Inactive Tutor answered 12/16/20
We find the transition matrix, then the eigenvector associated with the eigenvalue of 1. The eigenvector is "normalized" to sum to 1. Let poor be x, satisfactory be y, and preferred be z
P =
.85 .15 0
.05 .9 .05
0 .05 .95
Let z = 1
From the third column, .05y + .95 = 1; .05y = .05; y = 1
From the first column, .85x + .05 = x
.05 = .15x
x = 1/3
Then, we normalize. x + y + z = 7/3; therefore.
x = 1/3* 3/7 = 1/7
y = 1 * 3/7 = 3/7
z = 1 * 3/7 = 3/7
Poor = 1/7; Satisfactory = 3/7; Preferred = 3/7