Transcript Poisson distribution powerpoint

THE POISSON DISTRIBUTION
The Poisson random variable was first introduced by the
French mathematician Simeon-Denis Poisson (1781-1840). He
discovered it as a limit to the binomial distribution as the
number of trials n approaches infinity.
The Poisson distribution is used to determine the probability of
obtaining a certain number of successes within a certain
interval (of time or space).
Introduction to the Poisson Distribution
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
Poisson distribution is a distribution of a discrete
random variable X—if events happen at a
constant rate over time, the Poisson distribution
gives the probability of X number of events
occurring within a certain interval (of time or
space).
Possible examples are when describing the
number of flaws in a given length of material;
accidents on a particular stretch of road in a week,
telephone calls made to a switchboard in one day;
the number of misprints on a typical page of a
book; the number of fish caught in a large lake
per day.
X
Po (m)
Poisson Distribution, example
The probability distribution function for the discrete Poisson random
variable is
x m
me
Po ( X  x) 
x!
where m is the parameter of the distribution.
Formula Booklet
the parameter is called m.
Poisson distribution
x m
X
me
Po (m)  P ( X  x) 
, x  0,1, 2,...
x!
Poisson Mean and Variance


Mean
m
Variance and Standard
Deviation Var ( X )  m
For a Poisson
random variable,
the variance and
mean are the
same!
 m
where m = expected number of hits in a given time
period, it is called the parameter.
Example

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The number of accidents in a randomly
chosen day at a road crossing can be
modelled by a Poisson distribution with
parameter 0.5. Find the probability that,
on a randomly chosen day, there are:
a) exactly two accidents
b) at least two accidents
Using Poisson pdf on GDC, part a:
P(X=2)=0.0758 (3 sf)
Part b)
P( X  2)  1  P( X  1)
Using Poisson cdf on GDC:
Example

For example, if new cases of West Nile
Virus in New England are occurring at a
rate of about 2 per month, then these
are the probabilities that: 0,1, 2, 3, 4,
5, 6, to 1000 to 1 million to… cases will
occur in New England in the next
month:
Poisson Probability table
x
0
P(X=x)
2 0 e 2
=.135
0!
1
2 1 e 2
=.27
1!
2
2 2 e 2
2!
=.27
3
2 3 e 2
=.18
3!
4
=.09
5
…
…
Using GDC Poisson pdf with the list:
Example: Poisson distribution
Suppose that a rare disease has an incidence of 1 in 1000 personyears. Assuming that members of the population are affected
independently, find the probability of k cases in a population of
10,000 (followed over 1 year) for k=0,1,2.
The expected value (mean) = = .001*10,000 = 10
10 new cases expected in this population per year.
(10) 0 e  (10)
P ( X  0) 
 .0000454
0!
(10)1 e (10)
P( X  1) 
 .000454
1!
(10) 2 e (10)
P ( X  2) 
 .00227
2!
Finding the parameter

If X Po ( ) find the value of the
parameter correct to 4 decimal
places, given that P( X  1)  0.25
Using Numerical solve on GDC:
   0.3574
Mean and variance of Poisson distribution
X
Po (m), then E ( X )  Var ( X )  m
If X Po (m) and X is defined on an
interval of length l, to find the
probability in an interval of length
kl we use another Poisson
distribution with mean k m.
Example
In a book of 520 pages there are 135 misprints.
a)
What is the mean number of misprints per page?
b)
Find the probability that pages 444 and 445 do not
contain any misprints?
Let X
Po ( )
a) E ( X )   
135 27

 0.2596
520 104
Let X be the number of misprints found on 2 pages.
X
Po (2 )
We need to find P(X=0), using
GDC:
more on Poisson…
“Poisson Process” (rates)
Note that the Poisson parameter  can be given as
the mean number of events that occur in a defined
time period OR, equivalently,  can be given as a
rate, such as =2/month (2 events per 1 month) that
must be multiplied by t=time (called a “Poisson
Process”) 
X ~ Poisson ()
(t ) k e  t
P( X  k ) 
k!
E(X) = t
Var(X) = t
Example
For example, if new cases of West Nile in
New England are occurring at a rate of about
2 per month, then what’s the probability that
exactly 4 cases will occur in the next 3
months?
X ~ Poisson (=2/month)
(2*3) 4 e  (2*3) 64 e  (6)
P(X  4 in 3 months) 

 0.134
4!
4!
Exactly 5 cases?
On Sunday mornings, cars arrive at a petrol station at an average rate
of 30 per hour. Assuming that the number of cars arriving at the petrol
station follows a Poisson distribution, find the probability that:
a) in a half-hour period 12 cars arrive
b) no cars arrive during a particular 5-minute interval
c) more than 4 cars arrive in a 15 minute interval
We need to calculate a new parameter
for each time interval as follows:
a) 0.5 times 30 = 15
b) 5 times 30/60=2.5
c) 15 times 30/60=7.5
And find each probability using GDC:
Practice problems
If calls to your cell phone are a Poisson
process with a constant rate =2 calls per
hour, what’s the probability that, if you forget
to turn your phone off in a 1.5 hour movie,
your phone rings during that time?
Answer
If calls to your cell phone are a Poisson process with a constant
rate =2 calls per hour, what’s the probability that, if you forget
to turn your phone off in a 1.5 hour movie, your phone rings
during that time?
X ~ Poisson (=2 calls/hour)
P(X≥1)=1 – P(X=0)
The random variable X follows the Poisson distribution with mean m and it
satisfies the following equation:
P( X  0)  P( X  1)  P( X  4)  0
a) Find the value of m correct to 4 decimal places
b) Calculate
P(2  X  4).
a) Using the formula we can write the following equation:
e  m m0 e  m m1 e  m m 4


0
0!
1!
4!
4
m
e  m (1  m  )  0
24
Now use numerical solve on GDC:
 m  3.1613 (4 d . p.)
b) Using GDC with storing the value of m:
P(2  X  4)  0.611 (3 s. f .)