The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. if jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above jane?

Let m be the mean and s be the standard deviation and find the z score. 

z = (x - m) /s = (0.8 s + m - m) / s = 0.8 

The percentage of student who scored above Jane is (from table of normal distribution). 

1 - 0.7881 = 0.2119 = 21.19% 

The number of student who scored above Jane is (from table of normal distribution). 

21.19% 0f 500 = 106 

21.19% 0f 500 = 106