Respuesta :
Let m be the mean and s be the standard deviation and find the z score.Â
z = (x - m) /s = (0.8 s + m - m) / s = 0.8Â
The percentage of student who scored above Jane is (from table of normal distribution).Â
1 - 0.7881 = 0.2119 = 21.19%Â
The number of student who scored above Jane is (from table of normal distribution).Â
21.19% 0f 500 = 106Â
z = (x - m) /s = (0.8 s + m - m) / s = 0.8Â
The percentage of student who scored above Jane is (from table of normal distribution).Â
1 - 0.7881 = 0.2119 = 21.19%Â
The number of student who scored above Jane is (from table of normal distribution).Â
21.19% 0f 500 = 106Â