poisson distribution. poisson poisson 1781-1840

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.

Probability

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.

b) Find the probability that in a 10 ml sample there are no amoebas

c) Find the probability that in a 10 ml sample there are fewer than three amoebas.

Useful probabilities to note

Graphics Calc Poisson DistStats Mode from Calc

F5 Distribution

Then F6 next

F1 Poisn

For point dist select Ppd

Select Data: Variable

For P(X=1),

For P(X=1) = 0.1494

For cumulative select Pcd after Poisn

For P(X<3),

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

• Find the probability that in one day two babies are born hairy.

• Probability = 0.2240

Using formula sheet

• 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.

Mean and standard deviation of a Poisson distribution

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

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.

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

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.

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.

N must be 231

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.

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.

What is the probability that in a gathering of k people, at least two share the same birthday?

Poisson Approximation: the Birthday Problem.

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 probabilitycalculation shows that

pn;k = 1- p(all birthdays are different)

=

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:

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

In other words we are solving the simple quadratic equation

In the case n=365, p=0.5, this gives k=23.0

Notice as values ofincrease, the distribution becomes normally

distributed.

As lambda increase, the normal approximation gets better.