EXPLANATION
We need to use the Normal Approximation to the Binomial Distribution with the probability and the mean:
p= 0.45
n=200
P(X < 100) = P(X <= 99)
Mean = np = 200(0.45) = 90
Standard Deviation = √(npq) = √(200*0.45*0.55) = √(49.5) = 7.03
Now, we need to compute the value of z-score:
[tex]z=\frac{99-90}{7.03}=\frac{9}{7.03}=1.28[/tex]Now, computing the probability in a z-table:
[tex]P(z<1.28)=0.8997[/tex]In conclusion, the probability is 0.8997
