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 1.25 minutes.

P(greater than 1.25 minutes) = 0.8611 (Approx)

Step-by-step explanation:

Given:

Waiting time = 0 - 9 minutes

Find:

Probability that selected passenger has a waiting time greater than 1.25 minutes.

Computation:

⇒ The probability that a randomly selected passenger has a waiting time greater than 1.25 minutes =

⇒  P(greater than 1.25 minutes) = [9-1.25] / 9

⇒ P(greater than 1.25 minutes) = [7.75] / 9

⇒ P(greater than 1.25 minutes) = 0.8611 (Approx)

-50/3

step-by-step explanation:

c. 16 units

step-by-step explanation:

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