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Q.100

Expert-verified
Found in: Page 357

### Introductory Statistics

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 $0.0458$

See the 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,

$P\left(x>10\right)=1-P\left(x<10\right)$

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

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

## Part (b) Step 3: Explanation

We know,

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

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

$P\left(x

Using log,

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

$\mathrm{k}=1.4385$

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\left(20

$\mathrm{P}\left(20

## 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,

$⇒\mathrm{P}\left(x<7\right)=P\left(x\le 6\right)$

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

$Poissoncdf\left(12,6\right)=0.458$

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