Question

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

a. Find the probability that the duration between two successive visits to the website 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 website. What is the probability that the next visit will occur within the next 5 minutes?

d. Find the probability that less than 7 visits occur within a one-hour period.

Short answer

a. the probability that the duration between two successive visits to the website is more than ten minutes is $0.1353$

b.$1.438$minutes

c. the probability that the next visit will occur within the next$5$minutes is$0.0116$

d. the probability that fewer than $7$visits occur within a one-hour period is $0.0458$

Step-by-step solution

Part (a) Step 1: Given Information

Probability is just the way in which likely something is to occur. At the point when we're uncertain about the result of an occasion, we can discuss the probabilities of specific results — how likely they are. The investigation of occasions represented by likelihood is called insights.

 Part (a) Step 2: Explanation 

We know,

Time is x= $10$minutes

Decay rate m = $\frac{12}{60}=0.2$

Using,

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

The probability that the duration between two successive visits to the website is more than ten minutes is,

$$\begin{gathered} P(x>10)=1-\left(1-e^{-0.2\times 10}\right) \\ P(x>10)=0.1353 \end{gathered}$$

Part (b) Step 3: Explanation 

We know,

Decay rate $m=\frac{12}{60}=0.2$

The top $25\%$of durations between visits are calculated,

$$\begin{gathered} P(x<k)=1-e^{-0.2\times k} \\ 0.25=1-e^{-0.2\times k} \end{gathered}$$

Using log,

$\mathrm{In}\left({\mathrm{e}}^{-0.2\times \mathrm{k}}\right)=\mathrm{In}\left(0.75\right)$

$\mathrm{k}=1.4385$

Hence the duration is$1.4385min$

Part (c) Step 4: Explanation 

We know,

The probability that the next visit will occur within the next $5$minutes ,

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

$$\begin{gathered} \mathrm{P}(20<x<25)=1-e^{-0.2\times 25}-\left(1-e^{-0.2\times 20}\right) \\ \mathrm{P}(20<x<25)=0.0183-0.0067 \\ =0.0116 \end{gathered}$$

Part (d) Step 5: Explanation 

We know,

The probability that fewer than $7$visits occur within a one-hour period is calculated using the Poisson distribution,

$\Rightarrow \mathrm{P}(x<7)=P(x\le 6)$

Using the calculator we find $P(x\le 6)$,

$Poissoncdf(12,6)=0.458$