Calculating the Probability of Receiving At Least 2 Calls in an Hour Using the Poisson Distribution

In the context of a sales firm that receives on average 3 calls per hour on its toll-free number, understanding the probability of receiving at least 2 calls in any given hour is critical for managing call volumes and staffing appropriately. This can be effectively modeled using the Poisson distribution, a statistical framework ideal for scenarios involving a fixed number of discrete events occurring independently over a given time interval at a known average rate.

Introduction to Poisson Distribution

The Poisson distribution is a discrete probability distribution named after French mathematician Siméon Denis Poisson. It is used to model the number of times an event occurs in a set interval of time or space. This distribution is particularly applicable in scenarios where the events are rare but occur frequently enough to maintain a constant average rate. The sales firm's call volume scenario is a perfect fit for this distribution as the calls are expected to occur at a steady rate, independent of time.

Understanding the Problem

The problem statement gives us the following information:

The average rate of calls per hour ((lambda)) is 3. We need to find the probability of receiving at least 2 calls in an hour (i.e., (P(X geq 2))).

Using the Poisson distribution, we can express this probability as:

[P(X geq 2) 1 - P(X

Applying the Poisson Probability Mass Function

The Poisson probability mass function is defined as:

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

where:

(k) is the number of events (in this case, calls). (lambda) is the average rate (in this case, 3 calls per hour). (e) is Euler's number, approximately equal to 2.71828.

Step-by-Step Calculation

First, we calculate the probabilities of receiving 0 and 1 call in an hour:

For (k 0)

[P(X 0) frac{e^{-3} cdot 3^0}{0!} e^{-3} cdot 1 e^{-3} approx 0.0498]

For (k 1)

[P(X 1) frac{e^{-3} cdot 3^1}{1!} e^{-3} cdot 3 approx 0.1494]

Calculating the Desired Probability

Now, we calculate the probability of receiving at least 2 calls:

[P(X geq 2) 1 - P(X

[P(X geq 2) 1 - (0.0498 0.1494) 1 - 0.1992 0.8008]

Therefore, the probability that the sales firm will receive at least 2 calls in an hour is approximately 0.8008 or 80.08%.

Conclusion

In conclusion, by using the Poisson distribution, we can effectively analyze call volumes and predict the probability of receiving at least 2 calls in an hour for a sales firm receiving an average of 3 calls per hour. This information is invaluable for managing call centers and optimizing resources.

Key takeaways:

The Poisson distribution is ideal for modeling call volumes in fixed time intervals. The probability of receiving at least 2 calls in an hour is approximately 80.08%.