Question

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

Answer 1

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!!!

Answer 2

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