Transcript Poisson Distribution - Erwin Sitompul

Probability and Statistics
Lecture 7
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
2 0 1 3
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Erwin Sitompul
PBST 7/1
Chapter 5.5
Negative Binomial and Geometric Distributions
Negative Binomial Distribution
 Consider an experiment where the properties are the same as
those listed for a binomial experiment, with the exception that the
trials will be repeated until a fixed number of successes occur.
 We are interested in the probability that the kth success occurs on
the xth trial.
 This kind of experiment is called negative binomial experiment.
 The number X of trials to produce k successes in a negative
binomial experiment is called a negative binomial random
variable, and its probability distribution is called the negative
binomial distribution.
 |Negative Binomial Distribution| If repeated independent trials
can result in a success with probability p and a failure with
probability q = 1–p, then the probability distribution of the random
variable X, the number of the trial on which the kth success occurs,
is
b ( x; k , p ) 
*
k
C k 1 p q
x 1
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xk
, x  k , k  1, k  2, ...
Erwin Sitompul
PBST 7/2
Chapter 5.5
Negative Binomial and Geometric Distributions
Negative Binomial Distribution
In an NBA (National Basketball Association) championship series, the
team which wins four games out of seven will be the winner.
Suppose that team A has probability 0.55 of winning over the team
B and both teams A and B face each other in the championship
games.
(a) What is the probability that team A will win the series in six
games?
(b) What is the probability that team A will win the series?
(a)
b (6; 4, 0.55) 
*
(b) P ( te a m
4
C 4 1 (0.55) (0.45)
6 1
64
 0.1853
A w in s th e ch a m p io n sh ip se rie s )
 b (4; 4, 0.55)  b (5; 4, 0.55)  b (6; 4, 0.55)  b (7; 4, 0.55)
*
*
*
*
 0.0915  0.1647  0.1853  0.1668  0.6083
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Erwin Sitompul
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Chapter 5.5
Negative Binomial and Geometric Distributions
Negative Binomial Distribution
In an NBA (National Basketball Association) championship series, the
team which wins four games out of seven will be the winner.
Suppose that team A has probability 0.55 of winning over the team
B and both teams A and B face each other in the championship
games.
(c) If both teams face each other in a regional playoff series and the
winner is decided by winning three out of five games, what is the
probability that team A will win a playoff?
(c)
P ( te a m A w in s th e re g io n a l se rie s )
 b (3; 3, 0.55)  b (4; 3, 0.55)  b (5; 3, 0.55)
*
*
*
 0.1664  0.2246  0.2021  0.5931
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Erwin Sitompul
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Chapter 5.5
Negative Binomial and Geometric Distributions
Geometric Distribution
 If we consider the special case of the negative binomial distribution
where k = 1, we have a probability distribution for the number of
trials required for a single success.
 If repeated independent trials can result in a success with
probability p and a failure with probability q = 1–p, then the
probability distribution of the random variable X, the number of
the trial on which the first success occurs, is
g ( x ; p )  pq
x 1
, x  1, 2, 3, ...
 The mean and variance of a random variable following the
geometric distribution are
 
1
and
p
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
2

1 p
p
2
Erwin Sitompul
PBST 7/5
Chapter 5.5
Negative Binomial and Geometric Distributions
Geometric Distribution
In a certain manufacturing process it is known that, on the average,
1 in every 100 items is defective. What is the probability that the
fifth item inspected is the first defective item found?
g (5; 0.01)  (0.01)(0.99)
4
 0.0096
At “busy time” a telephone exchange is very near capacity, so callers
have difficulty placing their calls. It may be of interest to know the
number of attempts necessary in order to gain a connection.
Suppose that we let p = 0.05 be the probability of a connection
during busy time. We are interested in knowing the probability that 5
attempts are necessary for a successful call.
P ( X  5)  g (5; 0.05)  (0.05)(0.95)
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4
 0.041
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PBST 7/6
Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution and Poisson Process
 Experiments yielding numerical values of a random variable X, the
number of outcomes occurring during a given time interval or in a
specified region, are called Poisson experiments.
 The time interval may be given in any length, such as minute, day,
week, and month.
 The specified region may be a line segment, an area, a volume, or
a piece of material
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Erwin Sitompul
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Chapter 5.6
Poisson Distribution and Poisson Process
Properties of Poisson Process
 A Poisson experiment is derived from the Poisson process and
possesses the following properties:
1. The number of outcomes occurring in one time interval or
specified region is independent of the number that occurs in any
other disjoint time interval or region of space.
2. The probability that a single outcome will occur during a very
short time interval or in a small region is proportional to the
length of the time interval or the size of the region and does not
depend on the number of outcomes occurring outside this time
interval or region.
3. The probability that more than one outcome will occur in such a
short time interval or fall in such a small region is negligible.
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Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution and Poisson Process
 |Poisson Distribution| The probability distribution of the Poisson
random variable X, representing the number of outcomes
occurring in a given time interval or specified region denoted by t,
is
t
x
e ( t )
p ( x;  t ) 
,
x  0,1, 2, ...
x!
where λ is the average number of outcomes per unit time or
region, and e = 2.71828.... (natural number).
 The mean and variance of the Poisson distribution p(x;λt) both
have the value λt.
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Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution and Poisson Process
During a laboratory experiment the average number of radioactive
particles passing through a counter in 1 millisecond is 4. What is the
probability that 6 particles enter the counter in a given millisecond?
x  6,  t  4
4
p (6; 4) 
e (4)
6
6!
 0.1042
Ten is the average number of oil tankers arriving each day at a
certain port city. The facilities at the port can handle at most 15
tankers per day. What is the probability that on a given day tankers
have to be turned away?
15
P ( X  15)  1  P ( X  15)  1  
 Table A.2 gives help
p ( x ;10)
x0
 e  10 (10)1 e  10 (10) 2
 1 


1!
2!


e
 10
(10)
15 !
15



 1  0 .9 5 1 3  0.0487
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Chapter 5.6
Poisson Distribution and Poisson Process
Table A.2 Poisson Probability Sums
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Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution As a Limit of Binomial
 It should be clear from the three properties of the Poisson process
that the Poisson distribution relates to the binomial distribution.
 In the case of the binomial, if n is quite large and p is small, the
conditions begin to simulate the continuous space or time region
implications of the Poisson process.
 Poisson distribution can be taken as a limiting form of the binomial
distribution when n ∞ and p  0, and np remains constant.
 If the conditions are fulfilled, the Poisson distribution can be used
with μ = np, to approximate binomial distribution.
 Let X be a binomial random variable with probability distribution
b(x;n,p). When n ∞ and p  0, and μ = np remains constant,
b ( x; n, p )  p ( x,  )
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Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution and Poisson Process
In a certain industrial facility accidents occur infrequently. It is
known that the probability of an accident on any given day is 0.005
and accidents are independent of each other.
(a) What is the probability that in any given period of 400 days there
will be an accident on one day?
(b) What is the probability that there are at most three days with an
accident?
x  1,  t  (0.005)(400)  2
2
(a)
p (1; 2) 
e (2)
1
1!
b (1; 400, 0.005) 
(b)
 Considered as
Poisson process
 0.2707
3
P ( X  3) 

3
p ( x ; 2) 
x0
3
P ( X  3) 
1
C 1 (0.005) (0.095)
400

2
e (2)
x0
 b ( x ; 400, 0.005)
399
 0.2707
 Considered as
Bernoulli process
x
 0.8571
x!
 0.8571
x0
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Chapter 5.6
Poisson Distribution and Poisson Process
Poisson Distribution and Poisson Process
In a manufacturing process where glass products are produced,
defects or bubbles occur, occasionally rendering the piece
undesirable for marketing. It is known that, on average, 1 in every
1000 of these items produced has one or more bubbles. What is the
probability that a random sample of 8000 will yield fewer than 7
items possessing bubbles?
  (8000)(0.001)  8
6
6
P ( X  7) 
 b ( x ; 8000, 0.001)
x0
 Actually a
problem for
Binomial
Distribution
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

p ( x ; 8)  0.3134
x0
 Solved by
approximation
using Poisson
Distribution
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PBST 7/14
Probability and Statistics
Homework 6
1. A communications system consists of n components, each of which will,
independently, function with probability p. The total system will be able
to operate effectively if at least one-half of its components function. For
what values of p is a 5-component system more likely to operate
effectively than a 3-component system?
(Ro.E5.1c s144)
2. It has been established that the number of defective stereos produced
daily at a certain plant is Poisson distributed with mean 4. Over a 2-day
span, what is the probability that the number of defective stereos does
not exceed 3?
(Ro.E5.2f s+10)
3. The probability of hitting a target is 1/5 and ten shots are fired
independently.
(a) What is the probability of the target being hit at least twice?
(b) Find the conditional probability that the target is hit at least twice,
assuming that at least one hit is scored.
(Fe.VI.10.5-6 s16.9)
 This homework includes the materials from
Lecture 6 and Lecture 7.
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