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CT fastrak bus waiting times probability that a randomly selected passenger will wait the following times for a CTfastrak bus

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Answer #1

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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

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