Question

A website experiences traffic during normal working hours at a rate of 12 visits per hour. Assume that the duration between visits has the exponential distribution

1. Find the probability that the duration between two successive visits to the web site is more than ten 10 minutes
2. Suppose that 20 minutes have passed since the last visit to the web site. What is the probability that the next visit will occur within the next 5 minutes?
3. The top 25% of durations between visits are at least how long?
4. Find the probability that less than 7 visits occur within a one-hour period.

Answer 1

Exponential distribution

$rate = \frac{12}{60}=0.2 \\ 1. \;P(x>10) = 1-P(x<10) \\ = 1- (1- e^{-0.2 \times 10}) \\ = e^{-0.2 \times 10} \\ = 0.1353 \\ 2. \; P(20<x<25) =P(x<25) -P(x<20) \\ = 1- e^{-0.2 \times 25}- (1- e^{-0.2 \times 20}) \\ = 0.0183-0.0067 \\ = 0.0116 \\ 3. \; P(x<k) = 1-e^{-0.2 \times k} \\ 0.25= 1 -e^{-0.2 \times k} \\ e^{-0.2 \times k}= 0.75 \\ ln(-0.2 \times k) = ln(0.75) \\ -0.2k= -0.2877 \\ k= 1.4385 \;min$

4. Required probability will follow Poisson distribution with mean 12

P(x<7) = P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4) +P(x=5) +P(x=6)

By using Excel

=POISSON(6,12,1)

P(x<7) = 0.0458