THE POISSON DISTRIBUTION
Achievement Standard: 90646 (2.6) (part) external; credits 4
Key Words: random occurrences (per interval).
1. Definitions
The Poisson Distribution occurs when the number of “successes” in a given interval
(eg time, space) are being counted.
The properties of the Poisson distribution are:
(a) The average number of successes (, ) occurring in the interval is
known.
(b) Occurrences are independent.
(c) Occurrences are random and the probability of a single success is
proportional to the length of the interval.
(d) The probability of more than one success is small compared to the
probability of a single success in a small interval.
2. Examples when the Poisson distribution arises
(a) The number of phone call arriving at an exchange in a given time interval.
(b) Typing errors on a page of an author’s manuscript.
– x
(c) The number
e  of faults on a computer’s circuitry over the period of a month.
(d) The number of deaths from horse kicks in an army corps over the period of a
x!
year.
3. The formula for Poisson Probabilities
P(X = x) =
e
– x

x!
, x = 0, 1, 2, 3………..
Nulake p275
Sigma p 335, Ex 16.01,
Ex 16.02
Example: The average number of emergency calls that a hospital gets on any day is
4.
What is the probability that the hospital gets 6 calls on a particular day?
e
–4
6
 4
P(XLet
= X6)be=a random variable representing the number of calls to the
Answer:
6!
hospital on any day. Then X is a Poisson random variable with  = 4.
P(X = 6) =
e
–4
6
 4
6!
= 0.1042 (tables or calculator)
4. Adjusting 
Example: During weekdays a garage knows that on average 10 cars per
hour are served. What is the probability that during a particular quarter
hour interval, there will be some cars wanting service?
Answer: Let X be a random variable representing the number of cars wanting
service in a quarter hour interval then X is Poisson with  = 10  4 = 2.5
P( X  0) = 1 – P(X = 0)
= 1 – 0.0821
= 0.9179
5. Inverse problems
Example: In samples of milk taken from a bulk transportation vehicle, 40% proved
to have no bacterial spores. Estimate the mean number of spores per
sample, and hence find probability that a randomly selected sample will
contain 2 spores.
e
– 0
 Answer: Let X be a random variable representing the number of spores per
= 0.4
0!
e
–
sample.Then X is Poisson with  unknown.
= 0.4 (a) P(X = 0) = 0.4
e
– 0

0!
e
Sigma p 347, Ex 16.04
= 0.4
–
= 0.4
– 0.92
2
e
 0.92
P(X = 2) = -lne = ln 0.4
- = ln2!
0.4
 = 0.92
(b)
P(X = 2) =
e
– 0.92
= 0.17
2
 0.92
2!
6. Use of sample statistics to estimate population parameters
Example: A fire chief collects data on the number of fires attended per week, each
week for a year.
No. of fires
0
1
2
3
4
5
6
7
8
9
10
No. of weeks 4
10
13
9
7
4
2
1
1
0
1
The chief thinks that the number of fires per week might fit a Poisson probability
model. Make calculations to see if he is justified.
Answer: From the chief’s data the sample mean of X = 2.846 can be found.
So the estimate of  is 2.714. The table shows the Poisson probabilities
and expected values for a year (× 52).
No. of fires
0
Poisson prob .058
Expected no. 3
1
2
3
4
5
6
7
8
9
10
.165
.235
.223
.159
.090
.043
.017
.006
.002
.001
9
12
12
8
5
2
1
0
0
0
The pattern of expected values is very close to the sample data so it is reasonable
to assume that the distribution of fires is Poisson.
7. Combined events using Poisson probabilities.
Sigma p 349, Ex 16.05
Example: The number of fires attended per week by a fire brigade has a Poisson
distribution with  = 2.714.
(a) Calculate the probability that there are 4 fires on each of two successive
weeks.
Answer: Let X represent the number of fires attended in any week. Then X is
Poisson with  = 2.714. P(X=4) = 0.15 so the probability that there are 4
fires on each of 2 successive weeks is 0.152 = 0.0225.
(b) Calculate the probability that on 6 successive weeks there are 4 fires on less
than 3 of the weeks.
Answer: Let Y represent the number of weeks (from the 6) when there are 4 fires.
Y is binomial with n = 6, π = 0.15.
P(Y ≤ 2) = 0.953 (3DP) (calculator – less than 3 = less than or equal to 2)
8. The mean and variance for a Poisson distribution
Sigma p 344, Ex 16.03
The mean for a Poisson random variable is  and the variance is also  (s.d. = ).