# 6 Distribusi Probabilitas Diskrit

Business Statistics: A Decision-Making Approach 8th Edition



Chapter 5 Discrete Probability Distributions

Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-1

Chapter Goals After completing this chapter, you should be able to:  Calculate and interpret the expected value of a discrete

probability distribution  Apply the binomial distribution to business problems  Compute probabilities for the Poisson and

hypergeometric distributions  Recognize when to apply discrete probability

distributions to decision making situations


5-2

Introduction to Probability Distributions  Probability is the way decision makers express their

uncertainty about outcomes and events.  Random Variables  A variable that takes on different numerical values based on chance.

 Represents a possible numerical value from a random event.

 Can vary from trial to trial.

 Contoh :  Inspektor memeriksa 3 TV plasma dan menentukan ‘diterima’ atau

‘tidak diterima’.  Outcome/hasil dari eksperimen yang mungkin adalah : x = {0, 1, 2, 3}  random variable


5-3

Introduction to Probability Distributions

Random Variables

Ch. 5

Discrete Random Variable


Continuous Random Variable

Ch. 6

5-4

Discrete Random Variable  A discrete random variable is a variable that can assume only

a countable number of values Many possible outcomes:  number of complaints per day  number of TV’s in a household  number of rings before the phone is answered Only two possible outcomes:  gender: male or female  defective: yes or no  spreads peanut butter first vs. spreads jelly first


5-5

Contoh :  Kuis Statistik Industri 1 berupa soal pilihan ganda. Apabila x merepresentasikan jumlah pertanyaan yang dijawab dengan benar. Maka x adalah discrete random variable. Nilai yang mungkin untuk x adalah : x = {0, 1, 2, 3, 4, 5, 6}  Asuransi Prudential melakukan survey pada konsumennya untuk menentukan jumlah anak berusia di bawah 22 tahun yang tinggal di rumah. Maka random variable adalah x yaitu jumlah anak yang tinggal di rumah. x = {0, 1, 2, 3, 4, ...}  discrete random variable Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Continuous Random Variable  A continuous random variable is a variable that can assume

any value on a continuum (can assume an uncountable number of values)  thickness of an item  time required to complete a task  temperature of a solution  height, in inches  These can potentially take on any value, depending only on

the ability to measure accurately.


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Discrete Random Variables  Can only assume a countable number of values

Examples:  Roll a die twice

Let x be the number of times 4 comes up x = {0, 1, or 2 times}  Toss a coin 5 times

Let x be the number of heads x = {0, 1, 2, 3, 4, or 5}


5-8

Discrete Probability Distribution Experiment: Toss 2 Coins. Let x = # heads. 4 possible outcomes

T

H H

T H

x Value Probability 0

1/4 = 0.25

1

2/4 = 0.50

2

1/4 = 0.25

T H

Probability

T

Probability Distribution

0.5 0 0.2 5

0

1

2

x

5-9

Discrete Probability Distribution  A list of all possible [ xi , P(xi) ] pairs

xi = Value of Random Variable (Outcome) P(xi) = Probability Associated with Value  xi’s are mutually exclusive (no overlap)

 xi’s are collectively exhaustive (nothing left out)  0 ≤ P(xi) ≤ 1 for each xi ▪ The probability of xi is between 0 and 1

 S P(xi) = 1 ▪ The sum of all probabilities in the sample space = 1


5-10

Discrete Probability Distribution 3 possible outcomes

21 possible outcomes

The number of possible outcomes increase, the distribution besome smoother dan the individual probability of any particular value tends to be reduce Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Discrete Random Variable Mean Expected Value (or mean) of a discrete distribution (Weighted Average)  the weight are probability assigned ti the values.  The average value when the experiment that generates values for the random varianle is repeated over the long run.  = E(x) = xi P(xi) 

Where : E(x) = Expected value of x x = values of the random variable P(x) = probability of the randim variable taking on the value x Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Discrete Random Variable Mean  Example: Toss 2 coins,

x = # of heads, compute expected value of x: x

P(x)

0

0.25

1

0.50

2

0.25

E(x) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0 Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Discrete Random Variable Standard Deviation  Variace of a discrete random variable

σ 2   [X i  E(X)] 2 P(Xi )  Standard Deviation of a discrete distribution

σx 

{x

 E(x)} P(xi ) 2

i

where:

E(x) = Expected value of the random variable (x) x = Values of the random variable P(x) = Probability of the random variable having the value of x


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Discrete Random Variable Standard Deviation (continued)

 Example: Toss 2 coins, x = # heads,

compute standard deviation (recall E(x) = 1)

σx 

{x

 E(x)} P(xi ) 2

i

σ x  (0  1) 2 (0.25)  (1  1) 2 (0.50)  (2  1) 2 (0.25)  0.50  .707

Possible number of heads = 0, 1, or 2 Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Contoh  Karena cuaca buruk, jumlah hari pada minggu depan untuk

pelayaran menjadi tidak tentu. Misalkan x adalah hari pelayaran terjadi. Berikut ini adalah distribusi probabilitas dari variabel, x ditentukan berdasarkan data historis saat cuaca buruk : X

0

P(x) 0,05

1

2

3

4

0,1

0,1

0,2

0,2

5

6

7

0,15 0,15 0,05

 Berdasarkan distribusi probabilitas, tentukan expected dari

hari per minggu yang memungkinkan untuk pelayaran?  Tentukan pula standar deviasinya! Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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 Expected value : x

0

1

2

3

4

5

6

7

P(x)

0,05

0,1

0,1

0,2

0,2

0,15

0,15

0,05

Sum

x.P(x)

0

0,1

0,2

0,6

0,8

0,75

0,9

0,35

3,7

 Standar deviasi :


5-17

Probability Distributions Ch. 5

Ch. 6

Discrete Probability Distributions

Continuous Probability Distributions

Binomial* Poisson*

Normal

Hypergeometric*

Uniform

Binomial Negative

Exponential

Multinomial Geometric

5-18

The Binomial Distribution  A distribution that gives the probability of x successes in n

trial in a process that meets the following condition (characteristics the Binomial Distribution) :  A trial has only two possible outcomes – “success” or “failure”  There is a fixed number, n (finite), of identical trials  The trials of the experiment are independent of each other  The probability of a success, p, remains constant from trial to trial  If p represents the probability of a success, then (1-p) = q is the probability of a failure The sampling is performed with replacement from a finite population. Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Binomial Distribution Examples  A manufacturing plant labels items as either defective or

acceptable  A firm bidding for a contract will either get the contract

or not  A marketing research firm receives survey responses of

“yes I will buy” or “no I will not”  New job applicants either accept the offer or reject it


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Counting Rule for Combinations  A combination is an outcome of an experiment where x

objects are selected from a group of n objects.

n! C  x! (n  x )! n x

where: Cxn = number of combinations of x objects selected from n objects n! =n(n - 1)(n - 2) . . . (2)(1) Order does not matter x! = x(x - 1)(x - 2) . . . (2)(1) NOTE: 0! = 1 (by definition) Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

i.e. ABC = CBA only one outcome 5-21

Contoh  Dilakukan pemeriksaan terhadap 4 kelompok sepatu untuk

menentukan kualitas “good” dan “defect”.  Maka kombinasi cara untuk mendapatkan kemungkinan “defect” bersifat discrete dengan hasil sebagai berikut : x = {0, 1, 2, 3, 4}  x=0

4! 4 x3x2 x1 C   1 0!(4  0)! 1(4 x3x2 x1) 4 0

 x=1

4! 4 x3x2 x1 4 C1   4 1!(4  1)! 1(3x2 x1)


G, G, G, G G, G, G, D G, G, D, G

G, D, G, G D, G, G, G

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 x=2

4! 4 x3x2 x1 4 C2   6 2!(4  2)! 2 x1(2 x1)  x=3

4! 4 x3x2 x1 C   4 3!(4  3)! 3x2 x1(1) 4 3

 x=4

G, G, D, D

G, D, G, D

G, D, D, G

D, G, D, G

D, D, G, G

D, G, G, D

G, D, D, D D, G, D, D

D, D, G, D D, D, D, G

4! 4 x3x2 x1 C   1 4!(4  4)! 4 x3x2 x1(1) 4 4


D, D, D, D 5-23

Binomial Distribution Formula n! x nP(x) = p q x ! (n - x )!

x

P(x) = probability of x successes in n trials, with probability of success p on each trial Example: Flip a coin x = number of successes in sample, four times, let x = # (x = 0, 1, 2, ..., n) heads: p = probability of “success” per trial n=4 q = probability of “failure” = (1 – p) p = 0.5 n = number of trials (sample size) 1 - p = (1 - 0.5) = 0.5 x = 0, 1, 2, 3, 4 Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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 Dengan menggunakan persoalan pada pemeriksaan sepatu

diketahui n = 4, p = 0,1 dan x = 2 defects  Maka :

n! P( x)  p x q n x x!(n  x)! 4! P(2)  (0,12 )(0,9 2 )  0,0486 2!2!


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Distribution x = # of defects

P(x)

0

0,6561

1

0,2916

2

0,0486

3

0,0036

4

0,0001

Total

1,0000

Perhitungan dengan menggunakan rumus seringkali memerlukan waktu. Untuk itu dapat digunakan tabel Binomial. Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Tabel Binomial


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Example: Calculating a Binomial Probability What is the probability of one success in five observations if the probability of success is .1?

x = 1, n = 5, and p = 0.1

n! P(x  1)  p x (1  p)n x x!(n  x)! 5!  (0.1)1 (1  0.1) 51 1!(5  1)!  (5)(0.1)(0 .9) 4  0.32805 Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc.

5-28

Binomial Distribution Summary Measures  Mean

μ  E(x)  np 

Variance and Standard Deviation

σ  npq 2

σ  npq Where n = sample size p = probability of success q = (1 – p) = probability of failure Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

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Binomial Distribution  The shape of the binomial distribution depends on the

values of p and n

Mean 



Here, n = 5 and p = 0.1 (Binomial distribution skewed. The skeweness will most pronoinced when n is small and p approaches 0 or 1) Here, n = 5 and p = 0.5 (bell-shapes distribution)

.6 .4 .2 0

P(X)

X

0

.6 .4 .2 0


n = 5 p = 0.1

P(X)

1

2

3

4

5

n = 5 p = 0.5 X

0

1

2

3

4

5 5-30

Binomial Characteristics Examples

μ  np  (5)(0.1)  0.5 Mean σ  npq  (5)(0.1)(1  .1)  0.6708

μ  np  (5)(0.5)  2.5 σ  npq  (5)(0.5)(1  0.5)  1.118

.6 .4 .2 0

P(X)

X

0

.6 .4 .2 0


n = 5 p = 0.1

P(X)

1

2

3

4

5

n = 5 p = 0.5 X

0

1

2

3

4

5 5-31

Binomial Distribution Example  Example: 35% of all voters support Proposition A. If a

random sample of 10 voters is polled, what is the probability that exactly three of them support the proposition? i.e., find P(x = 3) if n = 10 and p = 0.35 :

n! 10! x n x P(x  3)  pq  (0.35) 3 (0.65) 7  0.2522 x!(n  x)! 3!7! There is a 25.22% chance that exactly 3 out of the 10 voters will support Proposition A Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-32

Using Cumulative Binomial Distribution Tables If a random sample of 10 voters is polled, what is the probability that three or fewer of them support the proposition? n = 10 x

p=.15

p=.20

p=.25

p=.30

p=.35

p=.40

p=.45

p=.50

0 1 2 3 4 5 6 7 8 9 10

0.1969 0.5443 0.8202 0.9500 0.9901 0.9986 0.9999 1.0000 1.0000 1.0000 1.0000

0.1074 0.3758 0.6778 0.8791 0.9672 0.9936 0.9991 0.9999 1.0000 1.0000 1.0000

0.0563 0.2440 0.5256 0.7759 0.9219 0.9803 0.9965 0.9996 1.0000 1.0000 1.0000

0.0282 0.1493 0.3828 0.6496 0.8497 0.9527 0.9894 0.9984 0.9999 1.0000 1.0000

0.0135 0.0860 0.2616 0.5138 0.7515 0.9051 0.9740 0.9952 0.9995 1.0000 1.0000

0.0060 0.0464 0.1673 0.3823 0.6331 0.8338 0.9452 0.9877 0.9983 0.9999 1.0000

0.0025 0.0233 0.0996 0.2660 0.5044 0.7384 0.8980 0.9726 0.9955 0.9997 1.0000

0.0010 0.0107 0.0547 0.1719 0.3770 0.6230 0.8281 0.9453 0.9893 0.9990 1.0000

Example: n = 10, p = 0.35, x = 3:

P(x = 3, n =10, p = 0.35) = 0.5138

5-33

Contoh  Manager PT Coca Cola Amatil mengirimkan 300 email kepada

konsumen untuk mengukur kepuasan terhadap pelayanan tahunan. Selama ini tingkat pengembalian email (membalas email) adalah p = 0,11.  Sehingga expected value (mean) pengembalian email adalah : E(x) = n.p = 300 x 0,11 = 33  Pada tahun ini ternyata da 44 konsumen yang membalas email sebanyak 44 orang. Pengembalian ini melebihi 33 orang. Selanjutnya manager melakukan pengujian terhadap data tersebut : P(x≥44) = 1 – P(x≤43)  Tabel binomial memiliki keterbatasan untuk sample > 300, sehingga perhitungan dilakukan dengan menggunakan software excel ataupun minitab. Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-34

Using EXCEL 1. Buka sheet kosong 2. Pilih Formulas 3. Klik pada fx (Function 4.

5. 6. 7.

wizard) Pilih Statistical Category Pilih BINOMDIST function Lengkapi informasi yang diminta True mengindikasikan probabilitas kumulatif


P(x≤43) = 0,97 P(x≥44) = 1 – 0,97 = 0,03 Hanya 3% peluang dari 44 atau lebih konsumen yang membalas email dari peluang 11%. 5-35

The Poisson Distribution  Difficult to count the number of failures, so diffivult to

know possible outcomes (successes + failures).  Characteristics of the Poisson Distribution:  The outcomes of interest are rare relative to the possible outcomes  The average number of outcomes of interest per time or space interval is   The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest  The probability that an outcome of interest occurs in a given segment is the same for all segments Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-36

Contoh :  Mendeskripsikan suatu situasi :  Jumlah pasien yang datang pada unit IGD per jam.  Jumlah panggilan pada customer service Telkom setiap 30 menit.  Mendeskripsikan random variable :  Jumlah ‘dot’ pada setuap lembar kertas.  Jumlah kontaminan dalam segalon air


5-37

Poisson Distribution Summary Measures  Mean

E(x)  μ x  λt

λ and t must be in compatibe unit. the average number in t segment, is not necessarily the number we eill see if we observe the process for t segments. 

Variance and Standard Deviation

σ  λt 2

σ  λt where  = number of successes in a segment of unit size t = the size of the segment of interest Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-38

Graph of Poisson Probabilities 0.70

Graphically:

0.60

 = .05 and t = 100

0 1 2 3 4 5 6 7

0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 0.0000

P(x)

X

t = 0.50

0.50 0.40 0.30 0.20 0.10 0.00 0

1

2

3

4

5

6

7

x

P(x = 2) = 0.0758


5-39

Poisson Distribution Shape  The shape of the Poisson Distribution depends on the

parameters  and t:

t = 0.50

t = 3.0

0.70

0.25

0.60

0.20

0.40

P(x)

P(x)

0.50

0.30

0.15

0.10

0.20

0.05 0.10

0.00

0.00 0

1

2

3

4

5

6

7

x


1

2

3

4

5

6

7

8

9

10

11

12

x

5-40

Poisson Distribution Formula

(t ) e P( x)  x! x

 t

where: t = size of the segment of interest x = number of successes in segment of interest  = expected number of successes in a segment of unit size e = base of the natural logarithm system (2.71828...) Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-41

Using Poisson Tables t X 0 1 2 3 4 5 6 7

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

0.9048 0.0905 0.0045 0.0002 0.0000 0.0000 0.0000 0.0000

0.8187 0.1637 0.0164 0.0011 0.0001 0.0000 0.0000 0.0000

0.7408 0.2222 0.0333 0.0033 0.0003 0.0000 0.0000 0.0000

0.6703 0.2681 0.0536 0.0072 0.0007 0.0001 0.0000 0.0000

0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 0.0000

0.5488 0.3293 0.0988 0.0198 0.0030 0.0004 0.0000 0.0000

0.4966 0.3476 0.1217 0.0284 0.0050 0.0007 0.0001 0.0000

0.4493 0.3595 0.1438 0.0383 0.0077 0.0012 0.0002 0.0000

0.4066 0.3659 0.1647 0.0494 0.0111 0.0020 0.0003 0.0000

Example: Find P(x = 2) if  = 0.05 and t = 100

(t ) x e t (0.50) 2 e 0.50 P( x  2)    0.0758 x! 2! Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-42

Contoh  Figur kedatangn dari konsumen Mc. D disajikan dlam

distribusi Poissson dengan λ = 16 konsume per jam. Berdasarkan data didapatkan waktu pelayanan cukup konstan pada 6 menit. Di restoran terdapat 3 kasir, sehingga dalam 6 menit 3 konsumen dapat terlayani. Manager bermaksud mengetahui probabilitas 1 atau 2 konsumen tidak terlayani dalam 6 menit.  Convert 6 menit menjadi 0,1 jam sehingga t = 0,1. sehingga λt = 16 (0,1) = 1,6 konsumen.  P(1 atau lebih konsumen menunggu) = P(4) + P(5) + P(6) + ... Atau P(1 atau lebih konsumen menunggu) = 1 – P(x≤3) Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-43

 P(x≤3) = 0,9212  Sehingga P(4 atau lebih konsumen) = 1 – 0,9212 = 0,0788  Manager bermaksud memenuhi target agar konsumen yang

menunggu tidak lebih dari 0,05 sebagai acuan untuk penambahan kasir. P(4 atau lebih konsumen) = 1 – P(x≤?) ≤0,05 Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-44

 Sehingga diperlukan P(x≤?) ≥0,95

 P(4 atau lebih konsumen) = 1 – P(x≤4) = 1 – 0,9763 = 0,0237  Karena 0,0237 < 0,05 maka sebaiknya manager menambah

satu kasir. Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-45

The Hypergeometric Distribution  “n” trials in a sample taken from a finite population of

size N  Sample taken without replacement  Trials are dependent

 The probability changes from trial to trial  Concerned with finding the probability of “x” successes

in the sample where there are “X” successes in the population


5-46

Hypergeometric Distribution Formula

(Two possible outcomes per trial: success or failure)

P( x) 

C

NX . n x N n

C

X x

C

Where N = population size X = number of successes in the population n = sample size x = number of successes in the sample n – x = number of failures in the sample Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-47

Hypergeometric Distribution Example ■ Example: 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective? N = 10 X=4

P(x  2) 

C

NX n x N n

C

n=3 x=2

C

X x

6 1

4 2

C C (6)(6)    0.3 10 C3 120


5-48

Contoh  Pada produksi Nokia bulan Juli didapatkan 20 produk dicurigai

defect karena kesalahan proses. Manager segera memerintahkan untuk memisahkan 20 produk tersebut. Namun bagian QC terlanjur meloloskan 10 produk ke bagian pengepakan. Peluang defect dari 20 product adalah 2.  Maka peluang produk tanpa defect yang dikirim adalah (x = 0)? 20- 2 2 C10 C C nNxX C Xx -0 0 P(x  0)   20 C nN C10

(43,758)(1 ) P(x  0)   0.2368 184,756 Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-49

 Hypergeometric dustribution untuk semua defect yang

mungkin : x

P(x)

0

0,2368

1

0,5264

2

0,2368

Total

1,0000


5-50

Hypergeometric Distribution (k possible outcomes per trial) P ( x1 , x2 ,..., xk )  k

X i 1 k

x i 1

i

i

 N  n

C

X1 x1

C

X2 x2 N n

...C

XK xk

C

N = Population size n = Total sample size Xi = Number of items in the population with outcome i xi = Number of items in the sample with outcome i


5-51

Contoh  Studi terhadapt pasar sepeda motor dengan 4 merek

yang ada pada suatu dealer. Pada dealer tersebut tersebut terdapat stok sebagai berikut :    

5 brand Honda 4 brand Suzuki 6 brand Yamaha 4 brand Kawasaki

 Sales manager melakukan penelitian pilihan konsumen

tanpa melihat harga. Sebanyak 6 konsumen dipilih secara acak dan 3 memilih Suzuki, 2 memilih Kawasaki dan 1 memilih Yamaha. Tidak satupun memilih Honda.  Propabilitas pemilihan secara acak dengan asumsi memilih tanpa pengembalian adalah : Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall

5-52

 Tentukan ukuran populasi dan sample :

N = 19 dan n = 6  Tentukan event of interest : P(x1 = 0, x2 = 3, x3 = 1, x4 =2) = .....  Tentukan jumlah tiap kategori tiap populasi dan sample : Honda

Suzuki

Yamaha

Kawasaki

Total

X1 = 5

X2= 4

X3 = 6

X4 = 4

N = 19

x1 = 0

X2 = 3

x3 = 1

x4 = 2

n=6


5-53

 Hitung probabilitas


5-54

Chapter Summary  Reviewed key discrete distributions  Binomial

 Poisson  Hypergeometric  Found probabilities using formulas and tables

 Recognized when to apply different distributions  Applied distributions to decision problems


5-55

# 6 Distribusi Probabilitas Diskrit

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