This document discusses discrete probability distributions. It defines key terms like probability distribution, random variables, and types of random variables. It also covers calculating the mean, variance, and standard deviation of discrete probability distributions. Specific discrete probability distributions covered include the binomial, hypergeometric, and Poisson distributions. Examples are provided to demonstrate calculating probabilities and distribution properties.

1. Chapter Six
Distribusi Probabilitas DDiisskkrreett
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.

2. 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 (c o lle c tive ly e xha us tive )

3.  Contoh:
Eksperimen melempar koin 3 kali. Keluarnya
He a d (H) menjadi fokus, misalnya X adalah
kejadian keluar He a d (H).
H: hasil lemparan he a d dan T: hasil lemparan
ta il.
Maka, akan ada 8 kemungkinan hasil.

4. Contoh: Eksperimen melempar koin
tiga kali
Possible
Result
Lemparan Koin Number
Pertama Kedua Ketiga of Heads
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

5. Distribusi Probabilitas
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

6. 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.

7. 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.

8. Types of Random Variables
 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.

9. 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.

10. 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.

11. The Mean of a Discrete Probability
Distribution
The mean is computed by the formula:
μ = Σ[xP(x)]
where m represents the mean and P(x ) is the
probability of the various outcomes x .
Similar to the formula for computing grouped
mean where P(x) is replaced by relative
frequency.

12. 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 s2 (sigma
squared).
 The standard deviation is the square root
of s2

13. The Variance & standard deviation
of a Discrete Probability
Distribution
 The variance of a discrete probability
distribution is computed from the formula:
σ2 = Σ[(x - μ)2P(x)]
 The stadard deviation is the square root of s2
σ = σ2
Similar to the formula for computing grouped
variance where P(x) is replaced by relative
frequency.

14. 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:
# 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
 Set the probability distribution
 Compute mean and variance

15. EXAMPLE 2 c o ntinue d
 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

16. EXAMPLE 2 c o ntinue d
 Compute the mean number of houses painted
per week:
μ = E(x) =
Σ[xP(x)]
(10)(.25) (11)(.30) (12)(.35) (13)(.10)
= + + +
11.3
=
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

17. EXAMPLE 2 c o ntinue d
 Compute the variance of the number of
houses painted per week:
σ = Σ[(x -
μ) P(x)]
(10 11.3) (.25) ... (13 11.3) (.10)
= - + + -
0.4225 0.0270 0.1715 0.2890
= + + +
0.91
2 2
2 2
=
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

18. Types of Probability Distributions
 Discrete Probability Distributions
 Binomial Probability Distributions
 Hypergeometric Probability Distributions
 Poisson Probability Distributions
 Continuous Probability Distributions
 Normal Probability Distributions

19. Binomial Probability Distribution
 The binomial distribution has the following
characteristics:
 An outcome of an experiment is classified into
o ne o f TWO m utua lly e x c lus ive c a te g o rie s ,
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.

20. Binomial Probability Distribution
 To construct a binomial distribution, let
 n be the number of trials
 x be the number of observed successes
 p be the probability of success on each trial
 The formula for the binomial probability
distribution is:
P(x) = nCx p x(1- p)n-x

21. Binomial Probability Distribution
 The formula for the binomial probability
distribution is:
P(x) = nCx p x(1- p)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.

22. 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.

23. EXAMPLE 3 c o ntinue d
The Department of Labor reports that 20% of the workforce in
Surabaya is unemployed. From a sample of 14 workers
 The probability of exactly 3:
(3) (.20)3 (1 .20)11
P = C -
14 3
(364)(.0080)(.0859)
.2501
=
=
 The probability of at least 3 is:
P x ³ = P + P + P + +
P
( 3) (3) (4) (5) ....... (14)
3 11 14 0
C C
= + +
= + + + =
(.20) (.80) ... (.20) (.80)
14 3 14 14
.250 .172 ... .000 0.551

24. Example 3 c o ntinue d
The Department of Labor reports that 20% of the workforce in
Surabaya is unemployed. From a sample of 14 workers
 The probability of at least one being
unemp³loye=d.P(1) + P(2) +....+ P(14)
= -
= - -
= - =
0 14
14 0
P(x 1)
1 P(0)
1 C (.20) (1 .20)
1 .044 .956

25. Mean & Variance of the Binomial
Distribution
 The mean is found by:
m =np
 The variance is found by:
s 2 = np (1-p )

26. EXAMPLE 4
 From EXAMPLE 3, recall that p =.2 and n=14.
 Hence, the mean is:
m= n p = 14(.2) = 2.8.
 The variance is:
s2 = n (1- p ) = (14)(.2)(.8) =2.24.

27. 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

28. 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 1
m = np = 6*0,10 = 0,60
2 (1 ) 6*0,10(1 0,9) 0,54
= =
s np p
= - = - =
0,54 0,73
s

29. Soal
 Berdasarkan data yang ada, probabilitas
mahasiswa lulus Mata Kuliah Statistik
adalah 70%. Jika diambil sampel random
10 mahasiswa, hitung probabilitas :
1. 6 mahasiswa lulus
2. 3 mahasiswa tidak lulus
3. Kurang dari 9 mahasiswa lulus
4. Paling banyak 2 mahasiswa tidak lulus

30. Soal
• Mahasiswa Lulus n=10; p=0.7
1. 6 mahasiswa lulus P(6)
2. 3 mahasiswa tidak lulus = 7 mahasiswa lulus
dengan p=0.7 gunakan x =10-3=7P(7)
atau dengan p=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 p=0.3 P(x ≤2)=P(2)+P(1)+P(0)
atau dengan p=0.7 P(x≤8)=P(8)+…+P(0) =1-P(9)+P(10)

31. 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 .

32. 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

33. EXAMPLE 8 o f la s t le c ture
In a bag containing 7 red chips and 5 blue chips you
select 2 chips one after the other without
replacement.
R1
B1
R2
B2
R2
B2
7/12
5/12
6/11
5/11
7/11
4/11
The probability of a success (red chip) is not the same on each trial.

34. Hypergeometric Distribution
 The formula for finding a probability using the
hypergeometric distribution is:
P(x) = ( C )( C )
S x N - S n - x
C
N n
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.

35. 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?
(3) = ( )( )
4 3 10 - 4 5 -
3
10 5
( )( ) 4(15) 4 3 6 2
.238
10 5
252
P C C
C
C C
C
= = =

36. 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

37. 0,431
P C C
(4) ( )( ) (91.390)(10)
C
= 40 4 50 40 5 - 4 = = 2.118.760
-50 5
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

38. POISSON DISTRIBUTION

39. Poisson Distribution
1. Number of events that occur in an interval
• events per unit
— Time, Length, Area, Space
1. 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

40. Poisson Process
1. Constant event
probability
 Average of 60/hr is
1/min for 60 1-minute
intervals
1. One event per interval
 Don’t arrive together
1. Independent events
 Arrival of 1 person does
not affect another’s
arrival
© 1984-1994 T/Maker Co.

41. Poisson Probability Distribution
Function
p x
lxe-l
x
( )
!
=
p(x) = Probability of x given l
l = Expected (mean) number of
‘successes’
e = 2.71828 (base of natural logarithm)
x = Number of ‘successes’ per unit

42. 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.

43. 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 p x e
x
e
p
l l =
= =

44. 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.

45. Poisson Distribution Solution: Finding l*
 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):
l = (.00133 errors/word )(255 words)
= .34 errors/255-word transaction

46. Poisson Distribution Solution:
Finding p(0)*
-
x p x e
( )
x
e
0 -.34
( )
!
.34
(0) .7118
0!
p
l l =
= =

47. Chapter Six
Discrete Probability DDiissttrriibbuuttiioonnss
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