Question

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 9 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes.

Answer 1

Answer:

52.78% probability that a randomly selected passenger has a waiting time greater than 4.25 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X greater than x is given by the following formula.

$P(X > x) = \frac{b-x}{b-a}$

Uniformly distributed between 0 and 9 minutes.

This means that $a = 0, b = 9$

Find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes.

$P(X > 4.25) = \frac{9 - 4.25}{9 - 0} = 0.5278$

52.78% probability that a randomly selected passenger has a waiting time greater than 4.25 minutes.