Mastering Poisson Distribution: Examples and Applications

Discover real-world Poisson distribution examples and learn when to apply this crucial statistical concept. Enhance your problem-solving skills with our comprehensive guide to rare event modeling.

Poisson distribution

What is a poisson distribution?

The Poisson probability distribution is the one concerned in modeling the probability behaviour of the number of events considered successes, occurring within a particular interval of time.
Under certain circumstances, the Poisson distribution can be used as an approximation to the binomial distribution. And for that, in order to mathematically define the Poisson distribution we have to bring back what we have already learnt in past lessons. Remember that the general probability formula of a binomial distribution is:

$P(x) = \, _{n}C_{x}p^{x} (1-p)^{n-x} =$ $\large \frac{n!}{x!(n-x)!}$ $p^{x}(1-p)^{n-x}$

Where:
$n$ = number of trials
$x$ = number of successes in n trials
$p$ = probability of success in each trial
$_{n}C_{x} =$ $\large \frac{n!}{x!(n-x)!}$ = number of success outcomes
$P(x)$ = probability of getting x successes out of n trials

We also learnt about the mean and standard deviation of the binomial distribution, where the mean is defined as the multiplication of the number of trials in the experiments, times the probability of success in each trial:

$\mu = np$

The mean provides the best approximation of the amount of times you can expect a certain outcome to occur per a certain number of trials. With that in mind we define the standard deviation and variance (the standard deviation squared):

$\sigma = \sqrt{np (1-p)} =$ standard dev.
$\sigma^{2} = np(1-p) =$ variance
With all of this in mind we finally define the Poisson distribution formula for the probability as:

$P(x) = e^{-np}$ $\large \frac{(np)^{x}}{x!}$
This equation is also called the probability mass function (pmf) of a Poisson distribution.
It is important to mention that if you look through other study materials you will find a definition of the Poisson distribution pmf using another notation for the mean:

$\mu = \lambda = np$
The reason to this notation substitution is that $\lambda$ allows us to see that this average could refer to other kinds of quantities besides that defined by the mean of a binomial distribution, which is the average number reflecting the expected amount of success outcomes per trials; for the case of the Poisson distribution, the mean reflects the value of the expectation of success per time rate, or the average number of successes per time interval. Having this substitution, our probability mass function equation, or the Poisson distribution equation for probability looks like:

$P(x) = e^{-np}$ $\large \frac{(np)^{x}}{x!} \,$ →$\, P(x) = e^{- \lambda}$ $\large \frac{(\lambda)^{x}}{x!}$

When to use Poisson distribution

On this section let us take a look at the conditions in which the Poisson distribution can be used to calculate approximations to binomial probabilities:

1. The number of trials ($n$) is greater than or equal to 20
2. The probability of a success in each trial ($p$) is less than or equal to 0.05

In simple words, the Poisson distribution can provide a good value for a binomial distribution when the number of trials are sufficiently big in comparison with the amount of successes that occur. So there must be a very high amount of trials and a very small chance of a success occurring (reason why multiple textbooks and study materials like to focus on examples putting the probability of Earth being hit by a meteorite as a Poisson distribution probability).

Poisson distribution examples

Example 1

Determining the Poisson Distribution
The number of meteorites that hit the earth in a given day is modelled by a Poisson Distribution with $\lambda = 4$. What is the probability that 5 meteoroids hit the earth in a day?

Using the Poisson distribution definition for probability, we calculate that:

$P(x) = e^{- \lambda }$ $\large \frac{(\lambda)^{x}}{x!} \,$ →$\, P(5) = e^{-4}$ $\large \frac{(4)^{5}}{5!}$
$P(5) = e^{-4}$ $\large \frac{(4)^{5}} {5!}$ $= e^{-4}$ $\large \frac{1024} {120}$ $e^{-4} (8.5333) = 0.156$
Notice how, according to this Poisson distribution example problem, we have a chance of about 15.6% that 5 meteorites hit the Earth in any day. You may think this is either unrealistic or alarming, but do not worry!
Even if this result was in fact realistic (this is a made up problem, but let us look at the reality of such events), and the probability of collisions with Earth from meteorites is large, the event of a meteoroid having a trajectory that would impact Earth can have a wide variety of characteristics, which make a grand difference on the result of such event.
For example, let us start with what a meteoroid is: a meteoroid is any object, small in comparison with an asteroid, that orbits the Sun; it can be thought of a debri from explosive events in the solar system such as collision or even coming from the formation of the solar system itself.
When such object falls towards earth, depending on its size, you can have one of two outcomes: it can burn due friction when entering the atmosphere (which is what happens with most rocky objects smaller than 10m) at which point is called a meteor; or, it can survive the friction with the atmosphere (usually if is big enough, or if is not only rock, but metal in composition) at which point their are called meteorites.
Thousands of meteorites fall to the surface of our planet daily, but they are too small when they reach the surface due to the burning up when entering through the atmosphere, and/or fall into areas that are not populated (such as the oceans, which cover most of the surface of our planet, or areas like the Arctic and Antarctic); therefore, meteorites falling down on Earth are not a cause to worry. The probability of a big one which would affect us, is much much smaller than the general for such events.

Example 2

When making a video, a StudyPug teacher makes 1 error for every 20 minutes of video time. If one teacher makes 45 minutes of video, what is the probability that he makes 3 errors?

For this case we have that:
$n$ = 45 minutes, which is the complete video time interval.
$x$ = 3 errors, which are the successes we want to check for in the interval.
$p =$ $\large \frac{1}{20}$ errors/minute for the error that is made every 20 minutes.
For that we have that the mean of Poisson distribution for this case is: $\lambda = np =$45/20
And so, we use equation 6 to calculate the Poisson distribution probability of the teacher making a mistake in 45 minutes of video time:

$P(x) = e^{- \lambda}$ $\large \frac{(\lambda)^{x}} {x!} \,$ →$\, P(3) =$ $\large e^{-\frac{45} {20} } \frac{(\frac{45} {20})^{3} }{3!}$
$P(3) =$ $\large e^{-\frac{45} {20} } \frac{(\frac{45} {20})^{3} } {3!} = e^{- \frac{45} {20} } \frac{11.39}{6} = e^{-\frac{45} {20} }$ $(1.8984) = 0.2$
Which means that there are about 20% chances of this particular StudyPug teacher to make 3 errors while recording a 45-minute lesson. Dont worry, the videos that are posted are checked continuously, so those mistakes made are cut off and stay in our backroom ;)

Example 3

In a particular community the average person survives to age 100 with probability 0.005 (which is equivalent to 0.5%). If this community has 2,000 people, then what is the probability that 15 people in this community survive to age 100 using;

On this case we have that:
$n =$ 2000 people in the community
$x =$ 15 people are successful in living until the age of 100
$p =$ 0.05

So, we use equation 1 for the probability of a binomial distribution to compute:

$P(x) =$ $\large \frac{n!} {x!(n-x)!}$ $p^{x}(1-p)^{n-x} \,$→$\, P(15) =$ $\large \frac{2000!} {15!(2000-15)!}$ $(0.05)^{15} (1-0.05)^{2000-15}$
$P(15) =$ $\large \frac{2000!} {(1.308 \times 10^{12}) (2000-15)! }$ $(0.05)^{15} (1-0.05)^{2000-15} = 0.0346$
Now in this problem we ran into the problem that numbers such as 2000! Are so big, that this problem cannot be done by hand (even the single result of 2000! in the calculator usually returns a value of infinity); and so, we use the calculator direct method to compute a binomial distribution probability and find the result. Please take a look at the corresponding video on this lesson so you can see the process to follow to solve it using the calculator.

Now we work through the same problem with the Poisson distribution approach, and so, again, we have that:
$n =$ 2000 people in the community
$x =$ 15 people are successful in living until the age of 100
$p =$ 0.05
$\lambda = np =$ (2000)(15) = 30,000
Using equation 6 we calculate that:

$P(x) = e^{-\lambda}$ $\large \frac{(\lambda)^{x}} {x!}$ $\,$→$\, P(15) = e^{-30,000}$ $\large \frac{(30,000)^{15}} {15!}$
$P(15) = e^{-30,000} (1.0973 \times 10^{40}) = 0.0347$
Once again, the values are so big or small, that some calculators wont be able to process those quantities if done through the basic operations as shown in the equation above; therefore, we recommend you to use the direct method for the calculation of a probability of a Poisson distribution with the calculator as shown in the videos for this lesson.

Example 4

A fair coin is flipped 10 times, what is the probability using the Poisson Distribution commands on your calculator find: (Notice that these calculations can be done much more efficiently in a calculator with the direct method. Still, we are showing the calculations in here so you know how the method is done by hand).

For this case we have that:
$n =$ 10
$x =$ 5
$p =$ 0.05 because the chance at every flip is half and half
$\lambda = np =$ (10)(0.5) = 5
Now we calculate the probability as follows:

$P(x) = e^{- \lambda}$ $\large \frac{(\lambda)^{x} } {x!}$ $\,$→$\, P(5) = e^{-5}$ $\large \frac{(5)^{5}}{5!}$
$P(5) = 6^{-5}$ $\large \frac{(5)^{5}} {5!}$ $= e^{-5}$ $\large \frac{3,125} {120}$ $= 0.17547$
To solve this problem we have to add the probabilities of heads coming up 0 times, 1 time, 2, times, 3 times, 4 times and 5 times. So adding up all of the probabilities of heads resulting from 0 to 5

$P(x \leq 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)$
We already have the value for P(5), we just need to calculate the others, so we have that:
$n=10, x = 0,1,2,3,4; \, p = 0.5$ and $\lambda = np =(10)(0.5)=5$
Now we calculate the probability as follows:

$P(0) = e^{-5}$ $\large \frac{(5)^{0}} {0!}$ $= e^{-5}$ $\large \frac{1}{1}$ $= 0.00674$
$P(1) = e^{-5}$ $\large \frac{(5)^{1}} {1!}$ $= e^{-5}$ $\large \frac{5}{1}$ $= 0.03369$
$P(2) = e^{-5}$ $\large \frac{(5)^{2}} {2!}$ $= e^{-5}$ $\large \frac{25}{2}$ $= 0.008422$
$P(3) = e^{-5}$ $\large \frac{(5)^{3}} {3!}$ $= e^{-5}$ $\large \frac{125}{6}$ $= 0.14037$
$P(4) = e^{-5}$ $\large \frac{(5)^{4}} {4!}$ $= e^{-5}$ $\large \frac{625}{24}$ $= 0.17547$
We add all of these results together to find the probability for the cumulative Poisson distribution:

$P(x \leq 5) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)$
$P(x \leq 5) = 0.00674+0.03369+0.08422+0.14037+0.17547+0.175=0.61596$
There are a few ways to solve this problem since we can define the probability of flipping a coin 10 times and resulting in heads more than seven times can be defined in different ways. For example, we could say that is the addition of the probabilities of having heads coming up 8, 9 and 10 times:

$P(x \leq 7) = P(8) + P(9) + P(10)$
Or we could say the probability of having heads come up more than seven times is equal to 1 minus the probability of the coin coming up from 0 to 7 times:

$P(x \leq 7) = 1 - \left\{ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) \right\}$
Let us solve it both ways!



Notice that although the results are a little bit different, they both are very close, the differences may come from systematic errors such as using the results up to a certain amount of decimal digits.
It is important to mention that you should check out all of the videos for these Poisson distribution examples so you can see the way to calculate them using the calculator commands.

To finalize this lesson we recommend you to visit this lesson on the Poisson distribution, where you will find from historical information, to well explained content and example problems; also, on this statistics handbook you will find a lesson on the topic that includes poisson distribution graph examples along with the concepts.

This is it for the lesson of today, see you in the next one!