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Suppose that \(X\) has a Poisson distribution with a mean of 0.4 . Determine the following probabilities: (a) \(P(X=0)\) (b) \(P(X \leq 2)\) (c) \(P(X=4)\) (d) \(P(X=8)\)

Short Answer

Expert verified
(a) 0.6703, (b) 0.9913, (c) 0.000862, (d) 2.037 脳 10鈦烩伔

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution is defined by the formula \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \) where \( \lambda \) is the mean of the distribution and \( k \) is the number of occurrences. In this exercise, \( \lambda = 0.4 \).
02

Calculate Probability for (a) P(X=0)

We use the Poisson formula for \( k = 0 \). Substitute \( \lambda = 0.4 \) and \( k = 0 \) into the formula: \[P(X = 0) = \frac{0.4^0 \cdot e^{-0.4}}{0!} = e^{-0.4}\]Evaluating this gives \[P(X = 0) \approx 0.6703.\]
03

Calculate Probability for (b) P(X鈮2)

For \( P(X \leq 2) \), calculate and sum \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \):\[P(X = 1) = \frac{0.4^1 \cdot e^{-0.4}}{1!} = 0.4 \cdot e^{-0.4},\]\[P(X = 2) = \frac{0.4^2 \cdot e^{-0.4}}{2!} = 0.08 \cdot e^{-0.4}\]Adding them together, \[P(X \leq 2) = e^{-0.4} + 0.4 \cdot e^{-0.4} + 0.08 \cdot e^{-0.4} \approx 0.9913.\]
04

Calculate Probability for (c) P(X=4)

Using \( k = 4 \) in the Poisson formula: \[P(X = 4) = \frac{0.4^4 \cdot e^{-0.4}}{4!} = 0.00256 \cdot e^{-0.4}\]Evaluate this probability: \[P(X = 4) \approx 0.000862.\]
05

Calculate Probability for (d) P(X=8)

Using \( k = 8 \) in the Poisson formula: \[P(X = 8) = \frac{0.4^8 \cdot e^{-0.4}}{8!} = 6.5536 \times 10^{-6} \cdot e^{-0.4} \]Evaluate this probability: \[P(X = 8) \approx 2.037 \times 10^{-7}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation is fundamental to understanding the Poisson distribution. In probability theory, the aim is to determine the likelihood of different outcomes occurring. When dealing with Poisson distribution, which describes the probability of a given number of events happening in a fixed interval of time or space, the calculation involves a few clear steps. For this distribution, probabilities are determined using the mean, known as \( \lambda \), and a specific number of occurrences, \( k \).

In this context, the probability that a specific number of events take place is calculated using the Poisson formula. For instance, to calculate the probability of 0 events occurring when the mean is 0.4, you substitute the values into a specific formula. This involves some computation but provides a precise chance of that event happening. By understanding this, you can calculate the probability for any number of occurrences in the given time or space interval.

The simplicity and effectiveness of the Poisson distribution make it widely useful in fields like telecommunications, traffic engineering, and ecology, where often, the goal is to understand how certain events cluster over a period or area.
Poisson Formula
The Poisson formula is the backbone of calculating probabilities within the Poisson distribution. It is given by the expression: \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \). In this formula:
  • \( \lambda \) represents the mean number of occurrences.
  • \( k \) is the specific number of occurrences you are calculating the probability for.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( k! \) ("k factorial") is the product of all positive integers up to \( k \), and is used to normalize the probability.

This formula is elegant and powerful, allowing you to calculate the likelihood of different \( k \) values under a Poisson process. For example, substituting the values \( \lambda = 0.4 \) and \( k = 0 \) into the formula gives the probability of no events occurring. The computations required initially appear complex, but with practice, they become manageable.

Different values of \( k \) can be handled in the same way to find the probability of those occurrences. The Poisson formula is not only pivotal in theoretical mathematics but is also applied extensively in real-world scenarios, helping quantify phenomena in varied domains.
Discrete Probability Distributions
Discrete probability distributions, like the Poisson distribution, focus on variables that take discrete values鈥攄istinct and separate numbers or events. In simpler terms, these distributions handle scenarios where outcomes are countable, for example, the number of emails received in an hour, or the number of calls a call center gets in a day.

The Poisson distribution is a key type of discrete probability distribution used when events occur independently over a fixed period. Each possible number of events \( k \) has an associated probability, derived using the Poisson formula. This specificity to countable occurrences makes it particularly useful in understanding and modeling natural phenomena or service industry operations where patterns or clustering of events are significant.

While there are other types of discrete distributions, such as binomial or geometric distributions, Poisson distributions are unique in that they do not require a fixed number of trials, but rather focus on the average rate (\( \lambda \)) of occurrence over a potential infinity of trials.

Understanding discrete probability distributions helps in decision-making processes, statistical analyses, and various forecasting models. By grasping how the Poisson distribution is just one form of such distributions, you'll be better equipped to identify and apply the appropriate methodology in relevant practical contexts.

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Most popular questions from this chapter

An electronic scale in an automated filling operation stops the manufacturing line after three underweight packages are detected. Suppose that the probability of an underweight package is 0.001 and each fill is independent. (a) What is the mean number of fills before the line is stopped? (b) What is the standard deviation of the number of fills before the line is stopped?

Suppose that \(X\) is a negative binomial random variable with \(p=0.2\) and \(r=4\). Determine the following: (a) \(E(X)\) (b) \(P(X=20)\) (c) \(P(X=19)\) (d) \(P(X=21)\) (e) The most likely value for \(X\)

Suppose that \(X\) has a Poisson distribution with a mean of \(4 .\) Determine the following probabilities: (a) \(P(X=0)\) (b) \(P(X \leq 2)\) (c) \(P(X=4)\) (d) \(P(X=8)\)

Heart failure is due to either natural occurrences \((87 \%)\) or outside factors \((13 \%) .\) Outside factors are related to induced substances or foreign objects. Natural occurrences are caused by arterial blockage, disease, and infection. Suppose that 20 patients will visit an emergency room with heart failure. Assume that causes of heart failure for the individuals are independent. (a) What is the probability that three individuals have conditions caused by outside factors? (b) What is the probability that three or more individuals have conditions caused by outside factors? (c) What are the mean and standard deviation of the number of individuals with conditions caused by outside factors?

The number of errors in a textbook follows a Poisson distribution with a mean of 0.01 error per page. What is the probability that there are three or fewer errors in 100 pages?

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