Answer:
11.7% probability that the time between requests is between 1 and 2 seconds.
Explanation:
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
![f(x) = \mu e^(-\mu x)](https://img.qammunity.org/2021/formulas/mathematics/college/dam9hldn5eii4iphfl0p3y8th5zcdwsk06.png)
In which
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](https://img.qammunity.org/2021/formulas/mathematics/college/e3wq4vesqfh4k7cpas1osi6h6zh6fbaxh9.png)
Which has the following solution:
![P(X \leq x) = 1 - e^(-\mu x)](https://img.qammunity.org/2021/formulas/mathematics/college/a6ylb0hy2ltvg7lomfj0epinygu41sl4cu.png)
In this problem, we have that:
![m = 0.5, \mu = (1)/(m) = 2](https://img.qammunity.org/2021/formulas/mathematics/college/gp0kx9yw1iaz43ff7rl5s8rffm2pdnr3x1.png)
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)](https://img.qammunity.org/2021/formulas/mathematics/college/lr0sv2kov40l259re9fd6ii9v31vcjli2u.png)
![P(X \leq 2) = 1 - e^(-2*2) = 0.9817](https://img.qammunity.org/2021/formulas/mathematics/college/wa4cmul1bjak9coam9w1fbbqx8hptepqem.png)
![P(X \leq 1) = 1 - e^(-2*1) = 0.8647](https://img.qammunity.org/2021/formulas/mathematics/college/pm587egjhq3e1mr7w2eeaizm6kjv6nj93p.png)
![P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1) = 0.9817 - 0.8647 = 0.117](https://img.qammunity.org/2021/formulas/mathematics/college/3b7qsfo88shrl0cnsjmyzaxoawe2h9lfn5.png)
11.7% probability that the time between requests is between 1 and 2 seconds.