You take the mean of 12.89 from marcus' score od 13.7 to get .81, then divide that by the standard deviation of 1.95. This gives you .415 and thats 65.9% on the z table. In the summary tell him he only scored better than 65.9% of students or average.
Marcus score is 13.7, the mean is 12.89, and the standard deviation is 1.95, so I’ll do ( 13.7 – 12.89)/ 1.95, Since Marcus' z-score is 0.415. The area to the left of the z-score is 65.9% so therefore Marcus did scored better than 65.9% of the students. So Marcus needs to score 98% better of all the other students in order for him to receive an actual certificate.
