Question 1170982
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.