Question

Hello, can you help me with a STANDARD DEVIATION problem, please!

Answer 1

We want to know the proportion of the students that their score is less than 66. So we have that

$P(X<66)=P(Z<\frac{66-81}{5})=P(Z<-3)=0.0013$

So the amount of student which score is less than 66 is:

$4502\cdot0.0013=5.86\approx6$