The scores on a certain math test were normally distributed with a mean score of 80 and a standard deviation of 5. What percent of the students scored between 80 and 90?

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.)

So the final answers would be (90 > z)-P(80<z) = .47725 or 47.725%