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Suppose that the number of customers who enter a bank in an hour is a Poisson random variable, and suppose that \(P(X=0)=0.05 .\) Determine the mean and variance of \(X .\)

Short Answer

Expert verified
The mean and variance of \(X\) are both approximately 2.9957.

Step by step solution

01

Understanding the Poisson Distribution

In a Poisson distribution, if the number of events (in this case, bank customers) in a fixed interval is represented by \(X\), then \(X\) is described by \(P(X=k)=\frac{e^{-\lambda} \lambda^k}{k!}\) for \(k=0,1,2,...\) where \(\lambda\) is the average number of events (mean). The variance of a Poisson distribution is equal to the mean \(\lambda\).
02

Given Probability and Equation Setup

We are given that \(P(X=0) = 0.05\). Using the formula for Poisson distribution \(P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!}\), which simplifies to \(P(X=0) = e^{-\lambda}\). Therefore, \(e^{-\lambda} = 0.05\).
03

Solving for \(\lambda\)

The equation \(e^{-\lambda} = 0.05\) can be solved by taking the natural logarithm of both sides: \(-\lambda = \ln(0.05)\). Thus, \(\lambda = -\ln(0.05)\).
04

Calculating \(\lambda\)

Calculate \(\lambda\) using a calculator: \(\lambda = -\ln(0.05) \approx 2.9957\).
05

Concluding Mean and Variance

In a Poisson distribution, both the mean and the variance are equal to \(\lambda\). Therefore, the mean is approximately 2.9957 and so is the variance.

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Key Concepts

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

Mean and Variance
When discussing the Poisson distribution, the notion of 'mean' and 'variance' plays a crucial role. The Poisson distribution measures how likely a particular number of events will occur in a fixed interval of time or space. Here, the mean, often represented as \( \lambda \), indicates the average number of occurrences (like bank customers) within that period.
The interesting part about the Poisson distribution is that the mean \( \lambda \) and the variance are identical. This feature is distinctive because in many other probability distributions, mean and variance can differ. The simplicity of this equality makes Poisson distribution handy in theoretical and practical applications such as predicting customer traffic, call frequency, or system failures.
  • Mean (\( \lambda \)): Represents the average number of occurrences in a given time/space.
  • Variance: Similar to mean, indicating how much the number of events can spread out from the average.
Probability Theory
Probability theory is the mathematical framework that allows us to predict the likelihood of different outcomes. In the case of the Poisson distribution, probability theory helps us determine the probability of a given number of events occurring in a specified interval.
For the Poisson distribution, the formula \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \) describes the probability of \( k \) events happening when the mean number is \( \lambda \). This formula shows that the probability depends on both the rate of events \( \lambda \) and the number of occurrences \( k \). Understanding this balance is key in applying Poisson probabilities accurately.
  • \( e^{-\lambda} \) represents the exponential decay function, reducing odds as \( \lambda \) increases.
  • \( \lambda^k \) signifies the impact of the mean on the probability of \( k \) events.
  • \( k! \) (factorial of \( k \)) normalizes the probabilities, accounting for the variety of possible combinations.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a crucial mathematical function encountered when working with exponential growth or decay processes. When dealing with Poisson distributions, the natural logarithm becomes important particularly when solving for \( \lambda \), the mean and variance.
In our exercise, we are given that \( P(X=0) = 0.05 \), which can be represented in the equation \( e^{-\lambda} = 0.05 \). To solve for \( \lambda \), we need to apply the natural logarithm to both sides: \( \ln(0.05) = -\lambda \).
The natural logarithm helps transform multiplicative processes into an additive formula, simplifying complicated equations. This transformation provides ease when isolating variables like \( \lambda \). Understanding how and when to use \( \ln \) efficiently is a valuable mathematical skill.
  • \( \ln \) is essential for solving equations involving exponential functions.
  • Allows the conversion of multiplicative relationships into manageable additive expressions.
  • Crucial for isolating variables within a formula.

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