You arrive at a bus stop to wait for a bus that comes by once every 30 minutes. You don't know what time the last bus came by. The time x that you wait before the bus arrives is uniformly distributed on the interval from 0 to 30 minutes. (Round your answers to four decimal places.)

(a) What is the probability that you will have to wait longer than 22 minutes?
(b) What is the probability that you will have to wait less than 8 minutes?
(c) What is the probability that you will wait between 5 and 13 minutes?

a) P(X > 22) = 0.2667

b) P(X < 8) = 0.2667

c) P(5 < X < 13) = 0.2667

Step-by-step explanation:

Since the data is uniformly distributed between 0 and 30 minutes,

The expected value or mean is given as

(b+a)/2 where b and a are the two extreme.

Mean = (30+0)/2 = 15 minutes.

And the probability for this type of uniform distribution is given as

P(X) = ∫ [1/(b-a)] dx for all (a ≤ x ≤ b)

With the definite integral evaluated between the interval whose probability is required.

For this question, note that b = 30 mins and a = 0 min

a) The probability that you will have to wait longer than 22 minutes.

P(x > 22)

Since the maximum wait time is 30 mins, the interval for this probability is (22 < x < 30)

P(X > 22) = ∫³⁰₂₂ [1/(b-a)] dx

P(X > 22) = [x/(b-a)]³⁰₂₂ = [(30-22)/(30-0)]

P(X > 22) = (8/30) = 0.2667

b) The probability that you will have to wait less than 8 minutes = P(X < 8)

Since the minimum wait time is 0 min, the interval for this probability is (0 < x < 8)

P(X < 8) = ∫⁸₀ [1/(b-a)] dx

P(X < 8) = [x/(b-a)]⁸₀ = [(8-0)/(30-0)]

P(X < 8) = (8/30) = 0.2667

c) The probability that you will wait between 5 and 13 minutes = P(5 < X < 13)

P(5 < X < 13) = ∫¹³₅ [1/(b-a)] dx

P(5 < X < 13) = [x/(b-a)]¹³₅ = [(13-5)/(30-0)]

P(5 < X < 13) = (8/30) = 0.2667

Hope this Helps!!!

the expected value of the random variable : 0, 1 and 2

step-by-step explanation: