Poisson Distribution | Definition, Formula, Table and Examples

The Poisson distribution is a discrete probability distribution that calculates the likelihood of a certain number of events happening in a fixed interval of time, assuming the events occur independently. Interval divided into subintervals such that

• The probability of more than one event in a subinterval is 0.
• The probability of one event in a subinterval is proportional to the length of the subinterval.
• The probability of an event in a subinterval is independent of the probability of the event in another subinterval.

It is characterized by a single parameter, λ (lambda), which represents the event's average occurrence rate in an interval(not subinterval).

The distribution is used when the events are rare, the number of occurrences is non-negative, and can take on integer values (0, 1, 2, 3,...).

The key assumptions of the Poisson distribution are:

1. Events occur independently of each other.
2. The average rate of occurrence (λ) is constant over the given interval.
3. The number of events can be any non-negative integer.

In summary, the Poisson distribution is used to model the likelihood of events happening at a certain rate within a fixed time or space, under the assumptions of independence and constant occurrence.

Poisson Distribution Formula

Poisson distribution is characterized by a single parameter, lambda (λ), which represents the average rate of occurrence of the events. The probability mass function of the Poisson distribution is given by:

P (X = r) = \frac {e^{−λ}λ^r}{r!}

Where,

• P(X = r) is the Probability of observing k Events
• e is the Base of the Natural Logarithm (approximately 2.71828)
• λ is the Average Rate of Occurrence of Events
• r is the Number of Events that Occur

Recurrence Relation for Poisson Probabilities

In the Poisson distribution, there is a special recursive relationship that allows you to compute the probability of getting r events based on the probability of getting r−1 events. This relation is given by:

P(X = r) =\frac{\lambda}{r}P(X = r − 1) \: for \: r \ge 1

Example:

Calculate the value for P(X = 8) using the recurrence relation and the value for P(X = 7), where P(X = 7) = 0.0346, λ = 3 and r = 6.

Solution:P(X = 7) =\frac{\lambda}{r}P(6) = \frac{3}{6} 0.345 = 0.1725

Poisson Distribution Characteristics

Lets discuss some charecteristics of Poisson Distributions here.

Expectation and Variance

In the Poisson distribution, both the Expectation(mean) and variance are equal and are denoted by the parameter λ (lambda). This property of equal mean and variance is a distinctive characteristic of the Poisson distribution and simplifies its statistical analysis.

Expectation(mean), E(X) = λ and

Variance, V(X) = λ

where

• λ = np, (n is the Number of Trails, p is the Probability of Success)

Standard Deviation of Poisson Distribution

Standard Deviation of a poisson distribution is a measure of the amount of variability or dispersion in the distribution. Mathematically, it is given by:

σ = \sqrt {\lambda}

where,

• λ (lambda) is the Average Rate of Occurrence of Events
• σ (sigma) is the Standard Deviation of the Distribution

Probability Mass Function and Cumulative Distribution Function

Probability Mass Function (PMF) describes the likelihood of observing a specific number of events in a fixed interval. It is given by:

PMF = \frac{(e^{-λ} × λ^{r})} {r!} , r=0,1,2,…

where,

• e is the Base of the Natural Logarithm (approximately 2.71828)
• λ is the Parameter, which is also equal to the Mean, and Variance
• r is the Number of times an event occurs

Some properties of PMF are:

• P ( X = k ) ≥ 0 for all k.
• The sum of all probabilities over possible values of k is 1.

Example:

Suppose a hospital receives an average of λ = 4 emergency cases per hour. What is the probability that exactly 2 cases occur in an hour?
Solution:

Using the Poisson formula:

P (X = 2) = e^-4 4² /2! = e^-4 ✕ 16/2 = 0.0183 ✕ 16 /2 = 0.1465

Cumulative Distribution Function (CDF): gives the probability that the random variable is less than or equal to a certain value. It is expressed as:
F(x) = \sum^{k=0}_{⌊x⌋}\frac{ (e^{-λ} × λ^k) }{ k!}
where ⌊x⌋ denotes the greatest integer less than or equal to x.

Poisson Distribution Graph

The following illustration shows the Graph of the Poisson Distribution or the Poisson Distribution Curve.

The Poisson distribution is positively skewed (Skewness > 0) and leptokurtic (Kurtosis > 0), meaning it has a longer tail on the right side and heavier tails than the normal distribution. However, for large values of λ, it becomes increasingly symmetric and bell-shaped, resembling a normal distribution.

Note: Leptokurtic refers to a distribution that has a higher kurtosis than the normal distribution. Kurtosis measures the "tailedness" or sharpness of the peak of a frequency distribution curve.

The event with the highest probability is represented by the peak of the distribution—the mode.

• When λ is a non-integer, the mode is the closest integer smaller than λ.
• When λ is an integer, there are two modes: λ and  λ−1.

When λ is low, the distribution is much more distributed on the right side of its peak than its left (right-skewed).

As λ increases, the distribution starts to appear more and more similar to a normal distribution. When λ is 10 or greater, a normal distribution is a good approximation of the Poisson distribution.

Binomial Distribution v/s Poisson Distribution

The key differences between Poisson Distribution and Binomial Distribution are listed in the following table:

Poisson Distribution Solved Examples

Example 1: If 4% of the total items made by a factory are defective. Find the probability that less than 2 items are defective in the sample of 50 items.

Solution:  

Here we have, n = 50, p = (4/100) = 0.04, q = (1-p) = 0.96,  λ = 2

Using Poisson's Distribution,
P(X = 0) = \frac{2^0e^{-2}}{0!} = 1/e² = 0.13534
P(X = 1) = \frac{2^1e^{-2}}{1!} = 2/e² = 0.27068

Hence the probability that less than 2 items are defective in sample of 50 items is given by:
P( X > 2 ) = P( X = 0 ) + P( X = 1 ) = 0.13534 + 0.27068 = 0.40602

Example 2: If the probability of a bad reaction from medicine is 0.002, determine the chance that out of 1000 persons more than 3 will suffer a bad reaction from medicine.

Solution:

Here we have, n = 1000, p = 0.002, λ = np = 2
X = Number of person suffer a bad reaction 

Using Poisson's Distribution

P(X > 3) = 1 - {P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)} 
P(X = 0) = \frac{2^0e^{-2}}{0!} = 1/e²
P(X = 1) =  \frac{2^1e^{-2}}{1!} = 2/e² 
P(X = 2) =  \frac{2^2e^{-2}}{2!} = 2/e²
P(X = 3) =  \frac{2^3e^{-2}}{3!} = 4/3e²
P(X > 3) = 1 - [19/3e²] = 1 - 0.85712 = 0.1428

Example 3: If 1% of the total screws made by a factory are defective. Find the probability that less than 3 screws are defective in a sample of 100 screws.

Solution:

Here we have, n = 100, p = 0.01, λ = np = 1
X = Number of defective screws

Using Poisson's Distribution

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) 
P(X = 0) =  \frac{1^0e^{-1}}{0!} = 1/e
P(X = 1) = \frac{1^1e^{-1}}{1!} =1/e
P(X = 2) = \frac{1^2e^{-1}}{2!} =1/2e

Thus, P(X < 3) = 1/e + 1/e +1/2e = 2.5/e = 0.919698

Example 4: If in an industry there is a chance that 5% of the employees will suffer from corona. What is the probability that in a group of 20 employees, more than 3 employees will suffer from corona virusus?

Solution:

Here we have, n = 20, p = 0.05, λ = np = 1
X = Number of employees who will suffer corona

Using Poisson's Distribution

P(X > 3) = 1-[P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
P(X = 0) = \frac{1^0e^{-1}}{0!} = 1/e
P(X = 1) = \frac{1^1e^{-1}}{1!} = 1/e
P(X = 2) =\frac{1^2e^{-1}}{2!} =1/2e
P(X = 3) =\frac{1^3e^{-1}}{3!} =1/6e
P(X > 3) = 1 - [1/e + 1/e + 1/2e + 1/6e]

⇒ P(X > 3) = 1 - [ 8/3e] = 0.018988

Related Articles

• Poisson Distribution Meaning
• Types of Frequency Distribution
• Probability Distribution
• Binomial Distribution

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