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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.

a. Exponential distribution with mean 12

b. Normal distribution with mean 12

c. Poisson distribution with mean 12

d. Exponential distribution with mean 1/12

User Jumpy
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Final answer:

The duration between visits to a website follows an exponential distribution with a mean of 12 visits per hour. Different probabilities related to the duration between visits can be calculated using the exponential and Poisson distribution formulas.

Step-by-step explanation:

The given scenario deals with the duration between visits to a website that follows an exponential distribution. In an exponential distribution, the time between events has a constant failure rate and is memoryless. Based on the information given, the correct answer would be:

a. Exponential distribution with mean 12

To find the probability that the duration between two successive visits is more than 10 minutes, we can use the exponential distribution formula: P(X > x) = e^(-λx), where λ is the rate parameter. In this case, λ = 1/12 visits per minute. Plugging in the values, we have P(X > 10) = e^(-1/12 * 10) = 0.4066.

b. To find the top 25 percent of durations between visits, we can use the exponential distribution formula: t = (-1/λ) * ln(1-p), where p is the desired percentile. In this case, p = 0.75, so t = (-1/(1/12)) * ln(1-0.75) = 1.7916.

c. To find the probability that the next visit will occur within the next five minutes, we can use the exponential distribution formula: P(X < x) = 1 - e^(-λx). Plugging in the values, we have P(X < 5) = 1 - e^(-1/12 * 5) = 0.3935.

d. To find the probability that fewer than seven visits occur within a one-hour period, we can use the Poisson distribution formula: P(X < x) = e^(-λ) * (λ^x / x!). In this case, λ = 12 visits per hour and x = 7. Plugging in the values, we have P(X < 7) = e^(-12) * (12^7 / 7!) = 0.0901.

User Barlow
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