peluang mendapatkan nilai kurang dari \(k\) adalah 28,4%, maka tentukan nilai \(k\)
(B) Dari 1000 orang yang mengikuti ujian fisika, maka berapa banyak yang mendapat nilai lebih dari 80?
\(\mu = 70\)
\(\sigma = 8\)
Penyelesaian (A)
Misalkan X adalah nilai seorang siswa, maka:
\(\text{P}(\text{X} < k) = 28,4 \%\)
\(\text{P}(\text{X} < k) = 0,2840\)
\(\text{P}(\text{Z} < -z) = 0,2840\)
\(1\:-\:\text{P}(\text{Z} < z) = 0,2840\)
\(\text{P}(\text{Z} < z) = 0,7160\)
0,7160 berasal dari 0,7157 + 0,0003
\(\text{P}(\text{Z} < 0,571) = 0,7160\)
\(\color{blue} \textbf{z} = \dfrac{k\:-\:\mu}{\sigma}\)
\(0,571 = \dfrac{k\:-\:7}{8}\)
\(k\:-\:7 = 4,568\)
\(k = 11,568\)
Jadi, nilai \(k\) adalah 11,568.
Penyelesaian (B)
\(\text{P}(\text{X} > 80)\)
\(\text{P}(\text{X} > 80) = \text{P}\left(\text{Z} > \dfrac{80\:-\:70}{8}\right)\)
\(\text{P}(\text{X} > 80) =\text{P}(\text{Z} > 1,25)\)
\(\text{P}(\text{X} > 80) = 1\:-\: \text{P}(\text{Z} < 1,25)\)
\(\text{P}(\text{X} > 80) = 1\:-\: 0,8944\)
\(\text{P}(\text{X} > 80) = 0,1056\)
Banyak siswa yang mendapat nilai lebih dari 80 = 0,1056 × 1000 = 106 orang.
