# Poisson distribution problem?

Help on my homework needed! Thank you!

An installation technician for a specialized communication system is dispatched to a city only when three or more orders have been placed. Suppose orders follow a Poisson distribution with a mean of 0.25 per week for a city with a population of 100,000 and suppose your city contains a population of 800,000. a. What is the probability that a technician is required after a one-week period? b. What is the probability that no less than six orders have been placed in a two-week period? c. If you are the first in the city to place an order, what is the probability that a technician will not be dispatched for that week>

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What have you tried? –  TakeS Feb 14 '13 at 23:38
I don't know where to start, honestly... –  user59117 Feb 14 '13 at 23:47

I have used excel formula = poisson.dist. Where x = number of events, mean = mean, and cumulative = True/False. True for cumulative distribution function, and false if not, which will be probability mass function.

Mean = 0.25 per 100000 Users. For a Population of 800000, mean will be = 8*0.25, Which is 1. Mean = 1.

a) What is the probability that a technician is required after a one-week period - If the technician is required than sale should have been > 3. Minimum 4. So lets find what is the probability for a sale = 4. Whcih is p(x=4). 1.5 % is the probability that a sale of 4 communication systems sale happens in a week This is Probablity Mass function.

But sale > 3 could be any number more than 3. Like P(X= 4,5,6....Infinite), which is not possible to calculate probability for each success and till infinitiy. So calculate probability for p(x<3), and negate it with Sample Space 1. Which will be 0.019. Hence its 1.9% probability that, a technician will be sent after a weeks time of your order. 1.9% is the Probablity of order placement more than 3 orders in a week. This is cumulative distribution Fucntion, probablity of sale happening any number more than 3.

b)What is the probability that no less than six orders have been placed in a two-week period? p(x > 6) = 1 - (p(x)<6)) = 0.008% (If mean is 1, considering week does not influence the mean, probability will be 45%, if mean is 2.

c. If you are the first in the city to place an order, what is the probability that a technician will not be dispatched for that week> = The sale has to be less than 3. p(x<3) = 98.10%

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.25*8 = 2, mean will be 2 for a population of 800,000

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