Question

Suppose an audit process is taking place at a small airport, and the head
auditor arrives randomly throughout the day and assesses the operation of the airport.
She shows up in one-hr intervals at a time. The time between aircraft arrivals at this
small airport is exponentially distributed with a mean of 75 minutes. You may assume
the operation of the small airport is independent of the time interval under consideration.
a) What is the probability that at least four aircrafts arrive within an hour?
b)What is the probability that more than three and less than six aircrafts arrive within an
hour?
C)What is the amount of time (in hours) such that the probability of aircraft arrivals in the
interval is 75%?
D) What is the probability that the third interval the auditor shows up for is the second
interval with at least four aircraft arrivals?
e) If the auditor decides to show up four different times in a day, what is the mean of the
number of intervals she will experience with less than four aircraft arrivals?

Answer 1

Answer:

a) Let random variable X: time between aircraft arrivals at this small airport

X is exponetially distributed with mean =

X is exponentially distributed with

If x follows exponential distribution with then,

   

Here we have to find

= 0.9481 (Round to 4 decimal), From excel using function, =EXP(-0.0533)

The probability that at least four aircrafts arrive within an hour is 0.9481

b)

Now we have to find P(3 < X < 6)

  

  

  

= 0.9609 - 0.9233    (From excel using function, =EXP(-0.0798)-EXP(-0.0399))

= 0.0376

The probability that more than three and less than six aircrafts arrive within an hour is 0.0376

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