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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) 0.1044, c) Expect 4 callers, P(none) ≈ 0.0183, d) ≈ 0.0907

Step by step solution

01

Determine the Rate for 5 Minutes

The arrival rate is 48 calls per hour. To find the rate per minute, divide by 60: \( \frac{48}{60} = 0.8 \) calls per minute. For a 5-minute interval, multiply: \( 0.8 \times 5 = 4 \) calls.
02

Use Poisson Formula for 3 Calls

The probability of receiving \( k \) calls in a fixed interval is given by the Poisson distribution: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \] Here, \( \lambda = 4 \) and \( k = 3 \). Substitute these into the formula: \[ P(X = 3) = \frac{4^3 e^{-4}}{3!} = \frac{64 e^{-4}}{6} \approx 0.1954 \].
03

Determine the Rate for 15 Minutes

For a 15-minute interval, use the same rate per minute: \( 0.8 \times 15 = 12 \) calls.
04

Use Poisson Formula for Exactly 10 Calls

Using the Poisson formula with \( \lambda = 12 \) and \( k = 10 \): \[ P(X = 10) = \frac{12^{10} e^{-12}}{10!} \approx 0.1044 \].
05

Compute Expected Callers Waiting After 5 Minutes

The expected number of calls in 5 minutes is \( \lambda = 4 \). Since no calls are currently on hold, after the call is completed, 4 calls are expected to have arrived.
06

Probability of No Callers Waiting

Using the Poisson formula for 0 calls with \( \lambda = 4 \): \[ P(X = 0) = e^{-4} \approx 0.0183 \].
07

Determine Probability of No Calls in 3 Minutes

For a 3-minute interval, the expected number of calls is \( \lambda = 2.4 \). The probability that no calls arrive is: \[ P(X = 0) = e^{-2.4} \approx 0.0907 \].

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Key Concepts

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

Probability Calculations
Probability calculations allow us to determine the likelihood of an event happening. In the context of Poisson distribution, these calculations are quite specific. The Poisson distribution helps in analyzing events that occur independently over a continuous interval of time or space.
For instance, when figuring out the likelihood of 3 calls in a 5-minute interval, you use \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \). Here,
  • \( X \) is the number of occurrences (calls in this example).
  • \( k \) is a specific number of events (3 calls).
  • \( \lambda \) represents the average number of occurrences in the given time period (4 calls for 5 minutes here).
Plug in these values to yield the probability, providing a clear answer to problems like the ones in the exercise.
This step-by-step method ensures you know not just how to reach a solution but why each number is used.
Arrival Rate
The arrival rate refers to the average number of events occurring in a fixed period. In this exercise, it's the number of phone calls expected per minute or hour. Starting with an hourly rate of 48 calls, it gets converted into more manageable units—per minute or per specific time frames (like 5 or 15 minutes).
The conversion is crucial. Divide by 60 to establish the per-minute rate \( \frac{48}{60} = 0.8 \). In other scenarios, you extend that calculation:
  • 5-minute rate: \( 0.8 \times 5 = 4 \)
  • 15-minute rate: \( 0.8 \times 15 = 12 \)
These rates become the foundation for plugging into distribution formulas. Understanding this concept allows you to predict activity flow better, handling real-world tasks smoothly without surprises.
Random Variables
Random variables are used in probability to symbolize an outcome subject to chance. In the Poisson distribution, these random variables often revolve around counts of event occurrences.
In our scenario, the number of calls arriving in a given time frame is a random variable. It's important to understand that while the average rate \( \lambda \) guides expectations (like 4 calls in 5 minutes), the actual observed number, such as 3 or 10, are random because call arrivals vary.
The beauty of random variables is their ability to model real-life randomness. They help calculate odds for various outcomes, providing insights and statistical backing to decision-making processes. These calculations make managing and expecting certain levels of activity far more manageable.

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Most popular questions from this chapter

A new automated production process averages 1.5 breakdowns per day. Because of the cost associated with a breakdown, management is concerned about the possibility of having three or more breakdowns during a day. Assume that breakdowns occur randomly, that the probability of a breakdown is the same for any two time intervals of equal length, and that breakdowns in one period are independent of breakdowns in other periods. What is the probability of having three or more breakdowns during a day?

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