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Phone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways. a. Compute the probability of receiving three calls in a 5-minute interval of time. b. Compute the probability of receiving exactly 10 calls in 15 minutes. c. Suppose no calls are currently on hold. If the agent takes 5 minutes to complete the current call, how many callers do you expect to be waiting by that time? What is the probability that none will be waiting? d. If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted by a call?

Short Answer

Expert verified
a. 0.1954 b. 0.1048 c. Expect 4 calls; probability of none waiting is 0.0183. d. 0.0907

Step by step solution

01

Identify the Poisson distribution parameter

The phone calls arrive at an average rate of 48 calls per hour, which is 0.8 calls per minute. Let's convert this rate to the context of each part of the question.
02

Step A: Calculate the 5-minute interval parameter

For 5 minutes, we calculate the average as follows: \[ \lambda = 0.8 \text{ calls/minute} \times 5 \text{ minutes} = 4 \text{ calls} \] This means over 5 minutes, the average arrival rate is 4 calls.
03

Step A: Probability of receiving three calls

Using the Poisson probability formula: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where \( k=3 \) and \( \lambda=4 \), we calculate:\[ P(X=3) = \frac{e^{-4} \cdot 4^3}{3!} \approx 0.1954 \]
04

Step B: Calculate the 15-minute interval parameter

For 15 minutes, we calculate the average as follows: \[ \lambda = 0.8 \text{ calls/minute} \times 15 \text{ minutes} = 12 \text{ calls} \] This means 12 calls are expected in 15 minutes.
05

Step B: Probability of receiving exactly 10 calls

Using the Poisson probability formula again with \( k=10 \) and \( \lambda=12 \): \[ P(X=10) = \frac{e^{-12} \cdot 12^{10}}{10!} \approx 0.1048 \]
06

Step C: Calculate the expected number of calls in 5 minutes

In 5 minutes, we already identified that the average number of calls (\( \lambda \)) is 4. This is the expected number of callers after 5 minutes.
07

Step C: Probability of no calls arriving

Using \( k=0 \) and \( \lambda=4 \):\[ P(X=0) = \frac{e^{-4} \cdot 4^0}{0!} = e^{-4} \approx 0.0183 \]
08

Step D: Probability of no interruptions in 3 minutes

In 3 minutes, \( \lambda = 0.8 \times 3 = 2.4 \). The probability that no calls arrive (\( k=0 \)) is:\[ P(X=0) = \frac{e^{-2.4} \cdot 2.4^0}{0!} = e^{-2.4} \approx 0.0907 \]

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Key Concepts

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

Probability Calculation
When dealing with events occurring randomly over a fixed period, we often use probability calculations to make predictions. One common probability distribution model used in this context is the Poisson distribution, which helps us determine the likelihood of a certain number of events, such as phone calls, happening in a specified timeframe.

To calculate the probability using the Poisson distribution, we apply the formula:
  • \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \)
Here, \( \lambda \) represents the average number of occurrences in the interval, \( k \) is the number of events we're interested in, and \( e \) is the base of the natural logarithm (approximately 2.71828).

This formula allows us to find probabilities for specific event counts within given periods, aiding in effective decision-making.
Arrival Rate
The arrival rate is crucial in understanding how often events, like phone calls, occur over time. It defines the average number of times an event occurs in a unit of time and is a vital parameter in statistical modeling, particularly with Poisson distribution.

For example, if phone calls arrive at an average rate of 48 per hour, converting this into shorter intervals becomes necessary. We might be interested in a 5-minute timeframe, for instance. To find the rate, multiply the hourly rate by the desired time in minutes:
  • Example: \( \lambda = 0.8 \text{ calls/minute} \times 5 \text{ minutes} = 4 \text{ calls} \)
This conversion helps us understand the expected frequency of events in different intervals, allowing us to use it for the relevant probability computations.
Statistical Modeling
Statistical modeling allows us to represent and analyze complex real-world processes mathematically. Through models like the Poisson distribution, we can make sense of random events' behavior over time and predict outcomes.

Using the Poisson distribution model involves setting up parameters based on real-world data—such as an arrival rate of events—and applying formulas to determine the probability of various outcomes. This type of modeling is not only useful for calculating probabilities but also helps in optimizing operations, forecasting demand, and assessing risk.

By examining how likely events are to occur, organizations can prepare and respond more effectively, increasing efficiency in scenarios that resemble the phone call arrival example.
Expected Value
The concept of expected value plays a fundamental role in assessing potential outcomes in probability and statistics. It represents the average or mean value we anticipate obtaining in a scenario after conducting many trials.

In Poisson distribution problems, the expected value is often denoted as \( \lambda \), essentially the average rate of event occurrence over a unit of time. For example, if in a given exercise, 4 calls are expected in a 5-minute interval, then the expected value is 4.

The expected value gives insights into what is most likely to occur over time. For practical applications, understanding this average helps in planning resources effectively, such as staffing to handle incoming phone calls, without wasting unnecessary resources or facing shortages. It offers a clearer picture of what a typical scenario will look like when random events are involved.

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