Amarket research firm conducts telephone surveys with a 50% historic response rate. suppose a sample of 400 is taken. what is the probability that the sample proportion will be at least 0.475?

N=400
sample proportion >= .475 => n>=.475N=190
mean response rate of population, p=0.5
Problem satisfies criteria for binomial distribution, where
P(x>=190)=sum P(x=i), i=190 to 400
This can be obtained from binomial distribution tables, calculators, or software as 
P(x>=190)=1-P(x<190)=1-0.14685=0.85315

Alternatively, normal approximation can be used.
mean=Np=400*.5=200
variance=Npq=400(.5)(1-.5)=100
standard deviation=sqrt(variance)=sqrt(100)=10
Calculate P(X>=190)=1-P(X<190)
X with continuity correction = 189.5
Z=(X-mean)/standard deviation=(189.5-200)/10=-1.05
P(z<Z)=P(z<-1.05)=0.14686 (from normal probability tables)
hence
P(X>=190)=1-P(X<190)=1-P(z<-1.05)=1-0.14686=0.85314 (approximately)

almost same as the exact binomial solution.

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