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The number of requests for assistance received by a towing service is a Poisson process with rate \(\alpha=4\) per hour. a. Compute the probability that exactly ten requests are received during a particular 2-hour period. b. If the operators of the towing service take a 30 -min break for lunch, what is the probability that they do not miss any calls for assistance? c. How many calls would you expect during their break?

Short Answer

Expert verified
a. The probability is approximately 0.099. b. The probability of no calls is about 0.135. c. Expect 2 calls during the break.

Step by step solution

01

Identify the Poisson Process Parameters

The problem states that the number of requests for assistance is modeled as a Poisson process with a rate \(\alpha = 4\) requests per hour.
02

Compute the Probability for 2-hour Window (Part a)

For part (a), we want to find the probability of receiving exactly ten requests during a 2-hour period. The rate for this period is \(\lambda = 4 \times 2 = 8\). The probability of receiving \(k\) requests is given by the Poisson probability formula: \[P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\] Plugging in \(k=10\) and \(\lambda=8\), we get: \[P(X=10) = \frac{8^{10} e^{-8}}{10!}\] Calculate this probability using a calculator.
03

Understand Break Duration in Hours (Part b)

The operators take a 30-minute break. Convert this duration to hours: \(0.5\) hours.
04

Compute the Probability of Zero Requests during Break (Part b)

For part (b), the rate during the break is \(\lambda = 4 \times 0.5 = 2\). We need to calculate the probability of receiving zero requests: \[P(X=0) = \frac{2^0 e^{-2}}{0!} = e^{-2}\]Calculate \(e^{-2}\) to find the probability of no calls during the break.
05

Calculate Expected Calls during Break (Part c)

The expected number of requests in any Poisson process over a time period is given by the rate \(\lambda\). For the 30-minute break, the expected number of calls is \(\lambda = 2\).

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Key Concepts

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

Poisson distribution
The Poisson distribution is a probability distribution that's often useful for modeling the number of events occurring in a fixed interval of time or space. In this exercise, it is used to model requests for assistance received by a towing service. The core parameter in a Poisson distribution is the average rate, denoted by \( \alpha \). This rate tells us how many events, on average, are expected to occur during a single time period, such as an hour.
In mathematical terms, if \( X \) represents the number of events in a given period, \( X \) is distributed as a Poisson random variable with parameter \( \lambda = \alpha \cdot t \), where \( t \) is the length of time you're considering. Here, \( \alpha = 4 \) requests per hour, which means on average, you receive four calls per hour.
probability calculations
Probability calculations in the context of a Poisson distribution involve using the Poisson formula: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]where \( \lambda \) represents the average number of events over the interval, \( k \) is the number of events you're interested in, and \( e \) is approximately 2.718, a fundamental constant.
For example, to reveal the probability of 10 requests in a 2-hour period when requests occur at a rate of 4 per hour, first calculate \( \lambda = 4 \times 2 = 8 \). Then, plug \( \lambda = 8 \) and \( k = 10 \) into the formula:
\[ P(X = 10) = \frac{8^{10} e^{-8}}{10!} \]
This probability helps to know how likely exactly 10 requests will be during that time frame.
expected value
The expected value in the context of a Poisson process refers to the average number of requests you can anticipate during any given time period. It's computed simply using the rate of the process over the desired time period. In mathematical terms, the expected number, \( E(X) \), is given by the formula:\[ E(X) = \lambda \]where \( \lambda = \alpha \cdot t \).
As seen in the exercise, during a 30-minute break (0.5 hours), the expected number of calls would be:\[ E(X) = 4 \times 0.5 = 2 \]
On average, you expect 2 calls during that 30-minute interval.
request rate
Request rate is a measure of how frequently requests occur over a given time period. In a Poisson process, this rate is symbolized by \( \alpha \), which is crucial because it indicates the likelihood of receiving requests over time.
In the towing service example, the rate is \( \alpha = 4 \) requests per hour. This constant helps determine the probability and expected value of requests over any desired timeframe. The request rate is also multiplied by the length of the time period \( t \) to compute the parameter \( \lambda \) used in both probability calculations and expected outcomes. This allows us to predict occurrences like how often operators will receive calls or miss none during breaks.

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Most popular questions from this chapter

An appliance dealer sells three different models of upright freezers having \(13.5,15.9\), and \(19.1\) cubic feet of storage space, respectively. Let \(X=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(X\) has pmf \begin{tabular}{l|ccc} \(x\) & \(13.5\) & \(15.9\) & \(19.1\) \\ \hline\(p(x)\) & \(.2\) & \(.5\) & \(.3\) \end{tabular} a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) cubic feet is \(25 X-8.5\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price \(25 X-8.5\) paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-01 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?

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An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let \(p=\) the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than \(80 \%\) of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let \(X\) denote the number of homes with detectors among the 25 sampled. Consider rejecting the claim that \(p \geq .8\) if \(x \leq 15\) a. What is the probability that the claim is rejected when the actual value of \(p\) is 8 ? b. What is the probability of not rejecting the claim when \(p=.7 ?\) When \(p=.6 ?\) c. How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14 ?

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