GOALS

1. Define the terms probability distribution and random variables.

2. Distinguish between a discrete and continuous probability distributions.

3. Calculate the mean, variance, and standard deviation of a discrete probability distribution.

4. Binomial probability distribution.5. Hypergeometric distribution. 6. Poisson distribution.

Chapter SixDistribusi Probabilitas DiskretDistribusi Probabilitas Diskret

Distribusi Probabilitas

Distribusi Probabilitas: daftar seluruh hasil percobaan beserta probabilitas untuk masing-masing hasil.

Karakteristik Distribusi Probabilitas: Probabilitas sebuah hasil adalah antara 0 dan 1 Semua kejadian (event) adalah mutually

exclusive Jumlah probabilitas semua kejadian (event)

yang mutually exclusive=1 (collectively exhaustive)

Contoh: Eksperimen melempar koin 3 kali. Keluarnya Head (H) menjadi fokus, misalnya X adalah kejadian keluar Head (H). H: hasil lemparan head dan T: hasil lemparan tail.Maka, akan ada 8 kemungkinan hasil.

Contoh: Eksperimen melempar koin tiga kali

Possible Result

Lemparan Koin Number of HeadsPertama Kedua Ketiga

1 T T T 0

2 T T H 1

3 T H T 1

4 H T T 1

5 T H H 2

6 H T H 2

7 H H T 2

8 H H H 3

Number of Heads

(X)

Probability of Outcomes

P(X)

0 1/8 = 0,125

1 3/8 = 0.375

2 3/8 = 0.375

3 1/8 = 0,125

Total 1

Distribusi Probabilitas

Random Variables (Variabel Acak)

Variable Acak adalah nilai numerik yang ditentukan oleh hasil suatu eksperimen. Nilainya bisa bermacam-macam.

Contoh: Jumlah siswa yang absen pada hari ini,

angkanya mungkin 0, 1, 2, 3… dll. angka absen adalah variabel acak

Berat tas yang dibawa mahasiswa, mungkin 2,5 kg; 3,2kg, … dll.berat tas adalah variabel acak.

Types of Random Variables

A discrete random variable can assume only certain outcomes. Usually data was obtained by counting.

A continuous random variable can assume an infinite number of values within a given range. Usually data was obtained by measuring.

Examples of a discrete random variable: The number of students in a class. The number of children in a family. The number of cars entering a carwash in a

hour. Number of home mortgages approved by

Coastal Federal Bank last week. Number of CDs you own. Number of trips made outside Hong Kong in

the past one year. The number of ten-cents coins in your pocket.

Types of Random Variables

Types of Probability Distributions Examples of a continuous random

variable: The distance students travel to class. The time it takes an executive to drive to

work. The length of an afternoon nap. The length of time of a particular phone

call. The amount of money spent on your last

haircut.

The Mean of a Discrete Probability Distribution The mean:

reports the central location of the data. is the long-run average value of the

random variable. is also referred to as its expected value,

E(X), in a probability distribution. is a weighted average.

The Mean of a Discrete Probability Distribution

The mean is computed by the formula:

where represents the mean and P(x) is the probability of the various outcomes x.

Σ[xP(x)]μ

Similar to the formula for computing grouped mean where P(x) is replaced by relative frequency.

The Variance of a Discrete Probability Distribution

The variance measures the amount of spread (variation) of a distribution.

The variance of a discrete distribution is denoted by the Greek letter 2 (sigma squared).

The standard deviation is the square root of 2

The Variance & standard deviation of a Discrete Probability Distribution The variance of a discrete probability

distribution is computed from the formula:

The stadard deviation is the square root of 2

P(x)]μ)Σ[(xσ 22

Similar to the formula for computing grouped variance where P(x) is replaced by relative frequency.

2σ σ

EXAMPLE 2 Arman, owner of College Painters, studied his

records for the past 20 weeks and reports the following number of houses painted per week:

Set the probability distribution Compute mean and variance

# o f H o u s e s P a i n t e d Weeks to finish

10 5

11 6

12 7

13 2

EXAMPLE 2 continued Probability Distribution:

Number of houses painted, x W e e k s Probability, P(x)

10 5 .25 11 6 .30

12 7 .35

13 2 .10 Total 20 1.00

EXAMPLE 2 continued Compute the mean number of houses

painted per week:

11.3

(13)(.10)(12)(.35)(11)(.30)(10)(.25)

Σ[xP(x)]E(x)μ

x Week P(x) x.P(x)

10 5 0.25 2.5

11 6 0.30 3.3

12 7 0.35 4.2

13 2 0.10 1.3

Total 20 1 11.3

EXAMPLE 2 continued Compute the variance of the number

of houses painted per week:

0.91

0.28900.17150.02700.4225

(.10)11.3)(13...(.25)11.3)(10

P(x)]μ)Σ[(xσ22

22

x Week

P(x) x.P(x) x-µ (x-µ)2 (x-µ)2.P(x)

10 5 0.25 2.5 -1.3 1.69 0.42

11 6 0.30 3.3 -0.3 0.09 0.03

12 7 0.35 4.2 0.7 0.49 0.17

13 2 0.10 1.3 1.7 2.89 0.29

Total 20 1 11.3 0.91

Types of Probability Distributions

Discrete Probability Distributions Binomial Probability Distributions Hypergeometric Probability

Distributions Poisson Probability Distributions

Continuous Probability Distributions Normal Probability Distributions

Binomial Probability Distribution

The binomial distribution has the following characteristics: An outcome of an experiment is classified

into one of TWO mutually exclusive categories, such as a success or failure.

The data collected are the results of counting the success event of some trial.

The probability of success stays the same for each trial.

The trials are independent.

Binomial Probability Distribution

To construct a binomial distribution, let n be the number of trials x be the number of observed successes be the probability of success on each

trial

The formula for the binomial probability distribution is:

P(x) = nCx x(1- )n-x

Binomial Probability Distribution

The formula for the binomial probability distribution is:P(x) = nCx x(1- )n-x

TTT, TTH, THT, THH, HTT, HTH, HHT, HHH.

X=number of heads The coin is fair, i.e., P(head) = 1/2.

P(x=0) = 3C0 0.5 0(1- 0.5)3-0 =0.125=1/8

P(x=1) = 3C1 0.5 1(1- 0.5)3-1 =0.375=3/8

P(x=2) = 3C2 0.5 2(1- 0.5)3-2 =0.375=3/8

P(x=3) = 3C3 0.5 3(1- 0.5)3-3 =0.125=1/8

When the coin is not fair, simple counting rule will not work.

EXAMPLE 3

The Department of Labor reports that 20% of the workforce in Surabaya is unemployed. From a sample of 14 workers, calculate the following probabilities: Exactly three are unemployed. At least three are unemployed. At least one are unemployed.

EXAMPLE 3 continued

The probability of exactly 3:

The probability of at least 3 is:

2501.

)0859)(.0080)(.364(

)20.1()20(.)3( 113314

CP

3 11 14 014 3 14 14

( 3) (3) (4) (5) ....... (14)

(.20) (.80) ... (.20) (.80)

.250 .172 ... .000 0.551

P x P P P P

C C

The Department of Labor reports that 20% of the workforce in Surabaya is unemployed. From a sample of 14 workers

Example 3 continued

The probability of at least one being unemployed.(1) (2) .... (14)P P P

0 1414 0

P(x 1)

1 P(0)

1 C (.20) (1 .20)

1 .044 .956

The Department of Labor reports that 20% of the workforce in Surabaya is unemployed. From a sample of 14 workers

Mean & Variance of the Binomial Distribution The mean is found by:

The variance is found by:

n

)1(2 n

EXAMPLE 4 From EXAMPLE 3, recall that =.2 and

n=14.

Hence, the mean is:= n = 14(.2) = 2.8.

The variance is:2 = n (1- ) = (14)(.2)(.8)

=2.24.

Contoh

Probabilitas kerusakan pada barang yang diproduksi Perusahaan “X” adalah 10%. Jika diambil 6 sampel random, maka : Buatlah distribusi probabilitas Hitung rata-rata dan standar deviasi

probabilitas tersebut

Jumlah Barang Rusak (X)

Probabilitas, P(X)

0 P(0)=6C0 0.10(1- 0.1)6-0 = 0,531

1 P(1)=6C1 0.11(1- 0.1)6-1 = 0,354

2 P(2)=6C2 0.12(1- 0.1)6-2 =0,098

3 0,015

4 0,001

5 0,000

6 0,000

Total 160,010,0*6 n

73,054,0

54,0)9,01(10,0*6)1(2

n

Soal

Berdasarkan data yang ada, probabilitas mahasiswa lulus Mata Kuliah Statistik adalah 70%. Jika diambil sampel random 10 mahasiswa, hitung probabilitas :

1. 6 mahasiswa lulus2. 3 mahasiswa tidak lulus3. Kurang dari 9 mahasiswa lulus4. Paling banyak 2 mahasiswa tidak

lulus

Soal• Mahasiswa Lulus n=10; =0.71. 6 mahasiswa lulus P(6)2. 3 mahasiswa tidak lulus = 7 mahasiswa lulus

dengan =0.7 gunakan x =10-3=7P(7)atau dengan =1-0.7=0.3 P(3)

3. Kurang dari 9 mahasiswa lulus P(x<9)=P(8)+P(7)+…+P(0)atau P(x<9)=1-P(9)+P(10)

4. Paling banyak 2 mahasiswa tidak lulus = paling banyak 8 mahasiswa lulus

dengan =0.3 P(x ≤2)=P(2)+P(1)+P(0)atau dengan =0.7 P(x≤8)=P(8)+…+P(0) =1-P(9)+P(10)

Hypergeometric Distribution Use the hypergeometric distribution to

find the probability of a specified number of successes or failures if: the sample is selected from a finite

population without replacement (recall that a criteria for the binomial distribution is that the probability of success remains the same from trial to trial)

the size of the sample n is greater than 5% of the size of the population N .

Hypergeometric Distribution

The hypergeometric distribution has the following characteristics: There are only 2 possible outcomes, eg.

Success or failure It results from a count of the number of

successes in a fixed number of trials (number of success is the Random variable)

The probability of a success is not the same on each trial without replacement, thus events are not independent

EXAMPLE 8 of last lecture

R1

B1

R2

B2

R2

B2

7/12

5/12

6/11

5/11

7/11

4/11

In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement.

The probability of a success (red chip) is not the same on each trial.

Hypergeometric Distribution The formula for finding a probability

using the hypergeometric distribution is:

where N is the size of the population, S is the number of successes in the population, x is the number of successes in a sample of n observations.

nN

xnSNxS

C

CCxP

))(()(

EXAMPLE 5 The National Air Safety Board has a list of 10

reported safety violations. Suppose only 4 of the reported violations are actual violations and the Safety Board will only be able to investigate five of the violations. What is the probability that three of five violations randomly selected to be investigated are actually violations?

4 3 10 4 5 3

10 5

4 3 6 2

10 5

( )( )(3)

( )( ) 4(15).238

252

C CP

C

C C

C

Contoh

Perusahaan “X” mempunyai 50 karyawan, 40 diantaranya bergabung dalam Serikat Kerja. Jika diambil 5 sampel random, maka :

1. Berapa probabilitas 4 karyawan bergabung dalam Serikat Kerja

2. Buat distribusi probabilitas

431,0760.118.2

)10)(390.91())(()4(

550

454050440

C

CCP

Jumlah Karyawan (X) Probabilitas, P(X)

0 0,000

1 0,004

2 0,044

3 0,210

4 0,431

5 0,311

Total 1

POISSON DISTRIBUTION

Poisson Distribution

1. Number of events that occur in an interval • events per unit

— Time, Length, Area, Space

2. Examples Number of customers arriving in 20

minutes Number of strikes per year in the U.S. Number of defects per lot (group) of

DVD’s

Poisson Process

1. Constant event probability

Average of 60/hr is1/min for 60 1-minuteintervals

2. One event per interval Don’t arrive together

3. Independent events Arrival of 1 person does

not affect another’sarrival

© 1984-1994 T/Maker Co.

Poisson Probability Distribution Function

p(x) = Probability of x given = Expected (mean) number of ‘successes’e = 2.71828 (base of natural logarithm)x = Number of ‘successes’ per unit

p xx

( )!

x e-

Poisson Distribution Example

Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes?

© 1995 Corel Corp.

Poisson Distribution Solution

72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

-

4 -3.6

( )!

3.6(4) .1912

4!

x ep x

x

ep

Thinking Challenge

You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction?

© 1984-1994 T/Maker Co.

Poisson Distribution Solution: Finding *

75 words/min = (75 words/min)(60 min/hr)

= 4500 words/hr

6 errors/hr = 6 errors/4500 words= .00133 errors/word

In a 255-word transaction (interval): = (.00133 errors/word )(255 words)

= .34 errors/255-word transaction

Poisson Distribution Solution: Finding p(0)*

-

0 -.34

( )!

.34(0) .7118

0!

x ep x

x

ep

- END -

Chapter SixDiscrete Probability DistributionsDiscrete Probability Distributions