Assuming scores (denoted by the random variable [tex]X[/tex]) are normally distributed, we can say
[tex]\mathbb P(X\ge80)=\mathbb P\left(\dfrac{X-72}7\ge\dfrac{80-72}7\right)\approx\mathbb P(Z\ge1.143)\approx0.1265[/tex]
where [tex]Z[/tex] has the standard normal distribution. So if there are [tex]n[/tex] students, we can expect approximately 12.65% of them to have a score of at least 80%.
