Question about poisson processes and markov chains Question

Question about poisson processes and markov chains

Question about poisson processes and markov chains (1) (10 points) During normal working hours (9am to 5pm), customers arrive to a queue as a Poisson process with rate 6 per hour. Between 9am and 11am, there were 10 arrivals. Compute the distribution (give the probability mass function) of the number of arrivals between the hours of 10am and noon.

Solution

It is given that 9 a.m. to 5 p.m. X - no of customers arrive is Poisson : 6

Between 9 and 11 a.m. there were 10 arrivals.

As no of arrivals in 9 - 11 a.m. is 20, we can take on average per hour it is 10

i.e. from 10 a.m. to 11 a.m. no of customers arriving -10

Between 9 a.m. and 5 p.m. first two hours customer arrive = 20

Total customers = 8x 6 =48

Thus between 11 and 5 p.m. no of customers arrive = 48-20 = 28

Or per hour average = 28/6 = 14/3

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Thus pdf of no of customers for no of arriavals from 10 a.m. to 12 noon

= average of 10 and 14/3

= 10+14/3 = 44/3

Thus pdf is Poisson with average 44/3