[tex]\mathbb P(X\le40)\approx\mathbb P(X<40.5)\approx\mathbb P(Y<40.5)[/tex]
where [tex]Y[/tex] follows a normal distribution with mean [tex]np=45[/tex] and standard deviation [tex]\sqrt{np(1-p)}\approx2.12[/tex].
[tex]\mathbb P(Y<40.5)=\mathbb P\left(\dfrac{Y-45}{2.12}<\dfrac{40.5-45}{2.12}\right)=\mathbb P(Z<-2.12)[/tex]
where [tex]Z[/tex] follows the standard normal distribution. This probability is roughly 0.017003.
