In a certain very large city, the Department of Transportation (D. O.T.) has organized a complex system of bus transportation. In an advertising campaign, citizens are encouraged to use the new "GO-D. O.T!" system and head for the nearest bus stop to be transported to and from the central city. Suppose that at one of the bus stops the amount of time (in minutes) that a commuter must wait for a bus is a uniformly distributed random variable, T.

The possible values of T run from 0 minutes to 20 minutes.

(a) What is the probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D. O.T?

(b) What is the standard deviation?

(c) What is the probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D. O.T?

(d) What is the average waiting time?

a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T = P(7 < x ≤ 20) = 0.65

b) Standard deviation of the uniform distribution = 5.77 minutes

c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17) = 0.35

d) average waiting time for the uniform distribution = 10 minutes.

Step-by-step explanation:

This is a uniform distribution problem with lower limit of 0 minute and upper limit of 20 minutes.

a = 0, b = 20

Probability = f(x) = [1/(b-a)] ∫ dx (with the definite integral evaluated between the two intervals whose probability is required.

a) Probability that a randomly selected commuter will spend more than 7 minutes waiting for GO-D.O.T

P(7 < x ≤ 20) = f(x) = [1/(b-a)] ∫²⁰₇ dx

P(7 < x ≤ 20) = (20-7)/(20-0) = (13/20) = 0.65

b) Standard deviation of the uniform distribution

Standard deviation of a uniform distribution is given as

σ = √[(b-a)²/12]

σ = √[(20-0)²/12]

σ = √[20²/12]

σ = 5.77 minutes

c) Probability that a randomly selected commuter will spend longer than 10 minutes but no more than 17 minutes waiting for the GO-D.O.T = P(10 < x < 17)

P(10 < x < 17) = (17-10)/(20-0)

P(10 < x < 17) = (7/20) = 0.35

d) The average waiting time.

The average of a uniform distribution = (b+a)/2

Average waiting time = (20+0)/2

Average waiting time = 10 minutes

Hope this Helps!!!

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