Summary: Solutions to Homework 9
Problem 1
(a) The rate of defects per meter of wire is = .01
(b) The expected number of defects in the first spool is E[N(100)] = 1
(c) Recall that the Poisson process has stationary increments , thus
E[N(1000) - N(900)] = E[N(100)] = 1
(d)
P[N(100) = 0] =
(1)0
e-1
0!
= e-1
(e)
P[N(1000) - N(900) = 1] = P[N(100) = 1] = e-1
(f) Here the Poisson process has independent increments, our desired solution is
e-1
(g)
P[N(1000) = 1] = 10e-10
(h) Let T1 be the time when the first defect occurs. Here
E[T1] =

Source: Ayhan, Hayriye - School of Industrial and Systems Engineering, Georgia Institute of Technology

Collections: Engineering