Question

let X represent the amount of time till the next student will arriv ein the library partking lot at the university. If we know that the dstubtion of arrivlal time can be modeled using an exponential distruibution with a mean of 4 minutes, find the probabiity that it will take between 2 and 132 minutes for the net sutedn to arrive at the library partking lot 0.606531

Answer 1

Answer:

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

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:

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

The probability of finding a value higher than x is:

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

Mean of 4 minutes

This means that $m = 4, \mu = \frac{1}{4} = 0.25$

Find the probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot:

This is:

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)$

In which

$P(X \leq 132) = 1 - e^{-0.25*132} = 1$

$P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469$

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531$

0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.