Veda works in an insurance call centre where she handles claim enquiries from customers whose surnames start with A to E. Studies have revealed that the volume of calls arriving in Veda’s queue follows a Poisson distribution with an average of 1.5 calls in 5 minutes. If Veda’s queue receives more than 2 calls in 5 minutes, Veda won’t be able to keep up, and customers will abandon their calls.  Determine the probability that Veda’s queue will have abandoned calls during a randomly chosen interval of five minutes.

We are tasked with determining the probability that Veda's queue will have **abandoned calls** during a randomly chosen interval of five minutes. This happens when the number of calls exceeds 2 in a 5-minute period. Given that the number of calls follows a **Poisson distribution** with a mean of 1.5 calls in 5 minutes, we need to find the probability that the number of calls in a 5-minute interval exceeds 2 (i.e., \( P(X > 2) \)).

### Poisson Distribution Formula
For a Poisson distribution, the probability of observing \( k \) calls in a given time interval is given by the formula:

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

where:
- \( \lambda = 1.5 \) is the average number of calls in 5 minutes,
- \( k \) is the number of calls (0, 1, 2, ...),
- \( e \) is Euler's number (\( \approx 2.718 \)).

### Step 1: Calculate \( P(X \leq 2) \)
We first need to calculate the probability that the number of calls is **less than or equal to 2** in 5 minutes, i.e., \( P(X \leq 2) \). This is the sum of the probabilities of observing 0, 1, or 2 calls.

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

Now, we compute each term using the Poisson formula.

#### For \( P(X = 0) \):

\[
P(X = 0) = \frac{e^{-1.5} \cdot 1.5^0}{0!} = e^{-1.5} \approx 0.2231
\]

#### For \( P(X = 1) \):

\[
P(X = 1) = \frac{e^{-1.5} \cdot 1.5^1}{1!} = e^{-1.5} \cdot 1.5 \approx 0.3346
\]

#### For \( P(X = 2) \):

\[
P(X = 2) = \frac{e^{-1.5} \cdot 1.5^2}{2!} = e^{-1.5} \cdot \frac{1.5^2}{2} \approx 0.2510
\]

### Step 2: Sum the probabilities

Now, we sum the probabilities for \( X = 0 \), \( X = 1 \), and \( X = 2 \):

\[
P(X \leq 2) = 0.2231 + 0.3346 + 0.2510 = 0.8087
\]

### Step 3: Find \( P(X > 2) \)

To find the probability of more than 2 calls (i.e., the probability of abandoned calls), we subtract \( P(X \leq 2) \) from 1:

\[
P(X > 2) = 1 - P(X \leq 2) = 1 - 0.8087 = 0.1913
\]

Thus, the probability that Veda's queue will have abandoned calls during a randomly chosen 5-minute interval is approximately **0.1913** or **19.13%**.

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Would you like more details on any part of this explanation?

### Related Questions:
1. What happens to the probability if the average number of calls increases to 2 per 5 minutes?
2. How would the probability change if Veda could handle up to 3 calls instead of 2?
3. What is the probability that Veda's queue will receive exactly 3 calls in a 5-minute period?
4. How would we calculate the probability of receiving fewer than 1 call in a 10-minute period?
5. How does the Poisson distribution compare to other probability distributions, like the binomial?

#### Tip: 
The Poisson distribution is very useful when dealing with rare events, especially when you know the average rate but not the exact timing of individual events.

Veda works in an insurance call centre where she handles claim enquiries from customers whose surnames start with A to E. Studies have revealed that the volume of calls arriving in Veda’s queue follows a Poisson distribution with an average of 1.5 calls in 5 minutes. If Veda’s queue receives more than 2 calls in 5 minutes, Veda won’t be able to keep up, and customers will abandon their calls. Determine the probability that Veda’s queue will have abandoned calls during a randomly chosen interval of five minutes.

Discover how to calculate the probability of abandoned calls in Veda's queue using the Poisson distribution. This problem explains how to use the formula P(X > 2) with an average rate of 1.5 calls in 5 minutes, providing a step-by-step guide to finding the solution, which shows a 19.13% chance of call abandonment. Suitable for advanced high school or undergraduate students studying probability.

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