## The Poisson Distribution The Poisson distribution describes the distribution of events which occur randomly in a continuous interval. e.g. Time, length, volume etc. The following conditions must apply:

• The events occur at random
• The events are independent of one another
• Two or more events do not occur simultaneously
• The probability that an event occurs in an interval is proportional to the size of that interval

The Poisson distribution has only one parameter and that is λ the average number of occurrences of an event in a given time interval. i.e. A rate.

### Examples of Poisson Distribution

Examples of events which could follow a Poisson distribution are:

the number of cars visiting a service station in one half day.
the number of nails in a metre of wood.
the number of telephone calls arriving at a company's switchboard

### Poisson Probability from Formula

The formula for Poisson probability is: for x = 0, 1, 2, 3, ... λ is the mean number of occurrences in a given time or space and can be any positive value.

Example

The average number of cars passing a speed camera in an hour, which are exceeding the speed limit is 5.
Find the probability that in the next hour 4 cars passing the camera will be speeding.

λ = 5 and x = 4 ### Poisson Probability from Tables

Rather than use the formula for finding Poisson probabilities, tables are available to look these up.

table of the Poisson Distribution for values of λ from 0.1 to 6 are given with the Bursary examination.

The part of the table for for λ = 5 and x = 4 is shown below:

 x\λ 4.2 4.4 4.6 4.8 5 5.2 5.4 5.6 5.8 6 0 0.015 0.0123 0.0101 0.0082 0.0067 0.0055 0.0045 0.0037 0.003 0.0025 1 0.063 0.054 0.0462 0.0395 0.0337 0.0287 0.0244 0.0207 0.0176 0.0149 2 0.1323 0.1188 0.1063 0.0948 0.0842 0.0746 0.0659 0.058 0.0509 0.0446 3 0.1852 0.1743 0.1631 0.1517 0.1404 0.1293 0.1185 0.1082 0.0985 0.0892 4 0.1944 0.1917 0.1875 0.182 0.1755 0.1681 0.16 0.1515 0.1528 0.1339 5 0.1633 0.1687 0.1725 0.1747 0.1755 0.1748 0.1728 0.1697 0.1656 0.1606 6 0.1143 0.1237 0.1323 0.1398 0.1462 0.1515 0.1555 0.1584 0.1601 0.1606 7 0.0686 0.0778 0.0869 0.0959 0.1044 0.1125 0.12 0.1267 0.1326 0.1377

If the above question asked for the probability that there are four or fewer speeding cars i.e. P(X ≤ 4) then the probabilities for x = 0, 1, 2, 3 and 4 would need to be added together.

P(X ≤ 4) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 = 0.4405

### Mean and Variance of the Poisson Distribution

There are formulae for working the mean, variance and standard deviation of a binomial distribution.

 Mean (expected value) μ = λ Variance σ2 = λ Standard deviation σ = √λ

In the example above:

E(X) = λ = 5

VAR(X) = λ = 5

SD(X) = √λ = √5 = 2.236 (to 4 sig. fig.)

Further Poisson Example

 Cars arrive at a service station at the rate of 24 per hour. Let X represent the Poisson random variable for the number of cars arriving. a. Find the probability that 8 cars arrive in a 15 minute period. If 24 cars arrive in an hour: λ = 6 (for a 15 minute period) P(X = 8) = 0.1033 (from tables) The probability that 8 cars arrive in 15 minutes is 0.1033 b. Find the probability that more than 3 cars arrive in a 15 minute period. P(X > 3) =1 − P(X ≤ 3)               = 1 − (0.0025 + 0.0149 + 0.0446 + 0.0892)               = 0.8488 The probability that more that 3 cars arrive in 15 minutes is 0.8488