Question

CT fastrak bus waiting times probability that a randomly selected passenger will wait the following times for a CTfastrak bus uniformly distributed from zero to 20 minutes. Find the are b. Between 5 and 10 minutes Exactly 7.5922 minutes d. Exactly 5 minutes e. Between 15 and 25 minutes C.

Answer 1

Ans:

Uniform distribution with a =0 and b=20 minutes:

f(x)=1/(20-0)=1/20=0.05

b)

P(5<x<10)=(10-5)/(20-0)=5/20=1/4=0.25

c)

P(x=7.5922)=1/20=0.05

d)

P(x=5)=1/20=0.05

e)

P(15<x<25)=(25-15)/(20-0)=10/20=0.5