Soal-Soal Matematika/Analisis Data

Distribusi

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 (rata-rata) dan E(x)=$\sqrt{npq}$ (simpangan baku)

contoh soal:

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

• cara 1

Jawaban

$$\begin{array}{c}n=4,x=2,p=\frac{1}{2},q=\frac{1}{2}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(2,4,\frac{1}{2})& ={(}_2^4)(\frac{1}{2}{)}^2\cdot (\frac{1}{2}{)}^{4-2}\\ & ={(}_2^4)(\frac{1}{2}{)}^2\cdot (\frac{1}{2}{)}^2\\ & =\frac{4!}{2!(4-2)!}\frac{1}{4}\cdot \frac{1}{4}\\ & =\frac{4!}{2!2!}\frac{1}{16}\\ & =\frac{4\cdot 3\cdot 2!}{2!\cdot 2\cdot 1}\frac{1}{16}\\ & =6\frac{1}{16}\\ & =\frac{3}{8}\end{array}$$

• 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{array}{c}\ast 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}\\ \ast 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}\\ \ast \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}\\ \ast 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}\\ \ast \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{array}$$

• 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

$$\begin{array}{c}n=2,x=2,p=\frac{3}{6},q=\frac{3}{6}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(2,2,\frac{3}{6})& ={(}_2^2)(\frac{3}{6}{)}^2\cdot (\frac{3}{6}{)}^{2-2}\\ & ={(}_2^2)(\frac{3}{6}{)}^2\cdot (\frac{3}{6}{)}^0\\ & =\frac{2!}{2!(2-2)!}\frac{3^2}{6^2}\\ & =\frac{2!}{2!0!}\frac{9}{36}\\ & =\frac{9}{36}\\ & =\frac{1}{4}\end{array}$$

• 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

$$\begin{array}{c}n=5,x=3,p=\frac{1}{2},q=\frac{1}{2}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(3,5,\frac{1}{2})& ={(}_3^5)(\frac{1}{2}{)}^3\cdot (\frac{1}{2}{)}^{5-3}\\ & ={(}_3^5)(\frac{1}{2}{)}^3\cdot (\frac{1}{2}{)}^2\\ & =\frac{5!}{3!(5-3)!}\frac{1}{2^5}\\ & =\frac{5!}{3!2!}\frac{1}{32}\\ & =\frac{5\cdot 4\cdot 3!}{3!\cdot 2\cdot 1}\frac{1}{32}\\ & =10\frac{1}{32}\\ & =\frac{5}{16}\end{array}$$

• 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

$$\begin{array}{c}n=4,x=3,p=\frac{4}{6},q=\frac{2}{6}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(3,4,\frac{4}{6})& ={(}_3^4)(\frac{4}{6}{)}^3\cdot (\frac{2}{6}{)}^{4-3}\\ & ={(}_3^4)(\frac{4}{6}{)}^3\cdot (\frac{2}{6}{)}^1\\ & =\frac{4!}{3!(4-3)!}\frac{64}{216}\cdot \frac{2}{6}\\ & =\frac{4!}{3!1!}\frac{128}{1296}\\ & =\frac{4\cdot 3!}{3!\cdot 1}\frac{128}{1296}\\ & =4\frac{128}{1296}\\ & =\frac{512}{1296}\end{array}$$

• 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

$$\begin{array}{c}n=3,x=2,p=\frac{4}{5},q=\frac{1}{5}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(2,3,\frac{4}{5})& ={(}_2^3)(\frac{4}{5}{)}^2\cdot (\frac{3}{5}{)}^{3-2}\\ & ={(}_2^3)(\frac{4}{5}{)}^2\cdot (\frac{1}{5}{)}^1\\ & =\frac{3!}{2!(3-2)!}\frac{16}{25}\cdot \frac{1}{5}\\ & =\frac{3\cdot 2!}{2!}\frac{16}{125}\\ & =3\frac{16}{125}\\ & =\frac{48}{125}\end{array}$$

• 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

$$\begin{array}{c}n=3,x=2,p=\frac{1}{3},q=\frac{2}{3}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(2,3,\frac{1}{3})& ={(}_2^3)(\frac{1}{3}{)}^2\cdot (\frac{2}{3}{)}^{3-2}\\ & ={(}_2^3)(\frac{1}{3}{)}^2\cdot (\frac{2}{3}{)}^1\\ & =\frac{3!}{2!(3-2)!}\frac{1}{9}\cdot \frac{2}{3}\\ & =\frac{3\cdot 2!}{2!}\frac{1}{9}\cdot \frac{2}{3}\\ & =3\frac{1}{9}\cdot \frac{2}{3}\\ & =\frac{2}{9}\end{array}$$

• 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

$$\begin{array}{c}n=12,x=4,p=\frac{2}{5},q=\frac{3}{5}\\ f(x;n,p)& ={(}_x^n)p^x\cdot q^{n-x}\\ f(4,12,\frac{2}{5})& ={(}_4^{12})(\frac{2}{5}{)}^4\cdot (\frac{3}{5}{)}^{12-4}\\ & ={(}_4^{12})(\frac{2}{5}{)}^4\cdot (\frac{3}{5}{)}^8\\ & =\frac{12!}{4!(12-4)!}\frac{2^4}{5^4}\cdot \frac{3^8}{5^8}\\ & =\frac{12!}{4!8!}\frac{2^4}{5^4}\cdot \frac{3^8}{5^8}\\ & =\frac{12\cdot 11\cdot 10\cdot 9\cdot 8!}{4\cdot 3\cdot 2\cdot 1\cdot 8!}\frac{2^4\cdot 3^8}{5^{12}}\\ & =495\frac{2^4\cdot 3^8}{5^{12}}\end{array}$$

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

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=$\frac{k}{N}$