Answer to Question #315703 in Statistics and Probability for Mirylle Anne

Question

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 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) If you are the first one in the city to place an order, what is the probability that you

have to wait more than two weeks from the time you place your order until a

technician is dispatched?

Answer 1

$X_{100}\left( t \right) -number\,\,of\,\,orders\,\,in\,\,period\,\,t\,\,weeks\\X_{100}\left( t \right) \sim Poiss\left( \lambda t \right) \\X_{100}\left( 1 \right) \sim Poiss\left( 0.25 \right) \Rightarrow \lambda =0.25\\For\,\,800 thousands\,\,citizens:\\X_{800}\left( t \right) \sim Poiss\left( 8\lambda t \right) =Poiss\left( 2t \right) \\a: A\,\,technician\,\,is\,\,not\,\,required\,\,before\,\,1 week, i.e. in\,\,one\,\,week\,\,there\,\,are\,\,less\,\,than\,\,3 orders\,\,\\P\left( X_{800}\left( 1 \right) <3 \right) =P\left( X_{800}\left( 1 \right) =0 \right) +P\left( X_{800}\left( 1 \right) =2 \right) +P\left( X_{800}\left( 1 \right) =3 \right) =\\=\sum_{i=0}^2{\frac{2^ie^{-2}}{i!}}=e^{-2}\left( 1+2+\frac{2^2}{2} \right) =0.676676\\b: In\,\,two\,\,weeks\,\,there\,\,are\,\,less\,\,than\,\,2 orders\\P\left( X_{800}\left( 2 \right) <2 \right) =P\left( X_{800}\left( 2 \right) =0 \right) +P\left( X_{800}\left( 2 \right) =1 \right) =\\=\frac{4^0e^{-4}}{0!}+\frac{4^1e^{-4}}{1!}=0.0915782$