ONLY parts D and E Suppose customer arrive to a hank according to a Poisson proc

Question

ONLY parts D and E

Suppose customer arrive to a hank according to a Poisson process but the arrival rate fluctuates over time. From the opening time at 9 a.m. until 11, customers arrive at a rate of 10 customers per hour. From 11 to noon, the arrival rate increases linearly until it reaches 20 customers per hour. From noon to 1pm, it decreases linearly to 15 customers per hour, and remains at 15 customers per hour until the bank closes at 5 p.m. For notation, let N(t) be the number of arrivals in the t hours since the bank opened and lambda (t) the arrival rate at t hours after opening. (a) What is the arrival rate at 12: 30pm? (b) What is the average number of customers per day? (c) What is the probability of k arrivals between 11: 30 and 11: 45? (d) What is the probability of k arrivals between 11: 30 and 11: 45 given that there were 7 arrivals between 11: 00 and 11: 30? (e) Consider the first customer that arrives after noon and let T be the length of time since noon to when that customer arrives. What is the probability that this customer arrives after 12: 10, after 12: 20? Is T exponentially distributed?

Explanation / Answer

From 11 to noon,

RATE=20 customers/hour.

Length of time interval=15 minuites=0.25 hour

So, number of occurences in time interval of length 0.25 hour follows Poisson distribution with mean=20*0.25=5.

So, Probability of k arrivals in between 11:30 and 11:45

={(e^(-5))*(5^k)}/k!, where k=0,1,2......

Rate=20 customers per hour from 11 to noon.

Length of time interval between 11:00 to 11:30=0.5 hour

So, Probability of 7 arrivals in between 11:00 and 11:30 will be={(e^-(20*0.5))*(20*0.5)^7}/7!={(e^-10)*(10^7)}/7!=0.09008

A: Event of k arrivals between 11:30 and 11:45

B: Event of 7 arrivals between 11:00 and 11:30

Hence, Reqd probability=P(A/B)=P(A and B)/P(B)=P(A)*P(B)/P(B)=P(A)/P(B)=[{e^(-5)*(5^k)}/k!]/0.09008=0.000606954*(5^k),k=0,1,2,......

d Ans:0.000606954*(5^k)

e. For customer to arrive after 12:10,

T has to be Exponentially distributed with mean (1/15)

and also T has to be greater than 10 minuites, that is 1/6 hour.

So, customer arrives after 12:10=P(T>1/6)=1-P(T1/6)=1-F(1/6)=1-[1-e^{-15*(1/6)}]=e^(-15/6)=0.082085

Customer arrives after 12:20=P(T>20 minuites)=P(T>1/3)=e^(-15/3)=0.006738.

20 minuites=1/3 hour.

10 minuites=1/6 hour.

Yes, T is exponentially distributed with mean(1/15)

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