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Chapter 3: Problem 87

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. 0.099. b. 0.135. c. 2 calls.

Step by step solution

01

Understanding Part A

To find the probability that exactly ten requests are received during a 2-hour period, we use the Poisson probability formula: \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Here, \(\lambda\) is the average number of events in the given time period, and \(k\) is the exact number of events we want to find the probability for.
02

Calculating \(\lambda\) for Part A

Since the request rate \(\alpha\) is 4 per hour, for a 2-hour period, \(\lambda = 4 \times 2 = 8\). So, in a 2-hour period, the expected number of requests is 8.
03

Applying the Poisson Formula for Part A

We need to compute \(P(X = 10)\) where \(\lambda = 8\) and \(k = 10\).\[P(X = 10) = \frac{e^{-8} \cdot 8^{10}}{10!}\]Calculating this gives the probability.
04

Understanding Part B

For part B, we need to find the probability that no calls are received during a 30-minute break, which is a Poisson process with a 0.5-hour time frame. Here, \(\lambda = 4 \times 0.5 = 2\) as the rate is 4 calls per hour and the break lasts for 0.5 hours.
05

Calculating Probability for Part B

Using the Poisson formula for \(k = 0\), we find the probability of no calls:\[P(X = 0) = \frac{e^{-2} \cdot 2^0}{0!} = e^{-2}\]Evaluate \(e^{-2}\) to find the probability they do not miss any calls.
06

Understanding Part C

In part C, we need to find the expected number of calls during their 30-minute break. The expectation in a Poisson process is simply \(\lambda\). Here, for a 0.5-hour period, that would be \(\lambda = 4 \times 0.5 = 2\).
07

Conclusion

Summarize the results for each part of the exercise to ensure understanding. For part A, calculate the numeric value from Step 3. For B, use the value from Step 5. For C, state that 2 calls are expected.

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Key Concepts

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

Probability Calculation
Calculating probabilities using the Poisson distribution can be quite straightforward once you understand the formula. To determine the probability of a specific number of events occurring, we use the formula:
  • \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \)
In this formula:
  • \( \lambda \) represents the average number of occurrences within a specific time frame.
  • \( k \) is the number of events for which we want to find the probability.
The first step in solving a Poisson probability problem is to pinpoint the correct \( \lambda \). For instance, if we know the event rate is \( 4 \) per hour and we have a span of \( 2 \) hours, our \( \lambda \) would be \( 8 \). Then, plugging in \( k \) as \( 10 \), find the likelihood of exactly ten occurrences happening. This handy calculation aids in foreseeing events over time with remarkable precision.
Expected Value
In a Poisson distribution, the expected value, which quantifies the average number of occurrences, aligns directly with \( \lambda \). This can be enlightening because it tells us the anticipated event count over the chosen duration without even doing any math.
For example, if we examine a half-hour break window within an hourly rate of \( 4 \), the expected value becomes \( \lambda = 4 \times 0.5 = 2 \). This insight allows operators, like those in the towing service, to prepare for the likely number of calls they might face during their breaks.
The simplicity of using \( \lambda \) as the expected number makes planning and anticipation significantly more accessible.
Rate Parameter
The rate parameter \( \alpha \) dictates the rhythm of the Poisson process. It's akin to setting the beat for how frequently an event occurs within a given timeframe.
In our example, \( \alpha \) is given as \( 4 \) calls per hour. This \( \alpha \) forms the foundation for all calculations, whether predicting the future number of calls or calculating precise probabilities.
To tailor \( \lambda \) to varying timescales, simply multiply \( \alpha \) by the desired number of hours. It seamlessly transitions this abstract rate into practical terms, applicable directly to real-world scenarios. For calculating events in less than an hour, such as a 30-minute break, multiply the hourly rate by the fraction of the hour (e.g., \( 0.5 \) for half an hour). This adaptability of \( \alpha \) ensures solutions are always grounded with clear, predictable outcomes.

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Most popular questions from this chapter

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let \(X\) be the number among the six who are taking the test for the first time. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a certain chemical, which it sells to customers in 5 -lb batches. Let \(X=\) the number of batches ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X_{\text {.] }}\)

Each time a component is tested, the trial is a success ( \(S\) ) or failure \((F)\). Suppose the component is tested repeatedly until a success occurs on three consecutive trials. Let \(Y\) denote the number of trials necessary to achieve this. List all outcomes corresponding to the five smallest possible values of \(Y\), and state which \(Y\) value is associated with each one.

The probability that a randomly selected box of a certain type of cereal has a particular prize is .2. Suppose you purchase box after box until you have obtained two of these prizes. a. What is the probability that you purchase \(x\) boxes that do not have the desired prize? b. What is the probability that you purchase four boxes? c. What is the probability that you purchase at most four boxes? d. How many boxes without the desired prize do you expect to purchase? How many boxes do you expect to purchase?

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is \begin{tabular}{l|cccc} \(y\) & 0 & 1 & 2 & 3 \\ \hline\(p(y)\) & \(.60\) & \(.25\) & \(.10\) & \(.05\) \end{tabular} a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.
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