We are looking to find P(X>60 students)
X is normally distributed with mean 50 and standard deviation 5
We need to find the z-score of 60 students
[tex]Z= \frac{60-50}{5}=2 [/tex]
To find the probability of P(Z>2), we can do 1 - P(Z<2)
So we read the probability when Z<2 which is 0.9772, then subtract from one we get 0.0228
The number of students that has score more than 60 is 0.0228 x 1000 = 228 students
