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Q.100

Expert-verifiedFound in: Page 357

Book edition
OER 2018

Author(s)
Barbara Illowsky, Susan Dean

Pages
902 pages

ISBN
9781938168208

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.

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 $0.0458$

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.

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,

$P(x>10)=1-\left(1-{e}^{-0.2\times 10}\right)\phantom{\rule{0ex}{0ex}}P(x>10)=0.1353$

We know,

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

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

$P(x<k)=1-{e}^{-0.2\times k}\phantom{\rule{0ex}{0ex}}0.25=1-{e}^{-0.2\times k}$

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 $1.4385min$

We know,

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

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

$\mathrm{P}(20<x<25)=1-{e}^{-0.2\times 25}-\left(1-{e}^{-0.2\times 20}\right)\phantom{\rule{0ex}{0ex}}\mathrm{P}(20<x<25)=0.0183-0.0067\phantom{\rule{0ex}{0ex}}=0.0116$

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 $P(x\le 6)$,

$Poissoncdf(12,6)=0.458$

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