Question

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between0 and 7 minutes. Find the probabilty hat a randomly selected passenger has a waiting time greater than 2.25 minutes. Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes. (Simplify your answer. Round to three decimal places as needed.)

Answer 1

Answer:

0.643 = 64.3% probability that a randomly selected passenger has a waiting time greater than 2.25 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 lower than x is given by the following formula.

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

For this problem, we have that:

Uniformily distributed between 0 and 7, so $a = 0, b = 7$

Find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

Either the waiting time is 2.25 minutes or less, or it is greater than 2.25 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 2.5) + P(X > 2.5) = 1$

$P(X > 2.5) = 1 - P(X \leq 2.5)$

In which

$P(X \leq 2.5) = \frac{2.5 - 0}{7 - 0} = 0.357$

Finally

$P(X > 2.5) = 1 - P(X \leq 2.5) = 1 - 0.357 = 0.643$

0.643 = 64.3% probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.