The scores on a standardized test are normally distributed with a mean of 90 and a standard deviation of 15. What is the percentage of scores that are greater than 69?

Answer 1

x = score percentage.

First compute the z-score = (x - mean) / standard deviation = (69 -90) / 15 = -1.4

Want to find P(z > -1.4) which is 1 - P(z < -1.4) = 1 - 0.0808 = 0.9192.