Question

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

a. Find the probability that the duration between two successive visits to the web site is more than ten minutes.

b. The top \(25%\) of durations between visits are at least how long?

c. 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?

Short answer

Part a. \(P(x>10)=0.1353\)

Part b. \(P(x>k)=1.4385\)

Part c. \(P(20<x<25)=0.0116\)

Step-by-step solution

Part a. Step 1. Explanation

The required probability is given by exponential distribution with decay rate \(\frac{12}{60}=0.2\)

Required probability is calculated by.

\(P(x>10)=1-P(x<10)\)

\(P(x>10)=1-(1-e^{-0.2\times 10}\)

\(P(x>10)=e^{-0.2\times 10}\)

\(P(x>10)=0.1353\)

Therefore, the value of \(P(x>10)=0.1353\)

Part b. Step 1. Explanation

Required probability can be calculated by,

\(P(x>k)=1-e^{-0.2\times k}\)

\(0.25=1-e^{-0.2\times k}\)

\(e^{-0.2\times k}=0.75\)

Take log on both sides,

\(ln(e^{-0.2\times k})=ln(0.75)\)

\(-0.2k=-0.2877\)

\(k=1.4385\)

Therefore, given value \(P(x<k)=1.4385\) minutes

Part c. Step 1. Explanation

The required probability is given by,

\(P(20<x<25)=P(x<25)-P(x<20)\)

\(P(20<x<25)=1-e^{-0.2\times 25}-(1-e^{-0.2\times 20})\)

\(P(20<x<25)=0.0183-0.0067\)

\(P(20<x<25)=0.0116\)

Therefore, the value is \(P(20<x<25)=0.0116\)