Poisson lecture

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The Poisson Experiment

• Used to find the probability of a rare event

• Randomly occurring in a specified interval

– Time

– Distance

– Volume

• Measure number of rare events occurred in the

specified interval

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The Poisson Experiment - Properties

• Counting the number of times a success occur in an interval

• Probability of success the same for all intervals of equal size

• Number of successes in interval independent of number of successes in other intervals

• Probability of success is proportional to the size of the interval

• Intervals do not overlap

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The Poisson Experiment - Properties

• Some of the properties for a Poisson experiment can be hard to identify.

• Generally only have to consider whether:- – The experiment is concerned with counting number of

RARE events occur in a specified interval

– These RARE events occur randomly

– Intervals within which RARE event occurs are independent and do not overlap

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Typical cases where the Poisson experiment

applies:

– Accidents in a day in Soweto

– Bacteria in a liter of water

– Dents per square meter on the body of a car

– Viruses on a computer per week

– Complaints of mishandled baggage per 1000

passengers

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( ) ( ) !

xeP X x P x

x

Determining x successes in the interval:

where, = mean number of successes in interval = base of natural logarithm 2.71828

e

Calculating the Poisson Probability


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• Time interval of one week

• μ = 2 per week

• Occurrence of viruses are independent

• Can calculate the probabilities of a certain number of viruses in the interval

Are the conditions required for the Poisson experiment met?

• On average the anti-virus program detects 2 viruses per week on a notebook

The Poisson Experiment - Example

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• Let X be the Poisson random variable indicating the

number of viruses found on a notebook per week

2 02

0!( 0) (0) 0.1353

eP X P


Calculate the probability that zero viruses occur Calculate the probability that one virus occur Calculate the probability that three viruses occur

2 12

1!( 1) (1) 0.2707

eP X P

2 32

3!( 3) (3) 0.1804

eP X P

2 72

7!( 7) (7) 0 0034

.

.

.

.e

P X P

( ) ( ) !

xeP X x P x

x

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2 02

0!( 0) (0) 0.1353

eP X P


2 12

1!( 1) (1) 0.2707

eP X P

2 32

3!( 3) (3) 0.1804

eP X P

2 72

7!( 7) (7) 0 0034

.

.

.

.e

P X P

X P(X)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

↓ ↓

7 0.0034

↓ ↓

∑P(X) ≈ 1

• Let X be the Poisson random variable indicating the

number of viruses found on a notebook per week

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• Calculate the probability that two or less than two

viruses will be found per week


X P(X)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

↓ ↓

7 0.0034

↓ ↓

∑P(X) ≈ 1

P(X ≤ 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0.1353 + 0.2707 + 0.2707

= 0.6767

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• Calculate the probability that less than three



X P(X)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

↓ ↓

7 0.0034

↓ ↓

∑P(X) ≈ 1

P(X < 3)

= P(X = 0) + P(X = 1) + P(X = 2)

= 0.1353 + 0.2707 + 0.2707

= 0.6767

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• Calculate the probability that more than three



X P(X)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

↓ ↓

7 0.0034

↓ ↓

∑P(X) ≈ 1

P(X > 3)

= P(X = 4) + P(X = 5) + ………

= 1 – P(X ≤ 3)

= 1 - 0.8571

=0.1429

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• Calculate the probability that four viruses will be found in four weeks

• μ = 2 x 4 = 8 in four weeks


8 4

( 4)

8

4!0.0573

P X

e

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• Calculate the probability that two or less than two viruses will be found in two weeks

• μ = 2 x 2 = 4 in two weeks


4 0 4 1 4 2

( 2)

( 0) ( 1) ( 2)

4 4 4

0! 1! 2!0.0183 0.0733 0.2381

0.2381

P X

P X P X P X

e e e

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• Mean and standard deviation of Poisson

random variable


( )E X

2 ( )Var X

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– What is the expected number of viruses on the

notebook per week?

– What is the standard deviation for the number of

viruses on the notebook per week?

( ) 2E X

2 ( ) 2 1.41Var X


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Poisson lecture