SOLUTION: The front office of the Graduate School receives external telephone calls. Due to certain difficulties, it is only one person that is taking care of the Front Office. Ms. Alia Kobi

Question 1170982: The front office of the Graduate School receives external telephone calls. Due to certain difficulties, it is only one person that is taking care of the Front Office. Ms. Alia Kobi, the receptionist wants to vacate her post for lunch between the hours of 2 and 3 pm. She knows that the average number of phone calls received on the switchboard is 2. However, what she does not know is the number of calls she will miss. You can help her to determine the probability that during that one particular hour
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
To solve this problem, we need to use the Poisson distribution, which models the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known average rate and independently of the time since the last event.
**1. Define the Parameters:**
* **Average rate (λ):** 2 calls per hour
* **Time interval:** 1 hour
**2. Poisson Probability Formula:**
The probability of observing *k* events in a given interval is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
* P(X = k) is the probability of k events occurring
* e is Euler's number (approximately 2.71828)
* λ is the average rate of events
* k is the number of events
* k! is the factorial of k
**3. Calculate Probabilities:**
We want to find the probability of missing different numbers of calls. We'll calculate the probabilities for a few values of k:
* **Probability of missing 0 calls (k = 0):**
P(X = 0) = (e^(-2) * 2^0) / 0! = (e^(-2) * 1) / 1 ≈ 0.1353
* **Probability of missing 1 call (k = 1):**
P(X = 1) = (e^(-2) * 2^1) / 1! = (e^(-2) * 2) / 1 ≈ 0.2707
* **Probability of missing 2 calls (k = 2):**
P(X = 2) = (e^(-2) * 2^2) / 2! = (e^(-2) * 4) / 2 ≈ 0.2707
* **Probability of missing 3 calls (k = 3):**
P(X = 3) = (e^(-2) * 2^3) / 3! = (e^(-2) * 8) / 6 ≈ 0.1804
* **Probability of missing 4 calls (k = 4):**
P(X = 4) = (e^(-2) * 2^4) / 4! = (e^(-2) * 16) / 24 ≈ 0.0902
* **Probability of missing 5 calls (k = 5):**
P(X = 5) = (e^(-2) * 2^5) / 5! = (e^(-2) * 32) / 120 ≈ 0.0361
**4. Interpretation**
Ms. Alia Kobi can use these probabilities to understand the likelihood of missing different numbers of calls during her lunch break. For example:
* There's about a 13.53% chance she'll miss no calls.
* There's about a 27.07% chance she'll miss one call.
* There is about a 27.07% chance she will miss 2 calls.
* There is about a 18.04% chance she will miss 3 calls.
Depending on the importance of the calls, she can decide whether or not to take her lunch break during that hour.

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