Bill takes the commuter train to work every day. During the morning commute, a train arrives every 15 min. If Bill arrives at the station at a random time for the morning commute, what is the probability that he will have to wait at least 5 min for a train

P(X ≥ 5) = 0.6667

Step-by-step explanation:

Waiting for a train that comes once in 0 to 15 minutes is a uniform probability distribution problem.

For a uniform probability distribution, the probability function is given as

P(X) = f(x) = [1/(b-a)] for (a≤x≤b)

where a = lower limit of the uniform distribution sample space = 0 mins

b = upper limit of the uniform distribution sample space = 15 mins

But P(X≥x) = 1/(b-a) ∫ dx (with the Integral evaluated from x to b)

So, the probability that a person would wait at least 5 minutes for a train = P(X ≥ 5)

P(X ≥ 5) = 1/(b-a) ∫¹⁵₅ dx = 1/(15-0) ∫¹⁵₅ dx

P(X ≥ 5) = (1/15) × [x]¹⁵₅ = (1/15) × (15 - 5) = (10/15) = (2/3) = 0.6667

Hope this Helps!!!

segment bisector

step-by-step explanation:

line segment bisector. definition: a line, ray or segment which cuts another line segment into two equal parts. try this drag one of the orange dots at a or b and note the the line ab always divides the line pq into two equal parts.

mark brainliest!