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.