For part (c), the problem asks to compute the probability that the time between the entry of two customers, $X$, is between 1 minute and 2.75 minutes. We are working with an exponential distribution because the time between events in a Poisson process follows an exponential distribution.

The cumulative distribution function (CDF) of an exponential distribution is given by:

$F(X) = 1 - e^{-\lambda X}$

Where $\lambda = 5.7$ is the rate parameter (mean number of customers per 10 minutes), so the rate in terms of minutes is:

$\lambda' = \frac{5.7}{10} = 0.57$

To find the probability that the time is between 1 minute and 2.75 minutes, we calculate:

$P(1 \leq X \leq 2.75) = F(2.75) - F(1)$

Step 1: Calculate $F(2.75)$

$F(2.75) = 1 - e^{-0.57 \times 2.75} = 1 - e^{-1.5675} \approx 1 - 0.2083 = 0.7917$

Step 2: Calculate $F(1)$

$F(1) = 1 - e^{-0.57 \times 1} = 1 - e^{-0.57} \approx 1 - 0.5657 = 0.4343$

Step 3: Find the probability between 1 and 2.75

$P(1 \leq X \leq 2.75) = 0.7917 - 0.4343 = 0.3574$

Thus, the correct probability is approximately 0.3574, not 0.2883.

Would you like any further details or clarifications?

Here are five related questions:

1. What is the relationship between the Poisson and exponential distributions?
2. How do you compute probabilities for an exponential distribution over an interval?
3. What is the significance of the rate parameter $\lambda$ in the exponential distribution?
4. How would the probability change if the interval was from 0.5 to 2.5 minutes?
5. How can you calculate the expected value and variance of an exponential distribution?

Tip: When working with exponential distributions, always ensure that you are using the correct rate parameter depending on the time scale involved.