First, identify if the normal approximation is valid. To do this, check that np>5 and n(1-p)>5. Since p<(1-p), I'll just check np>5: 80(.15) = 12 > 5. So the approximation is valid.
μ = np = 12
σ = √(np(1-p)) ≈ 2.9445.
Next find the z-score associated with this x value. Remember to adjust .5 to the left or right for probabilities above or below 18 respectively. Since you didn't include the direction, I'll give both probabilities.
zGT = (18-0.5-12)/2.9445 = 1.87 ⇒ P(z ≥ 1.87) = 3.07%
zLT = (18+0.5-12)/2.9445 = 2.21 ⇒ P(z ≤ 2.21) = 98.64%
NOTE: these probabilities don't add up to 100%; they shouldn't add up to that because they're being calculated from 2 different points on the continuous spectrum. Also, they are simply approximations of the true value.
