Calculating Call Probabilities in a Small Telephone Exchange Using Poisson Distribution

Understanding the call patterns in a small telephone exchange is crucial for effective resource allocation and efficient service provision. This article explores the application of the Poisson distribution to determine the probability of a small telephone exchange receiving 2 or fewer calls within a one-minute interval.

Introduction to Poisson Distribution

The Poisson distribution is a fundamental statistical tool used to model the number of events occurring in a fixed interval of time or space, given a known average rate of occurrence. In the context of a telephone exchange, the Poisson distribution can be employed to estimate the likelihood of receiving a certain number of calls within a one-minute time frame.

Modeling Call Patterns with Poisson Distribution

In this scenario, a small telephone exchange receives an average of 5 calls per minute. We aim to calculate the probability of receiving 2 or fewer calls (0, 1, or 2 calls) within this interval. The Poisson probability mass function is given by:

[ P(X k) frac{e^{-lambda} lambda^k}{k!} ]

Where:

(lambda) is the average number of calls per minute (in this case, 5). (k) is the number of calls we are considering. (e) is the base of the natural logarithm, approximately equal to 2.71828.

Calculating Probabilities

To find the probability of receiving 2 or fewer calls, we need to calculate the probabilities for receiving 0, 1, and 2 calls and then sum these probabilities.

Step 1: Calculate (P(X 0))

[begin{align*} P(X 0) frac{e^{-5} cdot 5^0}{0!} e^{-5} cdot 1 e^{-5} end{align*}]

Step 2: Calculate (P(X 1))

[begin{align*} P(X 1) frac{e^{-5} cdot 5^1}{1!} e^{-5} cdot 5 end{align*}]

Step 3: Calculate (P(X 2))

[begin{align*} P(X 2) frac{e^{-5} cdot 5^2}{2!} e^{-5} cdot frac{25}{2} frac{25}{2} e^{-5} end{align*}]

Step 4: Combine the Probabilities

[begin{align*} P(X leq 2) P(X 0) P(X 1) P(X 2) e^{-5} 5 e^{-5} frac{25}{2} e^{-5} e^{-5} left(1 5 frac{25}{2}right) e^{-5} left(1 5 12.5right) e^{-5} cdot 18.5 end{align*}]

Step 5: Calculate (e^{-5})

[begin{align*} e^{-5} approx 0.006737 end{align*}]

Step 6: Final Calculation

[begin{align*} P(X leq 2) approx 18.5 times 0.006737 approx 0.124 end{align*}]

Hence, the probability that the exchange receives 2 or fewer calls is approximately 0.124 or 12.4%.

Application and Real-world Considerations

While the Poisson distribution provides a useful approximation for call patterns in a small telephone exchange, it is important to note that the assumption of a constant rate of call arrival may not hold perfectly in real-world scenarios. For instance, during specific times like the "busy hour," the rate of calls can differ significantly from the average.

Further Exploration

Now, let us consider a similar problem where (X) follows a Poisson distribution with a rate of 4. Using the same steps as above, we can find the probabilities for (X 0), (X 1), and (X 2).

Conclusion

Understanding and applying the Poisson distribution can greatly enhance the management and planning of call volumes in a small telephone exchange. By calculating and analyzing these probabilities, network administrators can better allocate resources and optimize service delivery.