Answer:
probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG = 0.8668
Step-by-step explanation:
mean, μ = 80 mg/mi
Standard deviation, [tex]\sigma = 4 mg/mi[/tex]
probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG
[tex]P(X > x ) = P(z > \frac{x - \mu}{\sigma})[/tex]
[tex]P(X > x ) =1 - P(z < \frac{x - \mu}{\sigma})[/tex]
[tex]P(X > 86 ) =1 - P(z < \frac{86 - 80}{4})[/tex]
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