Intervals drinking coffee

Now I'm trying to understand the Poisson process.

If over 24 hours I drink 11 cups of coffee, what is the probability that the interval between cups is more than 1.5 hours but less than 2 hours?
purplesoupAsked:
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deightonprogCommented:
process is poisson with lamda = 11/24 (per hour)

the poisson process is related to exponential distribution, see
http://en.wikipedia.org/wiki/Exponential_distribution#Cumulative_distribution_function

prob of time between cups given by exponential distribution

t
P(x > X) =  e ^ (-lamda x)

P(x > 1.5)=  e ^ (- (3/2) lamda )

P(x > 2) =  e ^ (- 2 lamda )

P (x > 1.5 and x < 2) = P(x > 1.5) - P(x > 2)
  = e ^ (- (3/2) lamda ) - e ^ (- 2 lamda )
= e ^ (- (3/2) (11/24) ) - e ^ (- 2 (11/24) ) = .103
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aburrCommented:
This can only be answered if it is known that the coffee is consumed at random intervals (unlikely) and that each cup is consumed in a VERY short time (also unlikely, especially if it is hot)
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purplesoupAuthor Commented:
ok let's say I'm (a) drinking the coffee over random intervals and (b) the coffee is in a very small cup and is at drinkable temperature, so therefore in effect it is drunk instantly.
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aburrCommented:
You have stated the requisite assumptions nicely but I still have not formulated the problem correctly in my mine. That is a deficiency on my part. A more knowledgeable person would probably give you a straight answer.
My difficulty is with the eleven cups.
Are you looking for the average interval?
Do you start the 24 hours by drinking a cup?
You universe which you are sampling goes from 0 to 24 (decimals allowed). Each interval is equally likely. So the average is 2.4 hours. I am having difficulty with the fact that the 10 intervals must add up to 24. I imagine you have some difficulty there too. Hope that a more knowledgeable person comes along.
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