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An insurance salesperson sells an average of \(1.4\) policies per day. a. Using the Poisson formula, find the probability that this salesperson will sell no insurance policy on a certain day. b. Let \(x\) denote the number of insurance policies that this salesperson will sell on a given day. Using the Poisson probabilities table, write the probability distribution of \(x\). c. Find the mean, variance, and standard deviation of the probability distribution developed in part b.

Short Answer

Expert verified
The probability of the salesperson selling no policy is \(P(0;1.4) = \frac{1.4^0 e^{-1.4}}{0!}\). The probability distribution of \(x\) is given by \(P(x;1.4) = \frac{1.4^x e^{-1.4}}{x!}\) for \(x = 0, 1, 2, 3, ...\). The mean, variance and standard deviation of the distribution are \(1.4\), \(1.4\) and \(\sqrt{1.4}\) respectively.

Step by step solution

01

Using Poisson formula to find the probability

The probability mass function of a Poisson distribution is given by the formula \(P(x;\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}\). In this case, \(x = 0\) (the number of insurance policies sold) and \(λ = 1.4\) (the average number of policies sold per day). That gives \(P(0;1.4) = \frac{1.4^0 e^{-1.4}}{0!}\).
02

Deriving the Probability Distribution of \(x\)

The Poisson distribution is a discrete probability distribution, meaning that it defines the probabilities of a number of discrete outcomes (which are nonnegative integers). For a Poisson distribution with \(λ = 1.4\), the probability distribution is given by the function \(P(x;1.4) = \frac{1.4^x e^{-1.4}}{x!}\) for \(x = 0, 1, 2, 3, ...\). Therefore, the probability distribution of \(x\) can be tabulated using this function to calculate each probability for the value of \(x\).
03

Finding the Mean, Variance and Standard Deviation

The mean (\(μ\)), variance (\(σ^2\)) and standard deviation (\(σ\)) of a Poisson distribution with parameter \(λ\) are \(μ = λ\), \(σ^2 = λ\) and \(σ = \sqrt{λ}\) respectively. Substitute \(λ = 1.4\) into these formulas to calculate the corresponding mean, variance and standard deviation.

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is a fundamental concept in the Poisson distribution. It specifies the probability of obtaining a particular number of events in a fixed interval of time or space. For the Poisson distribution, the probability that there are exactly \(x\) events is given by the formula: \[ P(x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!} \] Here, \( \lambda \) is the average number of events occurring in a fixed interval, and \( e \) is the base of the natural logarithm, approximately equal to 2.71828. The variable \(x!\) represents the factorial of \(x\), which is the product of all positive integers up to \(x\).
  • In the given example, \(\lambda = 1.4\) indicates an average of 1.4 policy sales per day.
  • The PMF can calculate the likelihood of selling exactly \(x\) number of policies in a day.
Understanding the PMF helps in deriving the entire distribution of events over time, making it a powerful tool for applications where the number of events follows the Poisson process.
Mean of Poisson Distribution
The mean of a Poisson distribution is denoted by \( \mu \) and is simply equal to the rate parameter \(\lambda\). In the Poisson distribution, the mean essentially represents the expected number of occurrences of the event in a given interval.
  • For example, with \( \lambda = 1.4\), the mean number of policies sold per day also turns out to be 1.4.
This mean value is significant because it provides a central measure around which the values cluster in a Poisson distribution. It allows us to assess how often a typical outcome (here, selling a certain number of policies) occurs. The mean informs us about the long-term average number of events if the scenario is repeated numerous times.
Variance of Poisson Distribution
The variance of a Poisson distribution is a measure of the spread or dispersion of the distribution. Interestingly, for a Poisson distribution, the variance \(\sigma^2\) is equal to the mean: \( \sigma^2 = \lambda \).
  • This implies that with \(\lambda = 1.4\), the variance also equals 1.4.
The reason the variance equals the mean is due to the nature of the Poisson process, where events are independently occurring and randomly spaced in time or space. This relationship between the mean and variance exemplifies the distinct characteristics of the Poisson distribution. It shows that as the mean number of events increases, so does the variability in the number of events.
Standard Deviation of Poisson Distribution
The standard deviation is a critical measure of dispersion, indicating how much individual observations tend to differ from the mean. For the Poisson distribution, the standard deviation \(\sigma\) is derived from the variance \(\sigma^2\), and is given by the square root of \(\lambda\):\[ \sigma = \sqrt{\lambda} \]
  • With a \( \lambda \) of 1.4, the standard deviation is \(\sqrt{1.4} \approx 1.18\).
The standard deviation provides insights into the spread of the distribution and highlights the extent of deviation from the average number of occurrences. In practical terms, it's a valuable tool to understand the variability one might expect around the mean in real-world scenarios covered by the Poisson distribution.

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Most popular questions from this chapter

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