Poisson Distribution
Poisson Distribution is a type of discrete probability distribution that describes models the number of times an events occurring in a fixed interval/period of time or space, given that these events happen independently and at a constant average rate. It is commonly used in scenarios like counting phone calls, defects, or arrivals over time. Defined by a single parameter λ (mean rate), it helps calculate the probability of observing a specific number of occurrences within the given interval.
1.0What is Meant by Poisson Distribution?
The Poisson Distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, assuming the events occur independently and at a constant average rate. It's widely used in statistics, data science, operations research, and probability theory.
2.0Poisson Distribution Definition
The Poisson Distribution gives the probability of a number of events happening in a fixed time, distance, or space, when these events occur with a known constant rate and independently of the time since the last event.
3.0Poisson Distribution Formula
The probability of observing exactly x events in an interval is given by:
P(X=x)=x!e−λλx
Where:
- x: Number of occurrences (0, 1, 2, ...)
- λ: Mean number of occurrences in the interval
- e: Euler's number, approximately 2.71828
4.0Poisson Distribution Graph
The Poisson distribution graph is skewed right for small values of λ and becomes more symmetric as λ increases. It shows the probability mass function with bars centered at whole number values of x.
5.0Poisson Distribution Mean and Variance
- Mean (μ) = λ
- Variance (σ²) = λ
This implies that for a Poisson distribution, mean and variance are equal.
6.0Poisson Distribution Properties
- Discrete probability distribution.
- Describes count-based data/events.
- Applicable when events are independent.
- Mean = Variance = λ.
- Only one parameter (λ) governs the distribution.
7.0When to Use Poisson vs Binomial?
Use Poisson Distribution when:
- The number of trials is very large.
- The probability of occurrence per trial is very small.
- You know the average number of events (λ) per interval.
8.0Solved Examples on Poisson Distribution
Example 1: A website gets an average of 2 spam emails per hour. What is the probability that it receives exactly 3 spam emails in an hour?
Solution:
Given λ = 2, x = 3
P(X=3)=3!e−2.2⋅23=6e−2.8⋅8=60.1353⋅8≈0.1804
Probability = 0.1804
Example 2: On average, 5 patients arrive at a clinic every hour. What is the probability that exactly 7 patients arrive in an hour?
Solution:
P(X=7)=7!e−5⋅57≈50400.0067⋅78125≈0.1044
Probability = 0.1044
Example 3: A call center receives 10 calls per minute on average. Find the probability that it receives no calls in a minute.
Solution:
P(X=0)=0!e−10⋅100=e−10≈4.54×10−5
Probability = 0.0000454
Example 4: A call center receives 5 calls per minute on average. What is the probability that exactly 3 calls are received in a particular minute?
Solution:
Given:
- Λ = 5 (mean rate per minute)
- x = 3
Using the Poisson distribution formula:
P(X=x)=x!e−λλxP(X=3)=3!e−5⋅53=6e−5⋅125Using(e−5≈0.0067):P(X=3)=60.0067⋅125≈0.1396 Answer:P(X=3) ≈ 0.1396
Example 5: The average number of errors per page in a printed book is 0.3. What is the probability that a randomly selected page has no errors?
Solution:
Here,
λ = 0.3, x = 0
P(X=0)=0!e−0.3⋅0.30=e−0.3≈0.7408Answer:(P(X=0)≈0.7408)
Example 6: A machine fails on average 2 times per month. What is the probability that it fails at most once in a month?
Solution:
We need P(X ≤ 1) = P(X = 0) + P(X = 1)
Given λ = 2
P(X=0)=0!e−2⋅20=e−2≈0.1353P(X=1)=1!e−2⋅21=2e−2≈0.2707P(X≤1)=0.1353+0.2707=0.406Answer:(P(X≤1)≈0.406)
Example 7: A web server logs 3 errors per hour on average. What is the probability that it logs 4 errors in 2 hours?
Solution:
In 2 hours:λ=3×2=6,x=4P(X=4)=4!e−6⋅64=24e−6⋅1296Using(e−6≈0.00248):P(X=4)=240.00248⋅1296≈0.1341Answer:(P(X=4)≈0.1341)
Example 8: A radioactive source emits on average 1.5 particles per second. What is the probability of detecting at least 2 particles in one second?
Solution:
We wantP(X≥2)=1−P(X=0)−P(X=1)Given(λ=1.5)(P(X=0)=e−1.5≈0.2231)(P(X=1)=1!1.5⋅e−1.5=1.5⋅0.2231≈0.3346)P(X≥2)=1−(0.2231+0.3346)=1−0.5577=0.4423Answer:(P(X≥2)≈0.4423)
9.0Poisson Distribution Practice Problems
- A machine produces 4 defective parts per hour on average. Find the probability of exactly 6 defects in one hour.
- A customer care center receives 2 complaints per day. What is the probability that it receives 5 complaints in a day?
- Emails arrive at a rate of 3 per minute. What is the probability that no emails arrive in the next 2 minutes?
- A power outage occurs on average once per month in a city. What is the probability of exactly 2 outages in a month?
- What is the probability of at most 2 accidents in a week if the average rate is 1.5 accidents per week?