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Binomial Distribution

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DISTRIBUSI BINOMIAL

Distribusi binomial berasal dari percobaan binomial yaitu suatu proses Bernoulli yang diulang sebanyak n kali dan saling bebas. Distribusi Binomial merupakan distribusi peubah acak diskrit. Secara langsung, percobaan binomial memiliki ciri-ciri sebagai berikut: – – – – percobaan tersebut dilakukan berulang-ulang sebanyak n kali setiap percobaan menghasilkan keluaran yang dapat dikategorikan sebagai gagal dan sukses probabilitas sukses p tetap konstan dari satu percobaan ke percobaan lain percobaan yang berulang adalah saling bebas

Ruang sampel A untuk percobaan E yang terdiri dari himpunan tak hingga tetapi masih terhitung dari titik – titik sampel: Jika S = Sukses dan G = Gagal E1 : S (sukses pada percobaan pertama) E2 : GS (gagal pada percobaan pertama dan sukses pada percobaan kedua) E3 : SG (sukses pada percobaan pertama, gagal pada percobaan kedua) E4 : GGS (gagal pada percobaan 1 dan 2, sukses pada percobaan ketiga) E5 : GSG (gagal pada percobaan 1 dan 3, sukses pada percobaan kedua) E6 : SGG (gagal pada percobaan 2 dan 3, sukses pada percobaan pertama)

En : SSS ...S GGG...G (sukses sebanyak x kali, gagal sebanyak n – x kali) x n x

Jika peluang sukses dinotasikan dengan p maka peluang gagal adalah q = 1 – p. Peubah acak X menyatakan banyaknya sukses dari n percobaan yang saling bebas. Maka peluang X pada masing – masing percobaan E adalah:

P(X)  p P(X)  qp  pq P(X)  pq
2 2

untuk E1 untuk E2 untuk E3

P(X)  q p  pq untuk E4 P(X)  qpq  pq 2 untuk E5 P(X)  pqq  pq 2 untuk E6 P(X)  p x q n  x untuk En

 2 Dapat dilihat bahwa E2 dan E3 memberikan hasil yang sama. Jumlahnya   , yaitu 1 

jumlah semua titik sampel yang mungkin menghasilkan x = 1 yang sukses dan n – x = 2 – 1 = 1 yang gagal dari 2 percobaan. Begitupun untuk E4 , E5 , dan E6 juga
 3 memberikan hasil yang sama. Jumlahnya   , yaitu jumlah semua titik sampel yang 1 

mungkin yang menghasilkan x = 1 yang sukses dan n – x = 3 – 1 = 2 yang gagal dari 3 percobaan. Secara umum, jumlah titik sampel yang mungkin untuk menghasilkan x sukses dan n – x gagal dalam n percobaan adalah banyaknya cara yang berbeda dalam
n mendistribusikan x sukses dalam barisan n percobaan, sehingga terdapat   cara.  x

Dan distribusi peluang atau Probability Mass Function (PMF) X dinyatakan pada definisi berikut:
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

untuk x  1, 2,..., n dan 0  p  1 . Pembuktian distribusi Binomial merupakan suatu PMF. Bukti: Untuk membuktikan suatu peubah acak adalah PMF, maka harus ditunjukan: 1. 2. f ( x)  0

 f ( x)  1 x Akan ditunjukkan distribusi binomial memenuhi kedua syarat di atas: 1. f ( x)  0

Karena 0  p  1 dan nilai kombinasi pasti positif maka f(x) pasti positif. 2.

 f ( x)  1 x Menggunakan persamaan binomial Newton pada

 f ( x) , akan diperoleh: x  f ( x)    x  p (1  p) x x x 0

n

n  

n x

 ( p  (1  p)) n  1n  1

(1)

Dari 1 dan 2 dapat dikatakan bahwa distribusi binomial merupakan PMF.

Mean Jika X
B(n, p) ( X variabel random berdistribusi Binomial), maka nilai ekspektasi

dari X adalah
E ( X )   x.P( x) x   x.P( X  x) x 0 n

n

n   x.   p x (1  p) n  x x 0  x n n   x.   p x (1  p) n  x x 1  x n n!   x. p x (1  p) n  x x !(n  x)! x 1   x. x 1 n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)!

 np 

(n  1)! p x 1 (1  p) n  x x 1 ( x  1)!( n  x )! n Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X )  np m! p s (1  p)m s s  0 s !(m  s )! m Berdasarkan (1),

 s !(m  s)! p (1  p) s s 0

m

m!

m s

 1 , maka

E ( X )  np  s 0

m

m! p s (1  p) m  s s !(m  s )!

 np.1  np

Sehingga didapat mean dari X, E ( X )  np

Variansi
Var ( X )  E ( X 2 )  ( E( X ))2

Dalam meencari Var(X), kita harus tahu nilai ekspektasi dari X2 :

E ( X 2 )   x 2 .P( X  x)   x 2 . x 0 x 0

n

n

n! p x (1  p) n  x x !(n  x)!

E ( X 2 )   x 2 .P( x) x   x 2 .P ( X  x ) x 0 n n   x 2 .   p x (1  p) n  x x 0  x n n   x 2 .   p x (1  p) n  x x 1  x n n!   x2. p x (1  p) n  x x !(n  x)! x 1

n

  x2. x 1 n

n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)! (n  1)! p x 1 (1  p) n  x ( x  1)!(n  x)!

 np  x. x 1

Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X 2 )  np ( s  1) s 0 m m  m m! p s (1  p) m s  np  ( s  1)   p s (1  p) m s s !(m  s)! s 0 s 

E ( X 2 )  np  ( s  1) s 0

m

m! p s (1  p) m  s s !(m  s )!

m m  np  ( s  1)   p s (1  p) m s s 0 s  m  m m s  m m s  np.   s.   p (1  p)  1.   p s (1  p) m s  s 0 s   s 0  s    np[ mp  1]

 np[( n  1) p  1]  np[ np  p  1]

Var ( X )  E ( X 2 )  ( E ( X )) 2  np[np  p  1]  (np )2  (np )2  np 2  np  (np )2  np  np 2  np (1  p )
Sehingga didapat variansi dari X, Var ( X )  np(1  p)

Contoh: 1. Probabilitas bahwa sejenis komponen tertentu yang lolos uji kelayakan adalah ¾. Tentukan probabilitas dimana 2 dari 4 komponen yang selanjutnya diuji akan dinyatakan layak!
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

p = ¾, q = 1 – ¾ = ¼ untuk x = 2
42

 4 3   1  P( X  2)        2 4   4 
2

 6.

9 1 27 .  16 16 128

2. Berdasarkan data biro perjalanan PT Sentosa, yang khusus menangani perjalanan wisata turis mancanegara, 20% dari turis menyatakan sangat puas berkunjung ke Indonesia, 40% menyatakan puas, 25% menyatakan biasa saja, dan sisanya menyatakan kurang puas. Apabila kita bertemu dengan 5 orang dari peserta wisata turis mancanegara yang pernah menggunakan jasa biro perjalanan tersebut. Tentukan probabilitas: a. Tepat 2 diantaranya menyatakan biasa saja

b. Paling banyak 2 diantaranya menyatakan sangat puas

Jawab: n=5 a. p = 0.25, q = 1 – 0.25 = 0.75
5 2

5 1   3  P( X  2)        2 4   4 
2

 0.2637

b. p = 0.2, q = 1 – 0.2 = 0.8
P( X  2)  P( X  0)  P( X  1)  P( X  2) 5  5 5    (0.2) 0 (0.8)5    (0.2)1 (0.8)51    (0.2) 2 (0.8)52 0 1   2  0.32768  0.40960  0.20480  0.94208

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...Solution to selected exercises from chapter 3 3-14. f X (0)  P( X  0)  1/ 6  1/ 6  1/ 3 f X (1.5)  P( X  1.5)  1/ 3 f X (2)  1/ 6 f X (3)  1/ 6 a) P(X = 2) = 1/6 b) P(0.6 < X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P(0  X  2)  P( X  0)  P( X  1.5)  1/ 3  1/ 3  2 / 3 e) P(X = 0 or X = 2) = 1/3 + 1/6 = ½ --------------------------------------------------------------------------------------------------------------------3-15. All probabilities are greater than or equal to zero and sum to one. a) P(X  1) = 1/8 + 2/8 + 2/8 + 2/8 = 7/8 b) P(X > 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(1  X  1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X  1 or X = 2) = 1/8 + 2/8 +1/8 = 4/8 = ½ --------------------------------------------------------------------------------------------------------------------3-20. P(X = 0) = 0.053 = 1.25  104 P(X = 1) = 3[0.95(0.05)(0.05)] = 7.125  103 P(X = 2) = 3[0.95(0.95)(0.05)] = 0.1354 P(X = 3) = 0.953 = 0.8574 --------------------------------------------------------------------------------------------------------------------- 3-25. P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = 0.5 million) = 0.1 --------------------------------------------------------------------------------------------------------------------- 3-32. x0   0, 1/ 3 0  x  1.5      F ( x)  2 / 3 1.5  x  2  where 5 / 6 2  x  3    3 x   1   f X (0)  P( X  0)  1/ 6  1/ 6  1/ 3 f X (1.5)  P( X ...

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...statistics (proportions-population) - past events, Probability (sample) - future events. 1) A U B, A or B P(A or B) = P(A) + P(B) - P(A and B) Mutually exclusive => P(A and B)=0 2) P(B/A) Independent=P(B). Dependent=P(B/A)= P(A and B)/ P(A) 3) Intersection A ∩ B ///Dependent :P(A and B) = P(A) x P(B|A) Independent Events: P(A and B) = P(A) x P(B) 4) A stands for “not A” ; Complement Rule: P(A) + P(not A) = 1; P(A) = 1- P(not A); P(At least one) = 1 – P(none) ……………………………………………………………………………………………………………………………………………………………………… ……………………………………………………………………………. 1)UNIFORM DISTRIBUTION: The area under the uniform distribution: P( Mx1 ) z1= 1 - CI% x% (confidence interval dat) of the observation fall below X 3) BINOMIAL DISTRIBUTION: calculez : p(success); n= total no; x = number of successes in sample- ni se da in intrebare; BINOMIAL FORMULA: p(x)= [n!/ (X! (n-x)!)]*px(1-p)n-x ; P(x=x) =P(1)+ P(2)+………………..+P(n) BINOMIAL...

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