The mean is [tex] \mu=85 [/tex] and standard deviation is [tex] \sigma=5 [/tex]. Then the variable [tex] X\sim N(85,5) [/tex].
Use substitution [tex] Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-85}{5} [/tex], then the variable [tex] Z\sim N(0,1) [/tex].
The probability [tex] Pr(X>80) [/tex] can be calculated in the following way:
for X=80, [tex] Z=\dfrac{80-85}{5} =-1[/tex] and [tex] Pr(X>80)=Pr(Z>-1) [/tex]. From the diagram [tex]Pr(Z>-1)=0.34+0.34+0.135+0.025=0.84 [/tex].
Conclusion: if 188 students take the exam, then 188·0.84=157.92 (157) would receive a score above 80
