Hi, could someone help me explain how to get to 0.0487? Thanks.

The exercise:

The answer:

Customers bring cameras to an electronics shop for repair following a Poisson process with an average of five per week. Every camera goes directly into repair, and a camera's repair time is normally distributed with an expected value of two weeks and a standard deviation of one half week. Every customer who brings in a camera for repair is lent a spare camera until the repair is ready. There are sufficiently many spare cameras. (a) What is the equilibrium distribution of the number of spare cameras that have been lent out? (b) What is the fraction of the time that more than 15 cameras have been lent out? (c) Now, suppose that upon arrival, a broken camera's repair time is immediately determined and that spare cameras are lent out only if the repair time is more than two weeks. Answer the previous question for this situation. (a) The M/G/ queue is applicable, where we use the insensitivity property. We identify the repair time of a broken camera with the service time. The fraction of the time that more than 15 spare cameras have been lent out is equal to 15 Tj = 0.0487, where {T;} is a Poisson distribution with expected value 5 x 2 = 10. (b) The equilibrium distribution of the number of cameras that have been lent out is a Poisson distribution with expected value 1* E(X | X > 2). Here 1* = 5P(X > 2) = 2.5 and = ~ E(X|X > 2) =2 *