Answer :
Answer:
At least 3 processing booths should be open.
Step-by-step explanation:
1 vehicle spends an average of 15 seconds in a booth
In N booths , there will be N/15 vehicles in 1 second
In 1 hr, there will be N/15 * 3600 vehicles
Therefore, there will be a total of 240N vehicles/hour
Actual arrival rate of vehicles at the airport is 500 veh/h
Probability of arrival, P = [tex]\frac{Number of possible outcomes}{Total outcomes}[/tex]
P = 500/240N
P = 2.083/N
For the average time spends to be below 5 seconds
[tex]\frac{(\frac{P^{2} }{1-p} )}{500} = 5[/tex], since 500 is the average value.
[tex]\frac{P^{2} }{1-p} } = 2500\\P^{2} = 2500 (1-P)\\P^{2} = 2500 - 2500P\\(\frac{2.083}{N}) ^{2} = 2500 - 2500(\frac{2.083}{N})[/tex]
[tex]2.083^{2} = 2500N^{2} - 5208N\\2500N^{2} - 5208N - 4.34 = 0[/tex]
[tex]N^{2} -2.0832N - 0.0017 = 0[/tex]
Solving for N using the quadratic formula with a = 1, b = -2.0832, c = -0.0017
[tex]N = \frac{-b \pm \sqrt{b^{2} -4ac} }{2a} \\N = \frac{2.0832 \pm \sqrt{(-2.0832)^{2} -4*1*(-0.0017)} }{2}[/tex]
N = 1.0416 ± 1.0424
N = 2.084 or -0.0008
N = 2.084 is the only realistic value here
Therefore, the smallest number of processing booths that must be open should be 3