The Poisson Distribution (Group 17)

7/29/2019 The Poisson Distribution (Group 17)

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THE POISSON DISTRIBUTION

72. Mohamad Hanif Asyraf Bin Hussian UK27549

45. Wan Nurul Atiqah Binti Wan Azman UK27510

62. Nurul Haziqah Binti Jamal UK27529

111. Nur Aina Binti Mohd Azlan Jamal UK27612 86. Nuraini Binti Sapari UK27574

101.Tuan Masyitah Binti Tuan Sulong UK27600


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DISTRIBUTION



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In many practical situations we are interested in measuring

how many times a certain event occurs in a specific time

interval or in a specific length or area. For instance:

1 the number of phone calls received at an exchange

or call centre in an hour;

2 the number of customers arriving at a toll booth per

day; 3 the number of flaws on a length of cable;

4 the number of cars passing using a stretch of road

during a day.

The Poisson distribution plays a key role in modelling such

problems.



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The Poisson distribution is a discrete probability

distribution for the counts of events that occur randomly

in a given interval of time (or space). A Poisson random

variable takes on positive values (or zero).

If we let X = The number of events in a given

interval,

Then, if the mean number of events per interval is .

X has a Poisson Distribution with parameter and

P(X = x) =

!

= 0, 1, 2, 3, 4,

Note is a mathematical constant. 2.718282. There

should be a button on your calculator that calculates

powers of.



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EXAMPLES;

1. Consider, in an office 2 customers arrived today.Calculate the possibilities for exactly 3 customers to be

arrived on tomorrow.Step1: Find

.where, = 2and = 2.718

= 2.718 = 0.135.

Step2: Find .where, = 2 and = 3. = 23 = 8.

Step3: Find P(X=x) =

!

P(X=3) =(.)()

!= 0.18.

Hence there are 18% possibilities for 3 customers to be

arrived on tomorrow.THE POISSON DISTRIBUTION


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2. Births in a hospital occur randomly at an average rate of

1.8 births per hour. What is the probability of observing 4

births in a given hour at the hospital?

Let X = No. of births in a given hour = 4

(i) Events occur randomly

(ii) Mean rate = 1.8

We can now use the formula to calculate the

probability of observing exactly 4 births in a given hour

P(X = 4) =

1.8

.

4! = 0.0723



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b) What about the probability of observing more than or

equal to 2 births in a given hour at the hospital?

We want P(X 2) = P(X = 2) + P(X = 3) +

i.e. an infinite number of probabilities to calculate but

P(X 2) = P(X = 2) + P(X = 3) + ..

= 1 ( < 2)

= 1 (( = 0) + ( = 1))

= 1 ( ..

!+ .

.

!)

- = 1 (0.16529 + 0.29753)

= 0.537



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3. The number of visitors to a webserver per minute follows aPoisson distribution. If the average number of visitors perminute is 4, what is the probability that:

(a) There are two or fewer visitors in one minute?(b) There are exactly two visitors in 30 seconds?

For part (a), we need the average number of visitors in aminute. In this case the parameter = 4. We wish to calculate;

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

P(X=0) = 4

!

P(X=1) = 4

!

P(X=0) = 4

!

So the probability of two or fewer visitors in a minute is

+4+= 0.238THE POISSON DISTRIBUTION


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b) If the average number of visitors in 1 minute is 4, the

average in 30 seconds is 2.

So for this example, our parameter = 2.

P(X=2) =

!

=

= 0.271



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The mean and variance of a Poisson random

variable with parameter are both equal to :

(X) = , (X) =

Or known as


= = =


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Examples;

Suppose we know that births in a hospital occur randomly atan average rate of 1.8 births per hour. What is the probability

that we observe 5 births in a given 2 hour interval?

Let X = No. of births in a 2 hour period = 5

= *1.82 = 3.6

Then,

P(X=5) = ..

! = 0.13768


* Well, if births occur randomly at arate of 1.8 births per 1 hour interval.

Then births occur randomly at a

rate of 3.6 births per 2 hour interval.


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Where does the Poisson distribution come from?

Mathematical fact: The Poisson distribution is anapproximation for the binomial distribution Bin(n,p)

when:

n is large;

p is small; np is close to .

In other words, its like having lots of trials where the

expected number of successes is


* = np


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EXAMPLES;

1. It is known that 3% of the circuit boards from a production

line are defective. If a random sample of 120 circuit boards

is takenfrom this production line, use the Poissonapproximation to stimate the probability that the sample

contains:

(i) Exactly 2 defective boards.

(ii) At least 2 defective boards.

In this case, n 100 and np 10. Also,

= np = 120(0.03) = 3.6

(i) P(X = 2) = ..

!= 0.177

Binomial calculation also gives an answer of1202(0.03)

2(1 0.03)120 2= 0.1766


()


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(ii) P(X2) = 1 (P(X = 0) + P(X = 1))

= 1 [ (..

!) +(.

.

!)]

= 1 (0.027 + 0.098)

= 0.875

= 0.88

Binomial distribution gives an answer of1 [1200(. )

(.)+1201(. ) ( . )]

1 0.02585 + 0.09597 = 0.878= 0.88


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The Poisson Distribution (Group 17)