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Let's go through each part of this problem step by step. This is based on a Poisson process, and the exponential distribution will be key in solving each part.

### Given:
- The mean number of customers entering a store every 10 minutes is \( \lambda = 5.7 \) customers.
- We are using the exponential distribution, which has the probability density function (PDF):  
  \[
  f(x; \lambda) = \lambda e^{-\lambda x}, \text{ for } x \geq 0
  \]
  where \( \lambda \) is the rate (in this case, number of customers per minute).

Since 5.7 customers enter the store every 10 minutes, the rate \( \lambda \) per minute is:
\[
\lambda = \frac{5.7}{10} = 0.57 \text{ customers per minute.}
\]

Now, let's solve each part.

### (a) How many minutes can you expect to pass between successive customers entering the store?
The expected value (mean) of an exponential distribution is given by:
\[
E(X) = \frac{1}{\lambda}
\]
Substituting \( \lambda = 0.57 \):
\[
E(X) = \frac{1}{0.57} \approx 1.7544 \text{ minutes.}
\]

### (b) Compute the probability that at most 2.5 minutes will pass between the arrival of one customer and the next.
For an exponential distribution, the cumulative distribution function (CDF) is:
\[
P(X \leq x) = 1 - e^{-\lambda x}
\]
Substitute \( \lambda = 0.57 \) and \( x = 2.5 \):
\[
P(X \leq 2.5) = 1 - e^{-0.57 \times 2.5} \approx 1 - e^{-1.425} \approx 1 - 0.2403 = 0.7597
\]
So, the probability is approximately \( 0.7597 \).

### (c) Find the probability that at least 1 to at most 3 minutes pass between the entry of two customers.
This is the probability \( P(1 \leq X \leq 3) \), which is:
\[
P(1 \leq X \leq 3) = P(X \leq 3) - P(X \leq 1)
\]
First, find \( P(X \leq 3) \):
\[
P(X \leq 3) = 1 - e^{-0.57 \times 3} = 1 - e^{-1.71} \approx 1 - 0.1805 = 0.8195
\]
Now, find \( P(X \leq 1) \):
\[
P(X \leq 1) = 1 - e^{-0.57 \times 1} = 1 - e^{-0.57} \approx 1 - 0.5650 = 0.4350
\]
So, the probability is:
\[
P(1 \leq X \leq 3) = 0.8195 - 0.4350 = 0.3845
\]
Thus, the probability is approximately \( 0.3845 \).

### (d) What is the probability that at least 4 minutes will pass until the next customer enters the store, given that at least 1 minute has passed?
This is a conditional probability problem. We need to find:
\[
P(X \geq 4 \mid X \geq 1)
\]
Using the memoryless property of the exponential distribution, this simplifies to:
\[
P(X \geq 4 \mid X \geq 1) = P(X \geq 3)
\]
And \( P(X \geq 3) \) is:
\[
P(X \geq 3) = 1 - P(X \leq 3) = 1 - 0.8195 = 0.1805
\]
So, the probability is approximately \( 0.1805 \).

### Summary of Final Answers:
- **(a)**: \( 1.7544 \) minutes
- **(b)**: \( 0.7597 \)
- **(c)**: \( 0.3845 \)
- **(d)**: \( 0.1805 \)

Would you like further details or clarifications on any of these parts? Here are some related questions:

1. What is the relation between Poisson and exponential distributions in this context?
2. How does the memoryless property of exponential distribution apply to part (d)?
3. Can you explain the process to find expected values in an exponential distribution?
4. How would the answers change if the rate \( \lambda \) were different?
5. What is the intuition behind the exponential distribution's CDF?

**Tip**: For exponential distributions, the mean is the inverse of the rate \( \lambda \), and it models the time between events (like customer arrivals).

The number of customers entering a 24-hour convenience store every 10 minutes can be modeled by the Poisson distribution with a mean of λ = 5.7 customers. You are to look at the amount of time passing between successive customers entering the convenience store, represented by X.

Solve a problem involving the Poisson distribution with a mean of 5.7 customers per 10 minutes, and learn how to calculate the expected time between customers, the probability of certain intervals, and the memoryless property of the exponential distribution. Suitable for college-level statistics students.

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