Question

In: Statistics and Probability

Students arrive at a local bar at a mean rate of 30 students per hour. Assume...

Students arrive at a local bar at a mean rate of 30 students per hour. Assume that the bouncer waits X (minutes) to card the next student. That is, X is the time between two students arriving at the bar. Then we know that X has approximately an exponential distribution. What is the probability that nobody shows up within the 2 minutes after the previous customer? What is the probability that the next student arrives in the third minute, knowing that nobody has shown up in the 2 minutes since the previous student?

Solutions

Expert Solution

X is the random variable which gives us the time between two students arriving at a bar in minutes.

It follows exponential distribution that is

f( X= k) =

where is the mean of no. of students coming to the bar in a minute.

Thus, = 30/60 = 0.5

(X has a mean rate of 30 students per hour so 0.5 mean rate for a minute).

a) In the first part, we want to find the Probability such that the next student arrives after 2 minutes that is P ( X > 2) which is given by

b) Here, we want to find the probability that the next student arrives in third minute that is X < 3 given that X >2.

i.e. P( X >2 | X >3) = P ( 2 < X < 3)

which is given by

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