Probabilistic Analysis of Telephone Calls: Poisson and Binomial Distributions in a College Switchboard Scenario

The Poisson distribution is a fundamental tool in the study of random events over a period of time. Its application is prevalent in various fields, including telecommunications, where estimating call rates and their distribution is critical. This article explores the probability of a certain number of calls arriving in a given period using both Poisson and Binomial distributions. We will delve into the scenario of a college switchboard receiving calls.

The Scenario and Expected Call Rate

Telephone calls enter a college switchboard on an average of two every three minutes. This scenario involves understanding the probability of receiving 5 or more calls in a 9-minute period. Assuming that call arrivals are independent of each other and do not overlap within a 9-minute interval, we can utilize the Poisson distribution to solve this problem.

Data Collection and Calculation

To begin, let's calculate the average number of calls per 9-minute period. Given that calls arrive at an average rate of 2 calls every 3 minutes, we can derive the mean ((lambda)) for a 9-minute period as follows:

Poisson Distribution Application

The mean number of calls ((lambda)) in a 9-minute period is:

(lambda 2 times frac{9}{3} 6)

Using the Poisson distribution formula, we can find the probability of receiving a number of calls (X) in a given period:

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

To find the probability of 5 or more calls in a 9-minute period, we need to calculate the cumulative probability of receiving 0 to 4 calls and subtract it from 1:

Let (P(X leq 4) ) be the cumulative probability of receiving up to 4 calls. Then, the probability of receiving 5 or more calls is:

[ P(X geq 5) 1 - P(X leq 4) 1 - sum_{x0}^{4} P(X x) ]

Using the Poisson formula for each (x), we calculate:

[ P(X 0) frac{e^{-6} 6^0}{0!} e^{-6} approx 0.0025 ]

[ P(X 1) frac{e^{-6} 6^1}{1!} 6e^{-6} approx 0.0149 ]

[ P(X 2) frac{e^{-6} 6^2}{2!} 18e^{-6} approx 0.0447 ]

[ P(X 3) frac{e^{-6} 6^3}{3!} 36e^{-6} approx 0.0894 ]

[ P(X 4) frac{e^{-6} 6^4}{4!} frac{1296e^{-6}}{24} approx 0.1341 ]

Calculating (P(X leq 4))

[ P(X leq 4) P(X 0) P(X 1) P(X 2) P(X 3) P(X 4) ]

[ P(X leq 4) approx 0.0025 0.0149 0.0447 0.0894 0.1341 0.2856 ]

Final Probability Calculation

[ P(X geq 5) 1 - P(X leq 4) 1 - 0.2856 0.7144 ]

Therefore, the probability of receiving 5 or more calls in a 9-minute period is approximately 71.44%.

Binomial Distribution Application

Alternatively, we can use the Binomial distribution, which is suitable for situations where the number of trials is fixed and the probability of success is constant. In the context of telephone calls:

Binomial Probability Mass Function (PMF)

[ b_Xn , p^X (1-p)^{n-X} ]

where:

(X): Number of calls arriving (n): Total number of calls (9 in this case) (p): Probability of a call arriving in a 3-minute window ((frac{2}{3} times 3 2)) (b_Xn frac{n!}{X!(n-X)!})

To find the probability of 5 or more calls arriving, we need to sum up the probabilities from 5 to 9 calls:

[ P(X geq 5) 1 - left( b_09 , p^0 (1-p)^9 b_19 , p^1 (1-p)^8 b_29 , p^2 (1-p)^7 b_39 , p^3 (1-p)^6 b_49 , p^4 (1-p)^5 right) ]

Let's calculate each term:

[ b_09 1 , (2/3)^0 (1-2/3)^9 1 times 1 times (1/3)^9 approx 4.22e-05 ]

[ b_19 9 , (2/3)^1 (1-2/3)^8 9 times 2/3 times (1/3)^8 approx 2.51e-04 ]

[ b_29 frac{9!}{2! times 7!} , (2/3)^2 (1-2/3)^7 36 times (4/9) times (1/3)^7 approx 7.68e-04 ]

[ b_39 frac{9!}{3! times 6!} , (2/3)^3 (1-2/3)^6 84 times (8/27) times (1/3)^6 approx 1.63e-03 ]

[ b_49 frac{9!}{4! times 5!} , (2/3)^4 (1-2/3)^5 126 times (16/81) times (1/3)^5 approx 2.66e-03 ]

Final Calculation

[ P(X leq 4) b_09 b_19 b_29 b_39 b_49 approx 4.22e-05 0.000251 0.000768 0.00163 0.00266 approx 0.005371 ]

[ P(X geq 5) 1 - P(X leq 4) 1 - 0.005371 approx 0.99463 ]

Thus, using the Binomial distribution, the probability of receiving 5 or more calls in a 9-minute period is approximately 99.46%.

Conclusion

Both the Poisson and Binomial distributions provide similar probabilities for the scenario considered. The choice between the two distributions depends on the specific characteristics of the problem. In this case, the Binomial distribution yielded a higher probability due to its assumptions better aligning with the discrete nature of call arrivals.

Application and Implications

Understanding the probability of call arrivals helps in managing switchboard operations, optimizing staffing, and planning for capacity. This knowledge is crucial for maintaining a smooth flow of calls and ensuring timely responses, which positively impacts customer satisfaction and operational efficiency.

Further Reading and Resources

Poisson Distribution on Wikipedia Binomial Distribution on Wikipedia Binomial Distribution Tutorial Poisson Distribution vs Chi-Squared Distribution

For detailed calculations and further applications, please refer to the cited resources.