Answer:
65.54% (nearest hundredth)
Step-by-step explanation:
Given:
[tex]\sf X \sim N(\mu, \sigma^2)\implies X \sim N(9000,1000^2)[/tex]
P(X > 8600) = 1 - P(X ≤ 8600)
          = 1 - 0.3445782584
          = 0.6554217416
          = 65.54%