Let's solve each part of the problem step-by-step, with numerical values rounded to six decimal places.

Given Information:

• Customers arrive at a rate of $\lambda = 4.4$ per 10 minutes.
• We are dealing with the Poisson process, so the inter-arrival times follow an exponential distribution with rate parameter $\lambda = 4.4$.

Solution Steps:

Part (a)

How many minutes can you expect to pass between successive customers entering the convenience store?

Since the inter-arrival time follows an exponential distribution with rate $\lambda = 4.4$, the expected time between arrivals (mean of the exponential distribution) is: $\text{Expected time} = \frac{1}{\lambda} = \frac{1}{4.4} = 0.227273 \text{ minutes}$

Part (b)

Compute the probability that at most 1.5 minutes will pass between the arrival of one customer and the next.

To find $P(X \leq 1.5)$, where $X$ is the time between arrivals, we use the cumulative distribution function (CDF) of the exponential distribution: $P(X \leq x) = 1 - e^{-\lambda x}$ Substituting $\lambda = 4.4$ and $x = 1.5$: $P(X \leq 1.5) = 1 - e^{-4.4 \times 1.5} = 1 - e^{-6.6} = 1 - 0.001363 = 0.998637$

Part (c)

Find the probability that at least 1.25 to at most 3.5 minutes pass between the entry of two customers in the store.

To find $P(1.25 \leq X \leq 3.5)$, we calculate: $P(1.25 \leq X \leq 3.5) = P(X \leq 3.5) - P(X \leq 1.25)$

1. Calculate $P(X \leq 3.5)$: $P(X \leq 3.5) = 1 - e^{-4.4 \times 3.5} = 1 - e^{-15.4} \approx 1 - 0.000201 = 0.999799$

2. Calculate $P(X \leq 1.25)$: $P(X \leq 1.25) = 1 - e^{-4.4 \times 1.25} = 1 - e^{-5.5} = 1 - 0.004070 = 0.995930$

Thus: $P(1.25 \leq X \leq 3.5) = 0.999799 - 0.995930 = 0.003869$

Part (d)

What is the probability that, in total, at least 2.5 minutes will pass until the next customer enters the store?

To find $P(X \geq 2.5)$, we use the complementary probability: $P(X \geq 2.5) = 1 - P(X \leq 2.5)$

1. Calculate $P(X \leq 2.5)$: $P(X \leq 2.5) = 1 - e^{-4.4 \times 2.5} = 1 - e^{-11} = 1 - 0.000017 = 0.999983$

Thus: $P(X \geq 2.5) = 1 - 0.999983 = 0.000017$

Summary of Answers:

• (a) Expected time: $0.227273$ minutes
• (b) $P(X \leq 1.5) = 0.998637$
• (c) $P(1.25 \leq X \leq 3.5) = 0.003869$
• (d) $P(X \geq 2.5) = 0.000017$

Would you like more details on any of these steps or additional questions?

Additional Related Questions:

1. How would the probabilities change if $\lambda$ were increased?
2. What is the variance of the time between customer arrivals?
3. How does the Poisson rate affect the expected inter-arrival time?
4. How can we interpret the probability in part (d) in real-life terms?
5. How would we compute these probabilities if the arrivals did not follow an exponential distribution?

Tip:

Remember that for exponentially distributed times, the probability of waiting more than a given time is just the survival function $P(X > x) = e^{-\lambda x}$.