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Suppose that the inter-arrival time between consecutive buses is 15 minutes with probability 0.5 and 30 minutes with probability 0.5. You just arrived at a bus stop. What is your expected waiting time?


My attempt: my gut tells me (since no real information about the type of distribution is provided) that it should just be 22.5 by splitting them down the middle. Is this correct?

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  • $\begingroup$ The distribution has just two points (15 minutes & 30 mins) each with equal probability right? $\endgroup$ – Yaitzme Nov 20 '14 at 11:33
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The expected waiting time = 11.25 minutes, in my opinion.

Expected time if the inter-arrival time is 15 minutes with probability 1 = 7.5 mins
Expected time if the inter-arrival time is 30 minutes with probability 1 = 15 mins

Now, expected waiting time = $0.5 * (7.5 + 15)=11.25$

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Because the bus interval arrival time is 30-minute or 15-minute with equal probability,

the probability your arrival falls into a 30-minute interval is $\frac{30}{30+15}=\frac{2}{3}$

the probability your arrival falls into a 15-minute interval is $\frac{15}{30+15}=\frac{1}{3}$

if your arrival falls into a 30-minute interval, the expected waiting time is 15 minutes

if your arrival falls into a 15-minute interval, the expected waiting time is 7.5 minutes

So, the overall expected waiting time is $15\times\frac{2}{3}+7.5\times\frac{1}{3}=12.5$ minutes

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  • $\begingroup$ This i believe is indeed the correct answer since the other answers do not take the probability of arrival within each interval. $\endgroup$ – Faisal Vali Apr 27 '17 at 11:04

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