Practice Problem Set 2 – Poisson and Gamma | Probability and Statistics Problem Solve

This post presents exercises on gamma distribution and Poisson distribution, reinforcing the concepts discussed in this blog post in a companion blog and blog posts in another blog. Because the shape parameter of the gamma distribution in the following problems is a positive integer, the calculation of probabilities for the gamma distribution is based on Poisson distribution.

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Practice Problem 2-A

Suppose that $X$ is the useful working life (in years) of a brand new industrial machine. The following is the probability density function of $X$ .

$\displaystyle f(t)=\frac{1}{24} \ \biggl(\frac{1}{5}\biggr)^5 \ t^4 \ e^{-\frac{1}{5} \ t} \ \ \ \ \ \ t>0$

A manufacturing plant has just purchased such a new machine. Determine the probability that the machine will be in operation for the next 20 years.

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Practice Problem 2-B

The annual rainfall (in inches) in Western Colorado is modeled by a distribution with the following cumulative distribution function.

$\displaystyle F(x)=1-e^{-0.2 x}-0.2 \ x \ e^{-0.2 x}-0.02 \ x^2 \ e^{-0.2 x} \ \ \ \ \ \ \ 0<x<\infty$

In a year in which the annual rainfall is above 20 inches, determine the probability that the annual rainfall is above 30 inches.

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Practice Problem 2-C

The annual rainfall (in inches) in Western Colorado is modeled by a distribution with the following cumulative distribution function.

$\displaystyle F(x)=1-e^{-0.2 x}-0.2 \ x \ e^{-0.2 x}-0.02 \ x^2 \ e^{-0.2 x} \ \ \ \ \ \ \ 0<x<\infty$

Determine the mean and the variance of the annual rainfall in this region.

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Practice Problem 2-D

The repair time (in hours) for an industrial machine has a gamma distribution with mean 1.5 and variance 0.75.

Determine the probability that a repair time exceeds 2 hours.
Determine the probability that a repair time is at least 5 hours given that it already exceeds 2 hours.

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Practice Problem 2-E

Customers arriving at a jewelry store according to a Poisson process with an average rate of 2.5 per hours. The store opens its door at 9 AM.

What is the probability that the first customer arrives at the store before 11 AM?
What is the probability that the first two customers arrive at the store before 11 AM?
What is the probability that the first three customers arrive at the store before 11 AM?
What is the probability that the first five customers arrive at the store before 11 AM?

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Practice Problem 2-F

In a certain city, telephone calls to 911 emergency response system arrive on the average of two every 3 minutes. Suppose that the arrivals of 911 calls are modeled by a Poisson process.

What is the probability of four or more calls arriving in a 5-minute period?
A call to the 911 system just ended. What is the probability that the wait time for the next 6 calls is more than 10 minutes?

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Practice Problem 2-G

Customers arrive at a store at an average rate of 30 per hour according to a Poisson process.

Determine the probability that at least 5 customers arrive at the store in the first 10 minutes after opening on a given day.
Determine the probability that, after opening, it will take more than 15 minutes for the 6th customer to arrive at the store.

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Practice Problem 2-H

Cars arrive at a highway tollbooth at an average rate of 6 cars every 10 minutes according to a Poisson process.

Determine the probability that the toll collector will have to wait longer than 20 minutes before collecting the seventh toll.
A toll collector just starts his shift. Determine the median time (in minutes) until he collects the first toll.

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Practice Problem 2-I

The number of claims in a year for an insured from a large group of insureds is modeled by the following model.

$\displaystyle P(X=x \lvert \lambda)=\frac{e^{-\lambda} \lambda^x}{x!} \ \ \ \ \ x=0,1,2,3,\cdots$

The parameter $\lambda$ varies from insured to insured. However, it is known that $\lambda$ is modeled by the following density function.

$\displaystyle g(\lambda)=32 \ \lambda^2 \ e^{-4 \lambda} \ \ \ \ \ \ \lambda>0$

An insured is randomly selected from the large group of insureds. Determine the mean and the variance of the number of claims for this insured in the next year.

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Practice Problem 2-J

Suppose that the number of accidents per year per driver in a large group of insured drivers follows a Poisson distribution with mean $\lambda$ . The parameter $\lambda$ follows a gamma distribution with mean 0.9 and variance 0.27.

Given that a randomly selected insured has at least one claim, determine the probability that the insured has more than one claim.

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Practice Problem 2-K

Customers arrive at a shop according to a Poisson process. The waiting time (in minutes) until the 5th customer is modeled by the following density function.

$\displaystyle f(t)=324 \ t^4 \ e^{-6 \ t} \ \ \ \ \ \ t>0$

Determine the mean and variance of the time until the arrival of the 6th customer after the opening of the shop on a given day.
Determine the probability that the wait for the 7th customer is longer than 2 minutes.

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*Problem*	………..Answer
2-A	$\displaystyle \frac{103}{3} e^{-4}=0.6288$

2-B	$\displaystyle \frac{25}{13} e^{-2}=0.2603$

2-C	mean = 15 variance = 75

2-D	$\displaystyle P(X>2)=13 e^{-4}$ $\displaystyle P(X>5 \lvert X>2)=\frac{61}{13} \ e^{-6}=0.01163$

2-E	$\displaystyle 1-e^{-5}=0.99326$ $\displaystyle 1-6 e^{-5}=0.95957$ $\displaystyle 1-18.5 e^{-5}=0.87535$ $\displaystyle 1-\frac{1569}{24} e^{-5}=0.55951$

2-F	$\displaystyle 1-\frac{1301}{81} e^{-10/3}=0.4270$ $\displaystyle \frac{197789}{729} e^{-20/3}=0.345285277$

2-G	$\displaystyle 1-\frac{1569}{24} e^{-5}=0.5595$ $\displaystyle \frac{52383.28125}{120} e^{-7.5}=0.241436$

2-H	$\displaystyle 7457.8 e^{-12}=0.04582$ $\displaystyle \frac{\text{ln}(0.5)}{-0.6}=1.15525$ min

2-I	mean = 0.75 variance = 0.9375

2-J	$\displaystyle \frac{0.6561}{1.3^4-1.3}=0.421631$

2-K	mean = 1, variance = 1/6 $\displaystyle 7457.8 e^{-12}=0.04582$