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 the time between requests is between 1 and 2 seconds. Probabili

Answer 1

Answer:

11.7% probability that the time between requests is between 1 and 2 seconds.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

In this problem, we have that:

$m = 0.5, \mu = \frac{1}{m} = 2$

Find the probability that the time between requests is between 1 and 2 seconds.

$P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1)$

$P(X \leq 2) = 1 - e^{-2*2} = 0.9817$

$P(X \leq 1) = 1 - e^{-2*1} = 0.8647$

$P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1) = 0.9817 - 0.8647 = 0.117$

11.7% probability that the time between requests is between 1 and 2 seconds.