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Pani-rosa [81]
3 years ago
8

The time between requests to a web server is exponentially distributed with mean 0.5 seconds. NOTE: This is a multi-part questio

n. Once an answer is submitted, you will be unable to return to this part. Find the probability that there will be exactly 5 requests in a 2-second time interval.
Computers and Technology
2 answers:
oksian1 [2.3K]3 years ago
3 0

Answer:

0.8987

Explanation:

Exponential random variable

P(X=x) = λ e^(-λ.x)

A mean of a request every 0.5 seconds translates to 0.5 seconds between each request.

Rate parameter = (1/0.5) = 2

λ = 2

5 requests in a 2-second time interval translates to (2/5) seconds between requests, that is, x = 0.40

P(X=x) = λ e^(-λ.x)

P(X=0.4) = 2 e^(-2×0.4)

P(X=0.4) = 2 e⁻⁰•⁸ = 2 × 0.4493 = 0.8987

Hope this Helps!!!

igor_vitrenko [27]3 years ago
3 0

Answer:

Probability that there will be exactly 5 requests is 0.156

Explanation:

Poisson probability will be used to solve this problem

P(X=x) =\frac{\lambda^{x}e^{-\lambda}  }{x!}...............(1)

\lambda = \frac{2}{0.5} \\\lambda = 4

Probability that there will be exactly 5 requests in a 2 second time interval

x = 5

Substituting the values of x and λ into equation (1)

P(X = 5) = \frac{4^{5} e^{-4} }{5!}

P(X=5) = 0.156

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