Question 1168022
This is binomial with n= 300 p=0.05 
can use binomcdf(300, 0.05,19) to get 0.8810 as a probability




binomial and can use a normal approximation
mean is np=300*0.05=15
variance us np(1-p)=15(0.85)=12.75
sd is sqrt (V)=3.57

z=(19.5-15)/3.57
4.5/3.57 with continuity correction factor
<1.26
probability is 0.8962, close to the exact value,