The number of calls received per hour follows a Poisson distribution with a mean (or average) $\lambda = 6$. The Poisson probability mass function (PMF) is given by:

$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$ where:

• $\lambda = 6$ (the average number of calls per hour),
• $k$ is the number of calls,
• $e$ is the base of the natural logarithm ($\approx 2.718$).

We can use the cumulative distribution function (CDF) of the Poisson distribution to calculate probabilities for "at least" or "4 or more" scenarios.

a) Probability of receiving at least 6 calls:

To calculate $P(X \geq 6)$, we can use the complement rule:

$P(X \geq 6) = 1 - P(X < 6) = 1 - P(X \leq 5)$

Now, we need to calculate $P(X \leq 5)$ using the Poisson CDF:

$P(X \leq 5) = P(X = 0) + P(X = 1) + \cdots + P(X = 5)$

b) Probability of receiving 4 or more calls:

This is represented as $P(X \geq 4)$, and again using the complement rule:

$P(X \geq 4) = 1 - P(X < 4) = 1 - P(X \leq 3)$

Next, I'll calculate both probabilities using the formulas.The probabilities are as follows:

a) The probability that the company will receive at least 6 calls in an hour:

$P(X \geq 6) = 0.554$

b) The probability that the company will receive 4 or more calls in an hour:

$P(X \geq 4) = 0.849$

Let me know if you need further details or have any questions!

Here are 5 related questions for further exploration:

1. What is the probability of receiving exactly 6 calls in an hour?
2. What is the probability of receiving fewer than 6 calls?
3. What is the probability of receiving between 4 and 8 calls?
4. How would the probabilities change if the average number of calls per hour increased to 8?
5. What is the expected number of calls in a 2-hour period?

Tip: For Poisson-distributed events, increasing the observation period scales the mean $\lambda$ accordingly. For example, for 2 hours, $\lambda$ would double.