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lotsofmoxie

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## Homework Statement

Customers arrive in single server queue to be serviced according to Poisson process with intensity 5 customers an hour.

(a) If the customers begin to arrive at 8am, find the probability that at least 4 customers arrived between 9am and 10am.

(b) Find the probability that the 6th customer will arrive after 12 noon.

(c) Find the variance of the time of the arrival of the 3rd customer.

If the service times of the customers are independent uniform r.v.’s on [0, 4] and also are independent of the interarrival times:

(d) Find the probability that the second customer to arrive will have to wait before being served.

(e) Find the expected value of the time that the second customer will have to wait before being served.

## Homework Equations

P(x=x) = (

**λt)^(x)*e^(-**

**λt)/x!**where lambda = poisson parameter and t=ime interval

[/B]

## The Attempt at a Solution

a) P(at least 4 arrivals between 9-10am)=1-P(0,1,2 or 3 arrivals between 9-10am)**=1-e^(-5*1)**

**(5*1****)^(0)/****0!+****(5*1****)^(1)****/1!+****(5*1****)^(2)****/2!+****(5*1****)^(3)****/3!)**=0.7349

b)

P(T6>4) = P(number of arrivals between 8-12 is exactly 0, 1, 2, 3, 4, or 5)

**=**(5*4

**e^(-5*4****)*****)^(0)/**

**0!+****(5*4****)^(1)****/1!+****(5*4****)^(2)****/2!+****(5*4****)^(3)****/3!+****...)**=0.0000719

c) The 3rd arrival time is the sum of the first three individual arrival times. These are all distributed exponentially with Var=1/

**[/B]**

**λ^2=1/25. Since the three random variables are independent, we can sum their variances to get Var(T3)=1/25+1/25+1/25.**

d) Let U = time taken to serve 1st customer

T = time between arrival of 1st and second customer

f(U)=1/(4-0)=1/4 when 0<=u<=4

f(T) = 5e^-5t when t>0, 0 elsewhere

Because of independence, the joint probability density function f(U,T)=1/4*5e^(-5t)

customer has to wait ------> we want P(U>T)=1-P(T>U) because it is easier to integrate

d) Let U = time taken to serve 1st customer

T = time between arrival of 1st and second customer

f(U)=1/(4-0)=1/4 when 0<=u<=4

f(T) = 5e^-5t when t>0, 0 elsewhere

Because of independence, the joint probability density function f(U,T)=1/4*5e^(-5t)

customer has to wait ------> we want P(U>T)=1-P(T>U) because it is easier to integrate

∫ ∫

**1/4*5e^(-5t)dudt where inner integral is over (4, infinity) and outer is over (0,4)**

e) same as d but with the integrand as multiplied by u-t since that is the waiting time

e) same as d but with the integrand as multiplied by u-t since that is the waiting time

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