SOLUTION: ) A dispatcher for an airport shuttle will send a van to the airport on average twice per hour during the Soccer World Cup in 2010. The driver must take a 15- minute lunch brea

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Question 1192425: ) A dispatcher for an airport shuttle will send a van to the airport on average
twice per hour during the Soccer World Cup in 2010. The driver must take
a 15- minute lunch break. What is the probability that he can complete his
lunch break before receiving a call?
Answer by CPhill(1987) (Show Source):
You can put this solution on YOUR website!
**1. Determine the Average Calls During Lunch Break**
* **Average calls per hour:** 2 calls
* **Average calls during 15 minutes:**
* 15 minutes is 1/4 of an hour
* Average calls = 2 calls/hour * (1/4) hour = 0.5 calls
**2. Use the Poisson Distribution**
* The number of calls received during a specific time interval follows a Poisson distribution when the events (calls) are independent and occur at a constant average rate.
* **Poisson Distribution Formula:**
* P(X = k) = (λ^k * e^(-λ)) / k!
* where:
* X is the number of calls
* λ is the average number of calls during the interval (0.5 in this case)
* k is the number of calls we're interested in (0 in this case)
* e is the base of the natural logarithm (approximately 2.71828)
* **Probability of no calls during lunch break (k = 0):**
* P(X = 0) = (0.5^0 * e^(-0.5)) / 0!
* P(X = 0) = (1 * e^(-0.5)) / 1
* P(X = 0) = e^(-0.5)
* P(X = 0) ≈ 0.6065
**Therefore, the probability that the driver can complete his 15-minute lunch break before receiving a call is approximately 0.6065 or 60.65%.**