the poisson distribution (group 17)
TRANSCRIPT
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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
THE POISSON 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.
THE POISSON DISTRIBUTION
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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.
THE POISSON DISTRIBUTION
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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
THE POISSON DISTRIBUTION
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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
THE POISSON DISTRIBUTION
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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
THE POISSON DISTRIBUTION
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The mean and variance of a Poisson random
variable with parameter are both equal to :
(X) = , (X) =
Or known as
THE POISSON DISTRIBUTION
= = =
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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
THE POISSON DISTRIBUTION
* 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
THE POISSON DISTRIBUTION
* = 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
THE POISSON DISTRIBUTION
()
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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
THE POISSON DISTRIBUTION