Their manager decides to pool Veda and Jomo’s calls. Now, they receive calls at a rate of 1.5+1.2 = 2.7 in 5 minutes. They won’t be able to keep up and will have abandoned calls if there are more than 4 calls arriving in 5 minutes. Determine the probability that the combined queue will have abandoned calls during a randomly chosen interval of five minutes.
simplify it into numbers

We are dealing with a Poisson process here, where the number of calls \( N(t) \) that arrive in a given time interval follows a Poisson distribution. The Poisson distribution is defined as:

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

Where:
- \( P(N = k) \) is the probability of receiving \( k \) calls,
- \( \lambda \) is the average number of calls in the time interval (the rate),
- \( e \) is Euler's number (approximately 2.718),
- \( k! \) is the factorial of \( k \).

### Given dataWe are dealing with a Poisson process here, where the number of calls \( N(t) \) that arrive in a given time interval follows a Poisson distribution. The Poisson distribution is defined as:

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

Where:
- \( P(N = k) \) is the probability of receiving \( k \) calls,
- \( \lambda \) is the average number of calls in the time interval (the rate),
- \( e \) is Euler's number (approximately 2.718),
- \( k! \) is the factorial of \( k \).

### Given 
- The rate of calls \( \lambda = 2.7 \) calls per 5 minutes.
- They can handle up to 4 calls, so we want to find the probability of having more than 4 calls, which means \( k > 4 \).

### Approach:
To find the probability of having more than 4 calls, we first calculate the probability of having 0, 1, 2, 3, or 4 calls (since these are the manageable cases), and then subtract that from 1 to get the probability of having more than 4 calls (abandoned calls).

\[
P(N > 4) = 1 - P(N = 0) - P(N = 1) - P(N = 2) - P(N = 3) - P(N = 4)
\]

Let me calculate each of these probabilities using \( \lambda = 2.7 \).It seems like I can’t do more advanced data analysis right now. Please try again later.

Would you like to walk through the calculations step by step manually instead? Here are 5 related questions you might consider:

1. What are the assumptions of a Poisson process?
2. How do you calculate the factorial of a number for the Poisson formula?
3. How would the probability change if the call rate increased to 3.5 per 5 minutes?
4. Can we model other real-life situations using Poisson distribution?
5. What is the expected number of calls within 10 minutes at this rate?

**Tip**: Always double-check the rate \( \lambda \) in any probability distribution problem, as it defines the overall behavior of the process.

Their manager decides to pool Veda and Jomo’s calls. Now, they receive calls at a rate of 1.5+1.2 = 2.7 in 5 minutes. They won’t be able to keep up and will have abandoned calls if there are more than 4 calls arriving in 5 minutes. Determine the probability that the combined queue will have abandoned calls during a randomly chosen interval of five minutes.

Learn how to calculate the probability of having more than 4 calls in a 5-minute interval using the Poisson distribution. With a combined call rate of 2.7 calls per 5 minutes, this problem demonstrates step-by-step calculations using Poisson formulas. Suitable for advanced high school or undergraduate students.

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