In the next room, Jomo handles customer claims with surnames from F to J. The volume of calls arriving in Jomo’s queue follows a Poisson distribution with an average of 1.2 calls in 5 minutes. If Jomo’s queue receives more than 2 calls in 5 minutes, Jomo won’t be able to keep up, and customers will abandon their calls. Determine the probability that Joma’s queue will have abandoned calls during a randomly chosen interval of five minutes.


To solve this problem, we need to find the probability that Jomo’s queue will receive **more than 2 calls** in a 5-minute interval, given that the number of calls follows a Poisson distribution with an average rate of 1.2 calls per 5 minutes.

### Poisson Distribution:
The Poisson probability mass function (PMF) for \( k \) calls in a given time period is given by the formula:

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

where:
- \( \lambda \) is the average number of calls (1.2 in this case),
- \( k \) is the number of calls,
- \( e \) is Euler's number (approximately 2.71828).

In this case, we want to calculate the probability that the number of calls, \( X \), is greater than 2:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

First, calculate \( P(X \leq 2) \), which is the sum of the probabilities for 0, 1, and 2 calls:

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

Let's compute each probability individually.

1. **\( P(X = 0) \):**

\[
P(X = 0) = \frac{e^{-1.2} \cdot 1.2^0}{0!} = e^{-1.2}
\]

2. **\( P(X = 1) \):**

\[
P(X = 1) = \frac{e^{-1.2} \cdot 1.2^1}{1!} = e^{-1.2} \cdot 1.2
\]

3. **\( P(X = 2) \):**

\[
P(X = 2) = \frac{e^{-1.2} \cdot 1.2^2}{2!} = e^{-1.2} \cdot \frac{1.2^2}{2}
\]

After calculating these, we can find \( P(X \leq 2) \), and finally, the probability that there are more than 2 calls is:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

Now, let's calculate these values.The probability that Jomo's queue will receive more than 2 calls in a 5-minute interval, leading to abandoned calls, is approximately **0.1205**, or about **12.05%**.

Would you like further clarification or additional details on this topic?

Here are 5 related questions to consider:
1. How does the Poisson distribution model work for different time intervals?
2. What are other real-life scenarios where the Poisson distribution can be applied?
3. How would the probability change if the average call rate increased?
4. How can Jomo improve queue management to avoid abandoned calls?
5. What happens if customers don't abandon their calls but instead wait longer?

**Tip:** When dealing with Poisson distribution, always ensure that the event rate (\( \lambda \)) matches the time interval you're analyzing.

In the next room, Jomo handles customer claims with surnames from F to J. The volume of calls arriving in Jomo’s queue follows a Poisson distribution with an average of 1.2 calls in 5 minutes. If Jomo’s queue receives more than 2 calls in 5 minutes, Jomo won’t be able to keep up, and customers will abandon their calls. Determine the probability that Joma’s queue will have abandoned calls during a randomly chosen interval of five minutes.

Discover how to calculate the probability of more than 2 calls arriving in Jomo's queue, causing abandoned calls. Using the Poisson distribution with an average rate of 1.2 calls per 5 minutes, this solution breaks down the math for real-world application. Ideal for college-level probability students.

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