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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 Poisson Distribution

A Poisson distribution is characterized by a single parameter \( \lambda \), which represents both the mean and variance of the distribution. The probability mass function for a Poisson distribution is given by \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( k \) is the number of occurrences.
02

Given Condition for Zero Customers

We are given that \( P(X=0)=0.05 \). This implies \( \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} = 0.05 \). This equation allows us to solve for \( \lambda \).
03

Solving for Lambda

We solve the equation \( e^{-\lambda} = 0.05 \) to determine \( \lambda \). Taking the natural logarithm of both sides, we get \( -\lambda = \ln(0.05) \), thus \( \lambda = -\ln(0.05) \).
04

Calculating the Mean and Variance

In a Poisson distribution, the mean \( \mu \) is equal to \( \lambda \), and the variance \( \sigma^2 \) is also equal to \( \lambda \). Hence, both the mean and the variance of \( X \) are \( -\ln(0.05) \).
05

Simplifying \(-\ln(0.05)\)

To find the numerical value, calculate \( -\ln(0.05) \), which is approximately 2.9957. Thus, the mean and variance both approximate to 2.9957.

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Key Concepts

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

Mean of Poisson Distribution
In a Poisson distribution, the mean is a central concept that plays a crucial role in defining the distribution. It is denoted by the symbol \( \lambda \), which is the average rate at which events occur within a fixed interval of time or space. For example, if we have a value \( \lambda = 3 \), it indicates that on average, three events occur in the specified time frame. The beauty of the Poisson distribution lies in its simplicity, where the parameter \( \lambda \) itself is both the mean and also serves as the variance. This dual role sets Poisson apart from other distributions, offering a straightforward interpretation of the data.

When determining the mean, here's a practical approach:
  • Identify the average rate of occurrence (\( \lambda \)).
  • Recognize that the mean is directly equal to \( \lambda \).
  • Use this mean to predict the average outcome over time.
This underscores the distribution's applicability to real-world phenomena, such as the number of phone calls received at a call center over an hour. Understanding the mean enables you to have a foundational insight into the behavior of the distribution.
Variance of Poisson Distribution
The variance in a Poisson distribution also holds a pivotal position. In most distributions, variance is a separate calculation giving insight into the spread or variability of the data. However, Poisson distribution stands out because here the variance is equal to the mean. That's right—when dealing with a Poisson distribution, the variance is simply \( \lambda \), the same value as the mean.

This fascinating aspect means:
  • Every increase in \( \lambda \) results in a proportional increase in variance.
  • The data spread will increase at the same rate as the average occurrence rate (mean).
  • Examples where this concept applies include counting the number of cars passing through a toll both in an hour.
Thus, understanding variance in Poisson distribution helps us measure how much the events vary around the mean, giving insights into the reliability of the mean as a predictive measure. When planning or making decisions, grasping how variance behaves can provide clarity on whether the mean will fluctuate regularly or stay consistent over time.
Probability Mass Function
The probability mass function (PMF) in a Poisson distribution is a mathematical formulation that helps us calculate the probability of a certain number of events occurring in a fixed interval. It is expressed as:
\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Here, \( e \) is the base of the natural logarithm (approximately 2.71828), \( \lambda \) is the mean rate of occurrence, and \( k \) is the number of events you're interested in.

This equation helps by:
  • Providing a precise way to calculate probabilities for different numbers of occurrences.
  • Being versatile, as you can plug in different values of \( k \) to get different probabilities.
  • Allowing you to easily compute the likelihood of rare events based on the known average rate.
For instance, if you want to know the probability of exactly 2 customers arriving at a bank when \( \lambda = 3 \), you just substitute these values into the PMF formula. Understanding the PMF is essential for statistical modeling in various fields, such as finance, natural sciences, and traffic engineering, where event occurrence prediction is vital.

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Most popular questions from this chapter

Suppose \(X\) has a hypergeometric distribution with \(N=10, n=3,\) and \(K=4 .\) Sketch the probability mass function of \(X\). Determine the cumulative distribution function for \(X\)

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