If the mean (μ) is 80, and the standard deviation (σ) is 5, then all scores between 80 and 90 would fall between 0 and 2 standard deviations above the mean.
Using the equation for Z score (Z = (X-μ)/σ) for each X value (80 and 90) then the Z scores are 0 and 2, respectively.
Using a normal distribution table, it can be found that P(80 < z) = .5 (this is the probability that a random score would be greater than 80. It makes sense that it is .5 or 50% because 80 is the mean.)
And the P(90 > z) = .97725. (this is the probability that a random score would be less than 90.)
