Answer:
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!!!
