At Liberty High School, 140 students just completed a statistics course. The final exam percentages were normally distributed with a mean of 80 and a standard deviation of 6. What percentage of students scored greater than 86?

To find the percentage of students who scored greater than 86, we need to calculate the z-score for 86 and then find the area to the right of this z-score.

First, calculate the z-score using the formula:
z = \frac{x - \mu}{\sigma}
where:
x = score (86),
μ = mean (80),
σ = standard deviation (6).

z = \frac{86 - 80}{6} = \frac{6}{6} = 1

Next, we need to find the area to the right of z = 1 in the standard normal distribution table. This area represents the percentage of students who scored greater than 86.

Looking up z = 1 in the standard normal distribution table, we find that the area to the left of z = 1 is approximately 0.8413. Therefore, the area to the right of z = 1 (greater than 86) is:
1 - 0.8413 = 0.1587

So, approximately 15.87% of students scored greater than 86.

\boxed{15.87\%}