Answer:
probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG = 0.8668
Explanation:
mean, μ = 80 mg/mi
Standard deviation,

probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG



P(X > 86) = 1 - P(z < 1.5)
From the standard normal table, P(z < 1.5) = 0.9332
P(X > 86) = 1 - 0.9332
P(X > 86) = 0.0668