Transcript Example - Mags Maths

Poisson Distribution
Poisson
Poisson 1781-1840
Definition
• The Poisson distribution is a discrete
probability distribution.
• It expresses the probability of a number
of events occurring in a fixed time if
these events occur with a known
average rate and are independent of the
time since the last event.
Examples
• The number of emergency calls
received by an ambulance control in an
hour.
• The number of vehicles approaching a
motorway toll bridge in a five-minute
interval.
• The number of flaws in a metre length
of material.
Assumptions
• Each occurs randomly.
• Events occur singly in a given interval of
time or space.
• The mean number of occurrences in the
given interval, is known and finite.
l
Probability
e l
P( X = x) =
x!
-l x
Example 1
• A student finds that the average number of
amoebas in 10 ml of pond water from a
particular pond is four. Assuming that the
number of amoebas follows a Poisson
distribution.
• Find the probability that in a 10 ml sample
• a) there are exactly five amoebas.
• b) There are no amoebas
• c) There are fewer than three amoebas.
a) Find the probability that in a 10 ml sample
there are exactly five amoebas.
-4 5
e 4
P( X = 5) =
= 0.1563
5!
b) Find the probability that in a 10 ml sample
there are no amoebas
-4 0
e 4
P( X = 0) =
= 0.0183
0!
c) Find the probability that in a 10 ml sample
there are fewer than three amoebas.
P( X < 3) = P( X = 0) + P( X = 1) + P( X = 2)
-4 0
-4 1
-4 2
e 4 e 4 e 4
=
+
+
= 0.2381
0!
1!
2!
Useful probabilities to note
-4 0
e 4
-4
P( X = 0) =
=e
0!
-4 1
e 4
-4
P( X =1) =
= 4e
1!
Graphics Calc Poisson Dist
Stats Mode
from Calc
F5 Distribution
Then F6 next
F1 Poisn
For point dist
select Ppd
For cumulative
select Pcd after
Poisn
For P(X<3),
x=2
l or m = 3
Use P(x £ 2)
Select Data: Variable
For P(X=1),
l or m = 3
For P(X=1) = 0.1494
P(x < 3) = 0.4232
P(x £ 2) = 0.4232
Since tables only go to
4 dp, round to 4dp
Note the following:
• 1. The binomial distribution is affected by
the sample size and the probability while
the Poisson distribution is ONLY affected
by the mean.
• 2. The binomial distribution has values
from x = 0 to n but the Poisson distribution
has values from x = 0 to infinity.
Example: On average there are three
babies born a day with hairy backs.
• Find the probability
that in one day two
babies are born hairy.
• Find the probability
that in one day no
babies are born
hairy.
Using formula sheet
l=3
x =2
• Find the probability
that in one day two
babies are born hairy.
• Probability = 0.2240
Using formula sheet
l=3
x =0
• Find the probability
that in one day no
babies are born
hairy.
• Probability = 0.0498
Queue Theory
Example: Suppose a bank knows that on average 60
customers arrive between 10 A.M. and 11 A.M.
daily.
• Find the probability
that exactly two
customers arrive in a
given one-minute time
interval between 10
and 11 A.M.
60
l = =1
60
P( X = 2) = 0.1839
Mean and standard deviation
of a Poisson distribution
E( X ) = l
s= l
Compare Binomial to Poisson
Compare Binomial to Poisson
When probability is close to 1, Poisson
can approximate the Binomial.
Approximation:
If p is small, then the Binomial distribution
with parameters n and p is well
approximated by the Poisson distribution with
parameter np,
i.e. by the Poisson distribution with the same
mean
Example
Binomial situation, n= 100, p=0.075
Calculate the probability of fewer than
10 successes.
Probability = 0.7833
This would have been very tricky with
manual calculation as the factorials are
very large and the probabilities very small
The Poisson approximation to the Binomial
states that  will be equal to np, i.e. 100 x
0.075
so =7.5
Probability = 0.7764
So it is correct to 2 decimal places. Manually,
this would have been much simpler to do than
the Binomial.
Eggs are packed into boxes of 500. On average
0.7% of the eggs are found to be broken when
the eggs are unpacked. Find the probability that
in a box of 500 eggs
• Exactly three are
broken.
• At least two are
broken.
Eggs are packed into boxes of 500. On average
0.7% of the eggs are found to be broken when
the eggs are unpacked. Find the probability that
in a box of 500 eggs
• Mean = 500 x 0.007
= 3.5
e-3.5 3.53
P( X = 3) =
= 0.2158
3!
Eggs are packed into boxes of 500. On average
0.7% of the eggs are found to be broken when
the eggs are unpacked. Find the probability that
in a box of 500 eggs
At least two are broken.
P( X ³ 2) = 1- P( X = 0) - P( X = 1) =
= 1- e-3.5 - 3.5e-3.5 = 0.8641
A Christmas draw aims to sell 5000 tickets, 50
of which will win a prize. A syndicate buys 200
tickets.
Justify the use of a
Poisson distribution.
Probability is close to
zero
Strictly speaking, you don’t
have independent trials, but
n is very large.
A Christmas draw aims to sell 5000 tickets, 50
of which will win a prize. A syndicate buys 200
tickets.
Calculate
P( X £ 3)
P( X £ 3) = 0.8571
A Christmas draw aims to sell 5000 tickets, 50
of which will win a prize. A syndicate buys 200
tickets.
Calculate how many
tickets should be
bought in order for
there to be a 90%
probability of winning
at least one prize.
l = np = n ´ 0.01
P( X ³1) =1- P( X = 0)
= 1- e-0.01n = 0.9
A Christmas draw aims to sell 5000 tickets, 50
of which will win a prize. A syndicate buys 200
tickets.
Calculate how many
tickets should be
bought in order for
there to be a 90%
probability of winning
at least one prize.
P( X ³ 1) = 1- P( X = 0)
= 1- e-0.01n = 0.9
N must be 231
Þ e-0.01n = 0.1
Þ n = 230.25
Two identical racing cars are being tested on a circuit.
For each car, the number of mechanical breakdowns
can be modelled by the Poisson distribution with a
mean of one breakdown in 100 laps. If a car breaks
down it is attended and continues on the circuit. The
first car is tested for 20 laps and the second for 40 laps.
Find the probability
that the service
team is called out
to attend to
breakdowns.
a) Once
b) More than twice
Two identical racing cars are being tested on a circuit.
For each car, the number of mechanical breakdowns
can be modelled by the Poisson distribution with a
mean of one breakdown in 100 laps. If a car breaks
down it is attended and continues on the circuit. The
first car is tested for 20 laps and the second for 40 laps.
Since the average
number of
breakdowns in
100 laps is one,
the average in 20
laps is 0.2 and in
40 laps it is 0.4.
T = X + Y = 0.2 + 0.4
Two identical racing cars are being tested on a circuit.
For each car, the number of mechanical breakdowns
can be modelled by the Poisson distribution with a
mean of one breakdown in 100 laps. If a car breaks
down it is attended and continues on the circuit. The
first car is tested for 20 laps and the second for 40 laps.
T = X + Y = 0.2 + 0.4
P(T = 1) = 0.6e
= 0.3293
P(T > 2) = 1- P(0) - P(1) - P(2)
= 0.0231
-0.6
Poisson Approximation: the Birthday Problem.
What is the probability that
in a gathering of k people,
at least two share the same
birthday?
Suppose there are n days in the year
(on Earth we have n = 365)
Assume that each person has a birthday which
is equally likely to fall on any day of the year,
independently of the birthdays of the remaining
k - 1 persons (no sets of twins in the group).
Then a simple conditional probability
calculation shows that
pn;k = 1- p(all birthdays are different)
= 1 - n - 1 n - 2 n - 3 ........ n - k + 1
n
n
n
n
So that (on Earth) 23 is the minimum size of
gathering required for a better than evens
chance of two members sharing the same
birthday.
Proof of this
The mean number of birthday coincidences
in a sample of size k is:
1 æ k ö k ( k - 1)
=
ç
÷
n è2ø
2n
The number of birthday coincidences
should have an approximately Poisson
distribution with the above mean.
Thus, to determine the size of gathering
required for an approximate probability p
of at least one coincidence, we should
solve
æ k (k - 1) ö
1 - exp ç ÷= p
2n ø
è
In other words we are solving the
simple quadratic equation
k - k + 2n ln(1 - p) = 0
2
In the case n=365, p=0.5, this
gives k=23.0
Notice as values of l
increase, the distribution becomes normally
distributed.
l
l
l
l
=2
=3
=5
= 10
As lambda increase, the normal approximation
gets better.