Soal-Soal Matematika/Analisis Data

Ada tujuh jenis distribusi yaitu:

1. Distribusi seragam
   1. Fungsi: f(x,k) = ${\frac {1}{k}}$ untuk x=x₁,x₂,x₃, ……,x_k

1. Distribusi Bernoulli
   1. Fungsi: P(x=1) = p dan P(x=0) = 1-p untuk x=0,1

1. 1. Nilai harapan: E(x)=p

1. Distribusi binomial
   1. Fungsi: f(x;n,p) = ($_{x}^{n}$) p^xq^n-x untuk x=0,1,2,3,4, ….,n

1. 1. Nilai harapan: E(x)=np

contoh soal:

1. Pada pelemparan sebuah koin sebanyak empat kali. Berapa peluang muncul sisi dua gambar koin?

• cara 1

Jawaban

$${\begin{aligned}n=4,x=2,p={\frac {1}{2}},q={\frac {1}{2}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,4,{\frac {1}{2}})&=(_{2}^{4})({\frac {1}{2}})^{2}\cdot ({\frac {1}{2}})^{4-2}\\&=(_{2}^{4})({\frac {1}{2}})^{2}\cdot ({\frac {1}{2}})^{2}\\&={\frac {4!}{2!(4-2)!}}{\frac {1}{4}}\cdot {\frac {1}{4}}\\&={\frac {4!}{2!2!}}{\frac {1}{16}}\\&={\frac {4\cdot 3\cdot 2!}{2!\cdot 2\cdot 1}}{\frac {1}{16}}\\&=6{\frac {1}{16}}\\&={\frac {3}{8}}\\\end{aligned}}$$

• cara 2

jumlah seluruh sebuah koin muncul: 2⁴ = 16

muncul sisi dua gambar: GGAA, GAGA, GAAG, AGGA, AGAG, AAGG. jadi munculnya 6

P(x=2G) = ${\frac {n(A)}{n(S)}}={\frac {6}{16}}={\frac {3}{8}}$

1. Pada pelemparan sebuah dadu sebanyak empat kali. Berapa peluang:
   1. mata dadu 1 muncul satu kali?
   2. kelipatan 3 muncul dua kali?
   3. mata dadu 3 muncul paling sedikit tiga kali?
   4. mata dadu 4 atau 6 muncul empat kali?
   5. ganjil muncul dua kali dan mata dadu 2 atau 5 muncul empat kali?

• cara 1

Jawaban

$${\begin{aligned}*n=4,x=1,p={\frac {1}{6}},q={\frac {5}{6}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(1,4,{\frac {1}{6}})&=(_{1}^{4})({\frac {1}{6}})^{1}\cdot ({\frac {5}{6}})^{4-1}\\&=(_{1}^{4})({\frac {1}{6}})^{1}\cdot ({\frac {5}{6}})^{3}\\&={\frac {4!}{1!(4-1)!}}{\frac {1}{6}}\cdot {\frac {125}{216}}\\&={\frac {4!}{1!3!}}\cdot {\frac {125}{1296}}\\&=4\cdot {\frac {125}{1296}}\\&={\frac {125}{324}}\\*n=4,x=2,p={\frac {1}{3}},q={\frac {2}{3}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,4,{\frac {1}{3}})&=(_{2}^{4})({\frac {1}{3}})^{2}\cdot ({\frac {2}{3}})^{4-2}\\&=(_{2}^{4})({\frac {1}{3}})^{2}\cdot ({\frac {2}{3}})^{2}\\&={\frac {4!}{2!(4-2)!}}{\frac {1}{9}}\cdot {\frac {4}{9}}\\&={\frac {4!}{2!2!}}\cdot {\frac {4}{81}}\\&=6\cdot {\frac {4}{81}}\\&={\frac {24}{81}}\\*{\text{muncul 3 kali }}\\n=4,x=3,p={\frac {1}{6}},q={\frac {5}{6}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(3,4,{\frac {1}{6}})&=(_{3}^{4})({\frac {1}{6}})^{3}\cdot ({\frac {5}{6}})^{4-3}\\&=(_{3}^{4})({\frac {1}{6}})^{3}\cdot ({\frac {5}{6}})^{1}\\&={\frac {4!}{3!(4-3)!}}{\frac {1}{216}}\cdot {\frac {5}{6}}\\&={\frac {4!}{3!1!}}\cdot {\frac {5}{1296}}\\&=4\cdot {\frac {5}{1296}}\\&={\frac {20}{2196}}\\{\text{muncul 4 kali }}\\n=4,x=4,p={\frac {1}{6}},q={\frac {5}{6}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(4,4,{\frac {1}{6}})&=(_{4}^{4})({\frac {1}{6}})^{4}\cdot ({\frac {5}{6}})^{4-4}\\&=(_{4}^{4})({\frac {1}{6}})^{4}\cdot ({\frac {5}{6}})^{0}\\&={\frac {4!}{4!(4-4)!}}{\frac {1}{1296}}\\&={\frac {4!}{4!0!}}\cdot {\frac {1}{1296}}\\&={\frac {1}{1296}}\\{\text{total }}={\frac {20}{1296}}+{\frac {1}{1296}}={\frac {21}{1296}}\\*n=4,x=4,p={\frac {1}{3}},q={\frac {2}{3}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(4,4,{\frac {1}{3}})&=(_{4}^{4})({\frac {1}{3}})^{4}\cdot ({\frac {2}{3}})^{4-4}\\&=(_{4}^{4})({\frac {1}{3}})^{4}\cdot ({\frac {2}{3}})^{0}\\&={\frac {4!}{4!(4-4)!}}{\frac {1}{81}}\\&={\frac {4!}{4!0!}}\cdot {\frac {1}{81}}\\&={\frac {1}{81}}\\*{\text{ganjil muncul 2 kali }}\\n=4,x=2,p={\frac {1}{2}},q={\frac {1}{2}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,4,{\frac {1}{3}})&=(_{2}^{4})({\frac {1}{2}})^{2}\cdot ({\frac {1}{2}})^{4-2}\\&=(_{2}^{4})({\frac {1}{2}})^{2}\cdot ({\frac {1}{2}})^{2}\\&={\frac {4!}{2!(4-2)!}}{\frac {1}{4}}\cdot {\frac {1}{4}}\\&={\frac {4!}{2!2!}}\cdot {\frac {1}{16}}\\&=6\cdot {\frac {1}{16}}\\&={\frac {3}{8}}\\{\text{mata dadu 2 atau 5 muncul 4 kali }}\\n=4,x=4,p={\frac {1}{3}},q={\frac {2}{3}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(4,4,{\frac {1}{3}})&=(_{4}^{4})({\frac {1}{3}})^{4}\cdot ({\frac {2}{3}})^{4-4}\\&=(_{4}^{4})({\frac {1}{3}})^{4}\cdot ({\frac {2}{3}})^{0}\\&={\frac {4!}{4!(4-4)!}}{\frac {1}{81}}\\&={\frac {4!}{4!0!}}\cdot {\frac {1}{81}}\\&={\frac {1}{81}}\\{\text{total }}={\frac {3}{8}}\cdot {\frac {1}{81}}={\frac {3}{648}}\\\end{aligned}}$$

• cara 2

mata dadu 1 muncul 1 kali

p=1/6 dan q=5/6

beberapa kemungkinan yaitu 1/6.5/6.5/6.5/6 + 5/6.1/6.5/6.5/6 + 5/6.5/6.1/6.5/6 + 5/6.5/6.5/6.1/6 = 125/2196 + 125/2196 + 125/2196 + 125/2196 = 500/1296 = 125/324

P(x=1) = ${\frac {n(A)}{n(S)}}={\frac {125}{324}}$

kelipatan 3 muncul 2 kali

p=1/3 dan q=2/3

beberapa kemungkinan yaitu 1/3.1/3.2/3.2/3 + 1/3.2/3.1/3.2/3 + 1/3.2/3.2/3.1/3 + 2/3.1/3.1/3.2/3 + 2/3.1/3.2/3.1/3 + 2/3.2/3.1/3.1/3 = 4/81 + 4/81 + 4/81 + 4/81 + 4/81 + 4/81 = 24/81

P(x=2) = ${\frac {n(A)}{n(S)}}={\frac {24}{81}}$

mata dadu 3 muncul paling sedikit 3 kali

muncul paling sedikit 3 kali berarti 3 kali atau 4 kali

p=1/6 dan q=5/6

untuk muncul 3 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.5/6 + 1/6.1/6.5/6.1/6 + 1/6.5/6.1/6.1/6 + 5/6.1/6.1/6.1/6 = 5/2196 + 5/2196 + 5/2196 + 5/2196 = 20/1296 = 5/324

P(x=3) = ${\frac {n(A)}{n(S)}}={\frac {20}{1296}}$

untuk muncul 4 kali, beberapa kemungkinan yaitu 1/6.1/6.1/6.1/6 = 1/1296

P(x=4) = ${\frac {n(A)}{n(S)}}={\frac {1}{1296}}$

jadi P(x>=3) = P(x=3)+P(x=4) = ${\frac {20}{1296}}+{\frac {1}{1296}}={\frac {21}{1296}}$

mata dadu 4 atau 6 muncul 4 kali

p=1/3 dan q=2/3

beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81

P(x=4) = ${\frac {n(A)}{n(S)}}={\frac {1}{81}}$

ganjil muncul 2 kali dan mata dadu 2 atau 5 muncul 4 kali

ganjil muncul 2 kali

p=1/2 dan q=1/2

beberapa kemungkinan yaitu (1/2).(1/2).2/3.2/3 + (1/2).1/2.(1/2).1/2 + (1/2).1/2.1/2.(1/2) + 1/2.(1/2).(1/2).1/2 + 1/2.(1/2).1/2.(1/2) + 1/2.1/2.(1/2).(1/2) = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 = 6/16 = 3/8

P(x=2) = ${\frac {n(A)}{n(S)}}={\frac {3}{8}}$

mata dadu 2 atau 5 muncul 4 kali

p=1/3 dan q=2/3

beberapa kemungkinan yaitu 1/3.1/3.1/3.1/3 = 81

P(x=4) = ${\frac {n(A)}{n(S)}}={\frac {1}{81}}$

irisan untuk pernyataan independen: P(x=2) . P(x=4) = ${\frac {3}{8}}\cdot {\frac {1}{81}}={\frac {3}{648}}$

1. Pada pelemparan sebuah dadu sebanyak dua kali. Berapa peluang muncul dua angka bilangan prima?

• cara 1

Jawaban

$${\begin{aligned}n=2,x=2,p={\frac {3}{6}},q={\frac {3}{6}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,2,{\frac {3}{6}})&=(_{2}^{2})({\frac {3}{6}})^{2}\cdot ({\frac {3}{6}})^{2-2}\\&=(_{2}^{2})({\frac {3}{6}})^{2}\cdot ({\frac {3}{6}})^{0}\\&={\frac {2!}{2!(2-2)!}}{\frac {3^{2}}{6^{2}}}\\&={\frac {2!}{2!0!}}{\frac {9}{36}}\\&={\frac {9}{36}}\\&={\frac {1}{4}}\\\end{aligned}}$$

• cara 2

jumlah seluruh sebuah dadu muncul: 6² = 36

muncul dua angka bilangan prima: (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), (5,5). jadi munculnya 9

P(x=2P) = ${\frac {n(A)}{n(S)}}={\frac {9}{36}}={\frac {1}{4}}$

1. Pasangan suami-istri memiliki lima anak. Berapa peluang mereka memiliki tiga anak laki-laki?

• cara 1

Jawaban

$${\begin{aligned}n=5,x=3,p={\frac {1}{2}},q={\frac {1}{2}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(3,5,{\frac {1}{2}})&=(_{3}^{5})({\frac {1}{2}})^{3}\cdot ({\frac {1}{2}})^{5-3}\\&=(_{3}^{5})({\frac {1}{2}})^{3}\cdot ({\frac {1}{2}})^{2}\\&={\frac {5!}{3!(5-3)!}}{\frac {1}{2^{5}}}\\&={\frac {5!}{3!2!}}{\frac {1}{32}}\\&={\frac {5\cdot 4\cdot 3!}{3!\cdot 2\cdot 1}}{\frac {1}{32}}\\&=10{\frac {1}{32}}\\&={\frac {5}{16}}\\\end{aligned}}$$

• cara 2

jumlah muncul lima anak: 2⁵ = 32

muncul tiga anak laki-laki: (l,l,l,p,p), (l,l,p,l,p), (l,l,p,p,l), (l,p,l,l,p), (l,p,l,p,l), (l,p,p,l,l), (p,l,l,l,p), (p,l,l,p,l), (p,l,p,l,l), (p,p,l,l,l). jadi munculnya 10

P(x=3L) = ${\frac {n(A)}{n(S)}}={\frac {10}{32}}={\frac {5}{16}}$

1. Di dalam keranjang mangga terdapat 6 buah. Diambil 4 buah secara acak dan ternyata 2 buah busuk. Berapa peluang terambilnya 3 buah baik?

• cara 1

Jawaban

$${\begin{aligned}n=4,x=3,p={\frac {4}{6}},q={\frac {2}{6}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(3,4,{\frac {4}{6}})&=(_{3}^{4})({\frac {4}{6}})^{3}\cdot ({\frac {2}{6}})^{4-3}\\&=(_{3}^{4})({\frac {4}{6}})^{3}\cdot ({\frac {2}{6}})^{1}\\&={\frac {4!}{3!(4-3)!}}{\frac {64}{216}}\cdot {\frac {2}{6}}\\&={\frac {4!}{3!1!}}{\frac {128}{1296}}\\&={\frac {4\cdot 3!}{3!\cdot 1}}{\frac {128}{1296}}\\&=4{\frac {128}{1296}}\\&={\frac {512}{1296}}\\\end{aligned}}$$

• cara 2

jumlah muncul terambil empat buah: $C_{4}^{6}={\frac {6!}{4!2!}}={\frac {6\cdot 5\cdot 4!}{4!\cdot 2\cdot 1}}=15$

muncul terambil 3 buah baik: $C_{1}^{2}\cdot C_{3}^{4}={\frac {2!}{1!1!}}\cdot {\frac {4!}{3!1!}}=2\cdot 4=8$ atau (b1,b2,b3,B1), (b1,b2,b4,B1), (b1,b3,b4,B1), (b2,b3,b4,B1), (b1,b2,b3,B2), (b1,b2,b4,B2), (b1,b3,b4,B2), (b2,b3,b4,B2) jadi munculnya 8

P(x=3BB) = ${\frac {n(A)}{n(S)}}={\frac {8}{15}}$

1. Setia melakukan latihan tendangan penalti sebanyak 3 kali. Peluang sukses melakukan tendangan penalti sebanyak 4/5. Tentukan peluang Setia mencetak tepat 2 gol?

• cara 1

Jawaban

$${\begin{aligned}n=3,x=2,p={\frac {4}{5}},q={\frac {1}{5}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,3,{\frac {4}{5}})&=(_{2}^{3})({\frac {4}{5}})^{2}\cdot ({\frac {3}{5}})^{3-2}\\&=(_{2}^{3})({\frac {4}{5}})^{2}\cdot ({\frac {1}{5}})^{1}\\&={\frac {3!}{2!(3-2)!}}{\frac {16}{25}}\cdot {\frac {1}{5}}\\&={\frac {3\cdot 2!}{2!}}{\frac {16}{125}}\\&=3{\frac {16}{125}}\\&={\frac {48}{125}}\\\end{aligned}}$$

• cara 2

p=4/5 dan q=1/5

tepat 2 gol ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal)

jadi P(x=2) = P(MMG) + P(MGM) + P(GMM)

= 4/5 x 4/5 x 1/5 + 4/5 x 1/5 x 4/5 + 1/5 x 4/5 x 4/5 = 16/125 + 16/125 + 16/125 = 48/125

1. Peluang seorang penahan dapat menahan sasaran adalah 1/3. Dalam sebuah lomba penahanan setiap peserta diberi kesempatan memanah sebanyak 3 kali. Berapa peluang dua panah yang dilepaskan seorang atlet mengenai sasaran?

• cara 1

Jawaban

$${\begin{aligned}n=3,x=2,p={\frac {1}{3}},q={\frac {2}{3}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(2,3,{\frac {1}{3}})&=(_{2}^{3})({\frac {1}{3}})^{2}\cdot ({\frac {2}{3}})^{3-2}\\&=(_{2}^{3})({\frac {1}{3}})^{2}\cdot ({\frac {2}{3}})^{1}\\&={\frac {3!}{2!(3-2)!}}{\frac {1}{9}}\cdot {\frac {2}{3}}\\&={\frac {3\cdot 2!}{2!}}{\frac {1}{9}}\cdot {\frac {2}{3}}\\&=3{\frac {1}{9}}\cdot {\frac {2}{3}}\\&={\frac {2}{9}}\\\end{aligned}}$$

• cara 2

p=1/3 dan q=2/3

tepat 2 panah ada 3 kemungkinan adalah MMG, MGM, GMM (M=berhasil; G: gagal)

jadi P(x=2) = P(MMG) + P(MGM) + P(GMM)

= 1/3 x 1/3 x 2/3 + 1/3 x 2/3 x 1/3 + 2/3 x 1/3 x 1/3 = 2/27 + 2/27 + 2/27 = 6/27 = 2/9

1. Berdasarkan suatu survei kebersihan gigi diketahui 2 dari 5 orang beberapa bulan terakhir telah mengunjungi dokter gigi. Jika 12 orang dipilih secara acak, tentukan peluang 4 orang yang mengunjungi dokter dua bulan lalu!

• cara 1

Jawaban

$${\begin{aligned}n=12,x=4,p={\frac {2}{5}},q={\frac {3}{5}}\\f(x;n,p)&=(_{x}^{n})p^{x}\cdot q^{n-x}\\f(4,12,{\frac {2}{5}})&=(_{4}^{12})({\frac {2}{5}})^{4}\cdot ({\frac {3}{5}})^{12-4}\\&=(_{4}^{12})({\frac {2}{5}})^{4}\cdot ({\frac {3}{5}})^{8}\\&={\frac {12!}{4!(12-4)!}}{\frac {2^{4}}{5^{4}}}\cdot {\frac {3^{8}}{5^{8}}}\\&={\frac {12!}{4!8!}}{\frac {2^{4}}{5^{4}}}\cdot {\frac {3^{8}}{5^{8}}}\\&={\frac {12\cdot 11\cdot 10\cdot 9\cdot 8!}{4\cdot 3\cdot 2\cdot 1\cdot 8!}}{\frac {2^{4}\cdot 3^{8}}{5^{12}}}\\&=495{\frac {2^{4}\cdot 3^{8}}{5^{12}}}\\\end{aligned}}$$

1. Distribusi normal
   1. Fungsi: f(x;$\sigma ,\mu$) = ${\frac {1}{\sqrt {2\pi \sigma }}}({\frac {x-\mu }{\sigma }})^{2}$ untuk $\infty <x<\infty$

1. 1. Nilai harapan: E(x)=$\mu$

1. Distribusi Possion
   1. Fungsi: f(x;$\mu$) = ${\frac {e^{-\mu }\mu ^{x}}{x!}}$ untuk x=0,1,2,3,4, ….

1. 1. Nilai harapan: E(x)=$\mu$

1. Distribusi geometrik
   1. Fungsi: f(x;p) = pq^x-1 untuk x=1,2,3,4, ….

1. 1. Nilai harapan: E(x)=${\frac {1}{p}}$

1. Distribusi hipergeometrik
   1. Fungsi: f(x;N,n,k) = ${\frac {(_{x}^{k})(_{n-x}^{N-k})}{^{N}_{n}}}$ untuk x=0,1,2,3,4, ….,n

1. 1. Nilai harapan: E(x)=np dimana p=${\frac {k}{N}}$