Poisson Distribution Practice Problems

Poisson distribution is a probability distribution that models the number of events occurring within a fixed interval of time or space, where these events happen with a known constant mean rate and independently of the time since the last event. It is named after the French mathematician Siméon Denis Poisson.

Suppose a call center receives an average of 10 calls per hour. We can model the number of calls received in a given hour using a Poisson distribution with λ = 10. There are many scenarios that can be modelled with the help of Poisson Distribution. Some of these scenarios are:

• An intersection has an average of 3 cars passing through every minute.
• A factory produces 1000 widgets per day, with an average of 2 defective widgets.
• A website gets an average of 50 hits per minute.
• In a strand of DNA, an average of 0.3 mutations occur per unit length.

This article has covered practice questions on Poisson Distribution with solutions in detail.

The table below represents the important formulas of Poisson distribution.

Where,

• ƛ is Mean
• x is Number of Required Outcomes
• n is Total Number of Trials
• p is Probability of Success
• Var(X) is Variance
• σ is Standard Deviation

Q1. If 2% of the product made in a company are defective. Find the probability that less than 1 item is defective in the sample of 100 items.

Solution:

Here, n = 100, p = (2/100) = 0.02, q = 0.98, ƛ = np = 2

The formula for the required probability is given by:

P (X = x) = [ƛ^x × e^-ƛ] / x!

P(X > 1) = P(X = 0)

P(X > 1) = [2⁰ × e^-2] / 0!

P(X > 1) = e^-2

Q2. If the probability of choosing incorrect answer in an examination is 0.01, determine the chance that out of 500 students more than 1 will choose incorrect answer.

Solution:

Here, n = 500, p = 0.01, ƛ = np = 5

The formula for the required probability is given by:

P (X = x) = [ƛ^x × e^-ƛ] / x!

P(X > 1) = 1 – [P(X = 0) + P(X = 1)]

P(X = 0) = [5⁰ × e^-5] / 0! = e^-5

P(X = 1) = [5¹ × e^-5] / 1! = 5e^-5

P(X > 1) = 1 – e^-2 [ 1 + 5]

P(X > 1) = 1 – 6e^-2

P(X >1) = 1 – 0.812

P(X > 1) = 0.188

Q3. A man can send an email on average 2 emails per hour. What is the probability that the man sends no email in a given hour?

Solution:

Mean = 2

The formula for the required probability is given by:

P (X = x) = [ƛ^x × e^-ƛ] / x!

P(X = 0) = [2⁰ × e^-2] / 0! = e^-2

P(X = 0) = 0.135

Q4. Calculate the mean of the Poisson Distribution given that the number of trials is 20 and probability of success is 0.6.

Solution:

The mean is the Poisson distribution is given by:

Mean = np

Here, n = 20 and p = 0.6

Mean = 20 × 0.6

Mean = 12

Q5. Find the mean of the Poisson distribution given that the variance of the Poisson distribution is 4.

Solution:

We know that,

In Poisson distribution Mean = Variance = np

Mean = Variance

Mean = 4

Q6. Calculate the variance and standard deviation of the Poisson distribution given the number of trials as 50 and probability of failure as 0.3.

Solution:

Formula for the variance and standard deviation in Poisson distribution is given by:

Variance = np and Standard Deviation = √(np)

Here, n = 50 and p = 1- q = 0.7

Variance = n × p

Variance = 50 × 0.7 = 35

Standard Deviation = √Variance

Standard Deviation = √35

Q1. If 5% of the product made in a company are defective. Find the probability that less than 3 item is defective in the sample of 200 items.

Q2. If the probability of choosing incorrect answer in an examination is 0.001, determine the chance that out of 1000 students more than 3 will choose incorrect answer.

Q3. A shop has an average of 10 customers per hour. What is the probability that exactly 12 customers arrive in an hour?

Q4. Calculate the mean of the Poisson Distribution given that the number of trials is 100 and probability of success is 0.9.

Q5. Find the mean of the Poisson distribution given that the variance of the Poisson distribution is 20.

Q6. Calculate the variance and standard deviation of the Poisson distribution given the number of trials as 200 and probability of success as 0.8.

What is Poisson Mean Formula?

Poisson mean formula is given by:

E(X) = ƛ = n.p

What is the Formula for Poisson Distribution?

Formula for the Poisson distribution is given by:

P (X = x) = [ƛ^x × e^-ƛ] / x!

Is Poisson Discrete or Continuous?

Poisson distribution is a discrete probability distribution.

Can Poisson Distribution be zero?

Yes, Poisson distribution can be zero if its mean is 0.