Question

Using the normal approximation to the binomial to compute the probability of observing 460 or fewer in a sample of 1000 favoring consolidation if we assume that 50% of the entire population favor the change.

Answer 1

Normal approximation:

$\mu=np=1000*0.5=500,$

$\sigma=\sqrt{np(1-p)}=\sqrt{1000*0.5*0.5}=15.81.$

$P(X\le460)=P(X<460.5)=P(Z<\frac{460.5-500}{15.81})=P(Z<-2.50)=0.0062.$