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During the period of time that a local university takes phone-in registrations, calls come in at the rate of one every two minutes. a. What is the expected number of calls in one hour? b. What is the probability of three calls in five minutes? c. What is the probability of no calls in a five-minute period?

Short Answer

Expert verified
a: 30 calls, b: 0.2138, c: 0.0821 probability

Step by step solution

01

Identify the rate of calls

The problem states that calls come in at the rate of one every two minutes. This means the rate \( \lambda \) can be described as \( \lambda = \frac{1}{2} \text{ calls per minute} \).
02

Calculate the expected number of calls in one hour

The expected number of calls \( \mu \) in one hour (60 minutes) can be calculated by multiplying the rate \( \lambda \) by 60. Thus, \( \mu = \lambda \times 60 = \frac{1}{2} \times 60 = 30 \) calls per hour.
03

Determine the rate for a 5-minute period

For a 5-minute period, the rate \( \lambda' \) is \( \lambda' = \lambda \times 5 = \frac{1}{2} \times 5 = 2.5 \) calls.
04

Calculate the probability of three calls in five minutes using Poisson distribution

The probability \( P(X=k) \) of exactly \( k \) events in a Poisson distribution can be calculated with the formula:\[ P(X=k) = \frac{\lambda'^k \cdot e^{-\lambda'}}{k!} \]Plugging in \( k=3 \) and \( \lambda'=2.5 \), we have:\[ P(X=3) = \frac{(2.5)^3 \cdot e^{-2.5}}{3!} = \frac{15.625 \cdot e^{-2.5}}{6} \approx 0.2138 \]
05

Calculate the probability of no calls in a five-minute period using Poisson distribution

Using the same Poisson formula with \( k=0 \):\[ P(X=0) = \frac{(2.5)^0 \cdot e^{-2.5}}{0!} = e^{-2.5} \approx 0.0821 \]

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Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur. In the context of Poisson distribution, it helps us estimate the likelihood of a certain number of events happening within a fixed time frame. For example, with phone-in registrations, we are interested in predicting how many calls will be received in a certain period. The formula used in Poisson probability is:
  • \( P(X=k) = \frac{\lambda'^k \cdot e^{-\lambda'}}{k!} \)
Here, \( \lambda' \) represents the expected number of events in the given time interval, \( k \) is the number of occurrences you are interested in, and \( e \) is the base of the natural logarithm (approximately 2.71828).
The flexibility of the Poisson distribution makes it a valuable tool for handling relatively rare events occurring within a time or space, for instance, phone calls received by a university within an hour. This helps us not only predict occurrences but also plan and prepare for them accordingly.
Expected Value
The expected value represents the average number of occurrences over time. When using the Poisson distribution, the expected value gives us a base to understand what the typical case would look like. For instance, if calls arrive at a university at a rate of one every two minutes, the expected number of calls in an hour, calculated using the rate \( \lambda \), would be:
  • \( \mu = \lambda \times 60 = \frac{1}{2} \times 60 = 30 \) calls per hour.
To find this expected value, simply multiply the average rate of arrival by the duration of the period you are interested in. This gives you a middle ground or average number of calls you can expect. It helps in resource planning, such as how many staff might need to handle the anticipated volume of calls during any timeframe, like during registration periods at the university.
By understanding the expected value, organizations can efficiently manage their operations and ensure they are optimally equipped to handle expected workloads.
Rate of Events
The rate of events in a Poisson distribution is a crucial component, often symbolized as \( \lambda \). It defines how frequently the events you're studying occur. In our problem, this is the rate at which phone calls are received. Given that a call is received every two minutes, the rate \( \lambda \) is calculated como \( \frac{1}{2} \) calls per minute.
For other time periods like a 5-minute window, you multiply this rate by the time interval to find the rate for that specific period:
  • \( \lambda' = \frac{1}{2} \times 5 = 2.5 \) calls in 5 minutes.
The rate tells us how compact or spread out the events are over time. Knowing this helps identify patterns and make accurate predictions using the Poisson model. It aids in understanding not only what's expected but also in planning for rare but possible fluctuations, offering comprehensive insights into event management.

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