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)
Let p and h represent the price of the service call and the hours worked, respectively. The price charged is 35 dollars for each hour, so the time-based part of the price is 35h. The base fee of 55 dollars is added to that for a total price of ...
