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!!!

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5