An economics professor decides to curve the grades of his class. In doing so, he decides to make it such that the students who score in the top 11% receive an A.

Question

Assume a normal distribution among grades. How many standard deviations above the mean must a student get to receive an A? (Round to 2 decimal places, if needed)

Answer 1

z score = 1.227 for the 89th percentile (100-11 = 89)

1.227 standard deviations above the mean corresponds to the 89th percentile, with 11% of the grades above the mean + 1.227 standard deviations.

Round off to 2 decimal places, to get z = 1.23

go on line for a z score calculator. plug in 89 and it spits out 1.227 for the z score. the z score is the number of standard deviations from the mean that corresponds to the percentile.

or you could use z tables, which are also on line, if you don't have a textbook appendix with them.

or, for a rough approximation, use the empirical rule 68 - 95 - 99.7 for the percent of grades within 1, 2 and 3 standard deviations from the mean. 68% corresponds to the 50 + 68/2 = 50 +34 = 84th percentile. The 89th is only 5 percentiles higher, so since 68 corresponds to 1 standard deviation, the 89th percentile will be slightly higher than 1. 1.227 is slightly higher than 1. It's in the right range. For the 89th percentile, with 11% above that percentile, 100-2(11) = 100-22=78% of the grades will fall within 1.227 standard deviations from the mean. z score = (grade minus mean)/(standard deviation). if the grade is one standard deviation from the mean, the z score = 1. if the grade is 1.227 standard deviations from the mean, the z score = 1.227

Answer 2

The top 11% would be above the 89th percentile. Need to find z-score for 89th percentile using standard normal probability table = 1.22.

z-score is the number of standard deviations above or below the mean, so a score in the 89th percentile (or top 11 percent) is 1.22 standard deviations above the mean