GENERAL TOPIC:
DISCRETE PROBABILITY
DISTRIBUTIONS
SUB-TOPIC:
“POISSON DISTRIBUTION”
TRIXIE ELAINE MEDRANO
II-D1(BS MATH-AS)
INTRODUCTION:
A Poisson experiment is a statistical experiment that has the following properties:
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The experiment results in outcomes that can be classified as successes or
failures.
The average number of successes (μ) that occurs in a specified region is
known.
The probability that a success will occur is proportional to the size of the
region.
The probability that a success will occur in an extremely small region is
virtually zero.
Note that the specified region could take many forms. For instance, it could be a
length, an area, a volume, a period of time, etc.
Notation
The following notation is helpful, when we talk about the Poisson distribution.
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e: A constant equal to approximately 2.71828. (Actually, e is the base of
the natural logarithm system.)
μ: The mean number of successes that occur in a specified region.
x: The actual number of successes that occur in a specified region.
P(x; μ): The Poisson probability that exactly x successes occur in a
Poisson experiment, when the mean number of successes is μ.
DEFINITION OF TERMS:
Poisson Distribution
A Poisson random variable is the number of successes that result from a Poisson
experiment. The probability distribution of a Poisson random variable is called a
Poisson distribution.
Given the mean number of successes (μ) that occur in a specified region, we can
compute the Poisson probability based on the following formula:
Poisson Formula. Suppose we conduct a Poisson experiment, in which the
average number of successes within a given region is μ. Then, the Poisson
probability is:
P(x; μ) = (e-μ) (μx) / x!
where x is the actual number of successes that result from the experiment, and e
is approximately equal to 2.71828.
The Poisson distribution has the following properties:
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The mean of the distribution is equal to μ .
The variance is also equal to μ .
The Poisson Distribution was developed by the French mathematician Simeon
Denis Poisson in 1837.
The Poisson random variable satisfies the following conditions:
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Mean and variance of Poisson distribution
1. The number of successes in two disjoint time intervals is independent.
2. The probability of a success during a small time interval is proportional to
the entire length of the time interval.
Apart from disjoint time intervals, the Poisson random variable also applies to
disjoint regions of space.
Applications:
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the number of deaths by horse kicking in the Prussian army (first
application)
birth defects and genetic mutations
rare diseases (like Leukemia, but not AIDS because it is infectious and so
not independent) - especially in legal cases
car accidents
traffic flow and ideal gap distance
number of typing errors on a page
hairs found in McDonald's hamburgers
spread of an endangered animal in Africa
failure of a machine in one month
The probability distribution of a Poisson random variable X representing the
number of successes occurring in a given time interval or a specified region of
space is given by the formula:
where
x = 0, 1, 2, 3...
e = 2.71828 (but use your calculator's e button)
μ = mean number of successes in the given time interval or region of space
Mean and Variance of Poisson Distribution
If μ is the average number of successes occurring in a given time interval or
region in the Poisson distribution, then the mean and the variance of the Poisson
distribution are both equal to μ.
E(X) = μ
and
V(X) = σ2 = μ
Note: In a Poisson distribution, only one parameter, μ is needed to determine the
probability of an event.
EXAMPLE 1
A life insurance salesman sells on the average 3 life insurance policies per week.
Use Poisson's law to calculate the probability that in a given week he will sell
(a) some policies
(b) 2 or more policies but less than 5 policies.
(c) Assuming that there are 5 working days per week, what is the probability that
in a given day he will sell one policy?
Answer
Here, μ = 3
(a) "Some policies" means "1 or more policies". We can work this out by finding 1
minus the "zero policies" probability:
P(X > 0) = 1 − P(x0)
Now
so
So
(b)
(c) Average number of policies sold per day:
So on a given day,
Here is a neat Java applet which can find these values for you:
Poisson applet. (External site).
EXAMPLE 2
Twenty sheets of aluminum alloy were examined for surface flaws. The
frequency of the number of sheets with a given number of flaws per sheet was as
follows:
Number of flaws Frequency
0
4
1
3
2
5
3
2
4
4
5
1
6
1
What is the probability of finding a sheet chosen at random which contains 3 or more
surface flaws?
Answer
The total number of flaws is given by:
(0 × 4) + (1 × 3) + (2 × 5) + (3 × 2) + (4 × 4) + (5 × 1) + (6 × 1) = 46
So the average number of flaws for the 20 sheets is given by:
The required probability is:
Histogram of Probabilities
We can see the predicted probabilities for each of "No flaws", "1 flaw", "2 flaws",
etc on this histogram.
[The histogram was obtained by graphing the following function for integer values
of x only.
Then the horizontal axis was modified appropriately.]
EXAMPLE 3
If electricity power failures occur according to a Poisson distribution with an average of 3
failures every twenty weeks, calculate the probability that there will not be more than one
failure during a particular week.
Answer
The average number of failures per week is:
"Not more than one failure" means we need to include the probabilities for "0
failures" plus "1 failure".
EXAMPLE 4
Vehicles pass through a junction on a busy road at an average rate of 300 per
hour.
1. Find the probability that none passes in a given minute.
2. What is the expected number passing in two minutes?
3. Find the probability that this expected number actually pass through in a given
two-minute period.
Answer
The average number of cars per minute is:
(a)
(b) E(X) = 5 × 2 = 10
(c) Now, with μ = 10, we have:
Histogram of Probabilities
Based on the function
we can plot a histogram of the probabilities for the number of cars per minute:
EXAMPLE 5
A company makes electric motors. The probability an electric motor is defective
is 0.01. What is the probability that a sample of 300 electric motors will contain
exactly 5 defective motors?
Answer
The average number of defectives in 300 motors is μ = 0.01 × 300 = 3
The probability of getting 5 defectives is:
NOTE: This problem looks similar to a binomial distribution problem, that we met
in the last section.
If we do it using binomial, with n = 300, x = 5, p = 0.01 and q = 0.99, we get:
P(X = 5) = C(300,5)(0.01)5(0.99)295 = 0.10099
We see that the result is very similar. We can use binomial distribution to
approximate Poisson distribution (and vice-versa) under certain circumstances.
Histogram of Probabilities
Example 6
The average number of homes sold by the Acme Realty company is 2 homes per
day. What is the probability that exactly 3 homes will be sold tomorrow?
Solution: This is a Poisson experiment in which we know the following:
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μ = 2; since 2 homes are sold per day, on average.
x = 3; since we want to find the likelihood that 3 homes will be sold
tomorrow.
e = 2.71828; since e is a constant equal to approximately 2.71828.
We plug these values into the Poisson formula as follows:
P(x; μ) = (e-μ) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
Thus, the probability of selling 3 homes tomorrow is 0.180 .
Example 7
The average number of homes sold by the Acme Realty company is 2 homes per
day. What is the probability that exactly 3 homes will be sold tomorrow?
Solution: This is a Poisson experiment in which we know the following:
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μ = 2; since 2 homes are sold per day, on average.
x = 3; since we want to find the likelihood that 3 homes will be sold
tomorrow.
e = 2.71828; since e is a constant equal to approximately 2.71828.
We plug these values into the Poisson formula as follows:
P(x; μ) = (e-μ) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
Thus, the probability of selling 3 homes tomorrow is 0.180 .
Poisson Calculator
Clearly, the Poisson formula requires many time-consuming computations. The
Stat Trek Poisson Calculator can do this work for you - quickly, easily, and errorfree. Use the Poisson Calculator to compute Poisson probabilities and
cumulative Poisson probabilities. The calculator is free. It can be found under the
Stat Tables tab, which appears in the header of every Stat Trek web page.
Cumulative Poisson Probability
A cumulative Poisson probability refers to the probability that the Poisson random
variable is greater than some specified lower limit and less than some specified
upper limit.
Example 8
Suppose the average number of lions seen on a 1-day safari is 5. What is the
probability that tourists will see fewer than four lions on the next 1-day safari?
Solution: This is a Poisson experiment in which we know the following:
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μ = 5; since 5 lions are seen per safari, on average.
x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see
fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2,
or 3 lions.
e = 2.71828; since e is a constant equal to approximately 2.71828.
To solve this problem, we need to find the probability that tourists will see 0, 1, 2,
or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1;
5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula:
P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)
P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ]
P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [
(0.006738)(125) / 6 ]
P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
P(x < 3, 5) = 0.2650
Thus, the probability of seeing at no more than 3 lions is 0.2650.
SUMMARY:
The Poisson distribution mutually independent events, occurring at a known
and constant rate r per unit (of time or space), and observed through a certain
window: a unit of time or space.
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The probability of k occurrences in that unit can be calculated from p(k) =
r*k / (k!)(e*r).
The rate r is also the expected or most likely outcome (for whole number r
greater than 1, the outcome corresponding to r-1 is equally likely).
The frequency profile of Poisson outcomes for a given r is not
symmetrical; it is skewed more or less toward the high end.
For Binomial situations with p < 0.1 and reasonably many trials, the
Poisson Distribution can acceptably mimic the Binomial Distribution, and is
easier to calculate.