At a car rental kiosk at a busy airport, customers arrive and wait in a common l

Question

At a car rental kiosk at a busy airport, customers arrive and wait in a common line for the first available of two sales agents. The agents can serve a customer in an average of 2.5 minutes (assume negative exponential distribution). On average, 21 customers per hour will arrive to be served (assume Poisson arrivals). However, on this day, assume that one of the computers is not working, so the two sales agents work together helping one customer at a time. The new average service time per customer is now 1.9 minutes. What is the average time a customer spends waiting in line before being served (in minutes, rounded to two decimals)?

0.12 mins

0.24 mins

0.47 mins

3.77 mins

0.12 mins

0.24 mins

0.47 mins

3.77 mins

Explanation / Answer

Well, after calculating the above given information with the formula given under, we can conclude the answer here. So, it goes like this:

Formula:

P(x; ) = (e-) (x) / x!

here, x = random variable, = average service time, e = 2.71828

So, x = 168, = 1.9, e = 2.71828

After putting the values in this formula, we will get 0.476.

Hence, the correct answer from the above given options is option C, that is 0.47.

Regards.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.