The time between customer arrivals can be modeled using an exponential distribution, which is defined by the probability density function:

$f(t; \lambda) = \lambda e^{-\lambda t}, \quad t \geq 0$

where:

• $\lambda$ is the rate parameter (average number of arrivals per minute),
• $t$ is the time between arrivals.

The cumulative distribution function (CDF) is:

$P(T \leq t) = 1 - e^{-\lambda t}$

where $P(T \leq t)$ gives the probability that a customer will arrive within $t$ minutes.

Given:

• Average time between arrivals = 10 minutes → $\lambda = \frac{1}{10} = 0.1$ arrivals per minute.
• For the promotional event, $\lambda = 0.2$ arrivals per minute.

a) Probability that a customer will arrive within the next 5 minutes:

The formula for $P(T \leq t)$ is:

$P(T \leq t) = 1 - e^{-\lambda t}$

Substitute $\lambda = 0.1$ and $t = 5$:

$P(T \leq 5) = 1 - e^{-0.1 \times 5}$

$P(T \leq 5) = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$

So, the probability is approximately 39.35%.

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b) Probability that no customer will arrive within the next 15 minutes:

This is the probability $P(T > t)$, which is the complement of $P(T \leq t)$:

$P(T > t) = e^{-\lambda t}$

Substitute $\lambda = 0.1$ and $t = 15$:

$P(T > 15) = e^{-0.1 \times 15}$

$P(T > 15) = e^{-1.5} \approx 0.2231$

So, the probability is approximately 22.31%.

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c) New probability that a customer will arrive within the next 5 minutes during the promotional event:

The new rate parameter is $\lambda = 0.2$ (double the rate). Using the same formula for $P(T \leq t)$:

$P(T \leq 5) = 1 - e^{-\lambda t}$

Substitute $\lambda = 0.2$ and $t = 5$:

$P(T \leq 5) = 1 - e^{-0.2 \times 5}$

$P(T \leq 5) = 1 - e^{-1} \approx 1 - 0.3679 = 0.6321$

So, the probability is approximately 63.21%.

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Summary:

a) Probability that a customer will arrive within the next 5 minutes: 39.35%.
b) Probability that no customer will arrive within the next 15 minutes: 22.31%.
c) New probability that a customer will arrive within the next 5 minutes during the promotional event: 63.21%.

Would you like further clarification or calculations?

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Follow-up Questions:

1. How would the probabilities change for other time intervals (e.g., 2 minutes or 20 minutes)?
2. Can we use a different distribution to model customer arrivals? Why or why not?
3. How would increasing the arrival rate further impact customer service planning?
4. What is the expected number of customers arriving within the next hour before and during the promotional event?
5. How can this model be extended to consider multiple service desks?

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Tip:

When modeling arrival times, remember that the exponential distribution assumes independence between events, meaning the time until the next customer does not depend on previous arrivals.