SOLUTION: A bus arrives at a bus stop at 10am, 24 minutes past ten, and 11am. You arrive
at the bus stop at random times between 10:00am and 11:00am everyday, so all
arrival times are eq
Algebra ->
Probability-and-statistics
-> SOLUTION: A bus arrives at a bus stop at 10am, 24 minutes past ten, and 11am. You arrive
at the bus stop at random times between 10:00am and 11:00am everyday, so all
arrival times are eq
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Question 377540: A bus arrives at a bus stop at 10am, 24 minutes past ten, and 11am. You arrive
at the bus stop at random times between 10:00am and 11:00am everyday, so all
arrival times are equally likely. Find your expected waiting time for the bus.
(Hint: Find the probability that you will arrive at the bus stop between 10:00am
and 24 minutes past ten and find your mean waiting time in that case. Then find
the probability that you will arrive at the bus stop between 24 minutes past 10
and 11am and find your mean waiting time in that case.)
Please Help Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! Probability of 10:00 and 10:24 = 24/60 = .4 mean waiting time 24/2 = 12 min
Probability of 10:24 and 11:00 = 36/60 = .6 mean waiting time 36/2 = 18 min
Expected waiting time = .4*12 + .6*18 = 15.6 min (sum of the individual expectations)
.
Ed
